This problem was asked by Two Sigma.
Using a function rand7() that returns an integer from 1 to 7 (inclusive) with uniform probability, implement a function rand5() that returns an integer from 1 to 5 (inclusive).
Already asked in Problem #45
This problem was asked by Microsoft.
A number is considered perfect if its digits sum up to exactly 10.
Given a positive integer n, return the n-th perfect number.
For example, given 1, you should return 19. Given 2, you should return 28.
def n_perfect(n):
n_mod = sum([int(i) for i in str(n)]) % 10
return int(str(n)+str(10-n_mod))
if __name__ == '__main__':
n = [1, 2, 9, 15]
for i in n:
print(n, n_perfect(i))This problem was asked by Facebook.
Given a list of integers, return the largest product that can be made by multiplying any three integers.
For example, if the list is [-10, -10, 5, 2], we should return 500, since that's -10 * -10 * 5.
You can assume the list has at least three integers.
def max_product(arr):
n_max1, n_max2, n_max3 = 0, 0, 0
p_max1, p_max2 = 0, 0
for val in arr:
if val > 0:
if val > p_max1:
p_max3 = p_max2
p_max2 = p_max1
p_max1 = val
elif val > p_max2:
p_max3 = p_max2
p_max2 = val
elif val > p_max3:
p_max3 = val
else:
if val < n_max1:
n_max2 = n_max1
n_max1 = val
elif val < n_max2:
n_max2 = val
return max(p_max1*p_max2*p_max3, n_max1*n_max2*p_max1)
if __name__ == '__main__':
arr = [-10, -10, 5, 2]
print(arr, max_product(arr))
arr = [-10, 10, 5, 2, -20, 20]
print(arr, max_product(arr))This problem was asked by Google.
On our special chessboard, two bishops attack each other if they share the same diagonal. This includes bishops that have another bishop located between them, i.e. bishops can attack through pieces.
You are given N bishops, represented as (row, column) tuples on a M by M chessboard. Write a function to count the number of pairs of bishops that attack each other. The ordering of the pair doesn't matter: (1, 2) is considered the same as (2, 1).
For example, given M = 5 and the list of bishops:
(0, 0) (1, 2) (2, 2) (4, 0) The board would look like this:
[b 0 0 0 0] [0 0 b 0 0] [0 0 b 0 0] [0 0 0 0 0] [b 0 0 0 0] You should return 2, since bishops 1 and 3 attack each other, as well as bishops 3 and 4.
import numpy as np
def n_attacks_helper(board, cood, m):
r_skew, c_skew = cood
deltas = ((-1, -1), (-1, 1), (1, -1), (1, 1))
att_ct = 0
for i in range(1, m):
for c_r, c_c in deltas:
r_cood, c_cood = r_skew+(c_r*i), c_skew+(c_c*i)
if (0 <= r_cood < m) and (0 <= c_cood < m):
att_ct += board[r_cood][c_cood]
return att_ct
def n_attacks(m, bishops):
board = np.zeros((m,m))
for cood in bishops:
board[cood[0]][cood[1]] = 1
att_ct = 0
for cood in bishops:
att_ct += n_attacks_helper(board, cood, m)
print(board)
return att_ct/2
if __name__ == '__main__':
m = 5
bishops = [(0,0), (1,2), (2,2), (4,0)]
print(n_attacks(m, bishops))
m = 4
bishops = [(0,0), (2,0), (1,1), (3,3)]
print(n_attacks(m, bishops))""" Implement an LFU (Least Frequently Used) cache. It should be able to be initialized with a cache size n, and contain the following methods:
set(key, value): sets key to value. If there are already n items in the cache and we are adding a new item, then it should also remove the least frequently used item. If there is a tie, then the least recently used key should be removed. get(key): gets the value at key. If no such key exists, return null. Each operation should run in O(1) time. """
Similar to problem #52 (LRU cache), but with a catch that now we also store the number of times a node is accessed.
(65 wasn't mailed :/)
This problem was asked by Square.
Assume you have access to a function toss_biased() which returns 0 or 1 with a probability that's not 50-50 (but also not 0-100 or 100-0). You do not know the bias of the coin.
Write a function to simulate an unbiased coin toss.
import random
def toss_biased():
bias = 3
if random.randint(0, 9) < bias:
return False
return True
def toss_unbiased():
"""
We want a situation where prob of series of events is same
using the biased_func. If p is prob of biased_func,
(p)(1-p) = (1-p)(p) -->
So we can base our outputs using this information only.
"""
toss1 = toss_biased()
toss2 = toss_biased()
if toss1 and not toss2:
return True
if not toss1 and toss2:
return False
return toss_unbiased()
if __name__ == '__main__':
n_iter = 10000
counts = {'0':0, '1':0}
for i in range(n_iter):
if toss_unbiased():
counts['1'] += 1
else:
counts['0'] += 1
print(counts)This problem was asked by Google.
A knight's tour is a sequence of moves by a knight on a chessboard such that all squares are visited once.
Given N, write a function to return the number of knight's tours on an N by N chessboard.
def is_valid(board, move, n):
if 0 <= move[0] < n and 0 <= move[1] < n and \
board[move[0]][move[1]] is None:
return True
return False
def valid_moves(board, cur_r, cur_c, n):
moves = [(-2, -1), (-2, 1), (-1, 2), (1, 2),
(2, -1), (2, 1), (-1, -2), (1, -2)]
val_moves = [(cur_r+del_r, cur_c+del_c) for del_r, del_c in moves]
return [move for move in val_moves if is_valid(board, move, n)]
def n_tour_helper(board, cur_tour, n):
if len(cur_tour) == n*n:
return 1
count = 0
cur_r, cur_c = cur_tour[-1]
for r,c in valid_moves(board, cur_r, cur_c, n):
cur_tour.append((r,c))
board[r][c] = len(cur_tour)
count += n_tour_helper(board, cur_tour, n)
board[r][c] = None
cur_tour.pop()
return count
def get_n_tours(n):
count = 0
for i in range(n):
for j in range(n):
board = [[None for i in range(n)] for j in range(n)]
board[i][j] = 0
count += n_tour_helper(board, [(i, j)], n)
return count
if __name__ == '__main__':
ns = [1, 4]
for n in ns:
print(n, get_n_tours(n))Given a 2D matrix of characters and a target word, write a function that returns whether the word can be found in the matrix by going left-to-right, or up-to-down.
For example, given the following matrix:
[['F', 'A', 'C', 'I'], ['O', 'B', 'Q', 'P'], ['A', 'N', 'O', 'B'], ['M', 'A', 'S', 'S']] and the target word 'FOAM', you should return true, since it's the leftmost column. Similarly, given the target word 'MASS', you should return true, since it's the last row.
def has_word_util(matrix, i, j, word):
max_i = len(matrix)-1
max_j = len(matrix[0])-1
if not word:
return True
if (j<max_j and matrix[i][j+1] != word[0]) and (i<max_i and matrix[i+1][j] != word[0]):
return False
elif (j<max_j and matrix[i][j+1] == word[0]) and (i<max_i and matrix[i+1][j] == word[0]):
return has_word_util(matrix, i, j+1, word[1:]) or has_word_util(matrix, i+1, j, word[1:])
elif (j<max_j and matrix[i][j+1] == word[0]):
return has_word_util(matrix, i, j+1, word[1:])
elif (i<max_i and matrix[i+1][j] == word[0]):
return has_word_util(matrix, i+1, j, word[1:])
else:
return False
def has_word(matrix, word):
for i in range(len(matrix)):
for j in range(len(matrix[i])):
if matrix[i][j] == word[0]:
if has_word_util(matrix, i, j, word[1:]):
return True
return False
if __name__ == '__main__':
matrix = [['F', 'A', 'C', 'I'], ['O', 'B', 'Q', 'P'], ['A', 'N', 'O', 'B'], ['M', 'A', 'S', 'S']]
words = ['FOAM', 'MASS', 'QOBS', 'FOAB']
mat_str = """[['F', 'A', 'C', 'I'],
['O', 'B', 'Q', 'P'],
['A', 'N', 'O', 'B'],
['M', 'A', 'S', 'S']]"""
print(mat_str)
for word in words:
print(word, has_word(matrix, word))This problem was asked by Facebook.
There is an N by M matrix of zeroes. Given N and M, write a function to count the number of ways of starting at the top-left corner and getting to the bottom-right corner. You can only move right or down.
For example, given a 2 by 2 matrix, you should return 2, since there are two ways to get to the bottom-right:
Right, then down Down, then right Given a 5 by 5 matrix, there are 70 ways to get to the bottom-right.
def n_ways(n, m):
if n == 1 or m == 1:
return 1
return n_ways(n-1, m) + n_ways(n, m-1)
if __name__ == '__main__':
ns = [1, 2, 5]
ms = [2, 3, 5]
for n in ns:
for m in ms:
print(n, m, n_ways(n, m))This problem was asked by Google.
Implement integer exponentiation. That is, implement the pow(x, y) function, where x and y are integers and returns x^y.
Do this faster than the naive method of repeated multiplication.
For example, pow(2, 10) should return 1024.
# Log(Y) multiplications instead of Y.
def pow(x, y):
if y == 0:
return 1
if y%2 == 0:
return pow(x, y/2)*pow(x, y/2)
else:
return x*pow(x, y-1)
if __name__ == '__main__':
xs = [2, 3, 4, 5]
ys = [2, 3, 4, 5]
for x in xs:
for y in ys:
print(str(x)+'^'+str(y)+' = '+str(pow(x, y)))This problem was asked by Facebook.
Given a multiset of integers, return whether it can be partitioned into two subsets whose sums are the same.
For example, given the multiset {15, 5, 20, 10, 35, 15, 10}, it would return true, since we can split it up into {15, 5, 10, 15, 10} and {20, 35}, which both add up to 55.
Given the multiset {15, 5, 20, 10, 35}, it would return false, since we can't split it up into two subsets that add up to the same sum.
def sub_sum(arr, k):
if k == 0:
return True
if len(arr) == 0:
return False
return sub_sum(arr[1:], k-arr[0]) or sub_sum(arr[1:], k)
def set_split(arr):
sum_arr = sum(arr)
if sum_arr % 2 != 0:
return False
# Now find subset of elements whose sum is sum_arr/2
return sub_sum(arr, sum_arr/2)
if __name__ == '__main__':
arr = [15, 5, 20, 10, 35, 15, 10]
print(arr, set_split(arr))
arr = [15, 5, 20, 10, 35]
print(arr, set_split(arr))This problem was asked by Google.
Implement a file syncing algorithm for two computers over a low-bandwidth network. What if we know the files in the two computers are mostly the same?
Solution : https://scammingthecodinginterview.com/coding_problems/005.html
This problem was asked by Amazon.
An sorted array of integers was rotated an unknown number of times.
Given such an array, find the index of the element in the array in faster than linear time. If the element doesn't exist in the array, return null.
For example, given the array [13, 18, 25, 2, 8, 10] and the element 8, return 4 (the index of 8 in the array).
You can assume all the integers in the array are unique.
def log_search(arr, ele, st=0, end=None):
if not end:
end = len(arr)-1
if st > end:
return 'null'
mid = (st+end)/2
if ele == arr[mid]:
return mid
if ele > arr[mid]:
return log_search(arr, ele, mid+1, end)
return log_search(arr[:mid], ele, st, mid-1)
def get_index(arr, ele):
pivot = len(arr)/2
cliff = 0
if pivot != 0:
while arr[pivot] > arr[pivot-1] and arr[pivot] < arr[pivot+1]:
iterations += 1
if arr[pivot] > arr[0]:
pivot = (len(arr)+pivot)/2
if pivot == len(arr)-1:
pivot = 0
break
else:
pivot = pivot/2
cliff = pivot
if ele == arr[cliff]:
return cliff
elif ele > arr[0]:
return log_search(arr[0:cliff], ele)
else:
idx = log_search(arr[cliff:], ele)
if idx == 'null':
return idx
return cliff+idx
if __name__ == '__main__':
arr = [13, 18, 25, 2, 8, 10]
ele = 8
print(str(ele)+' occurs at index '+str(get_index(arr, ele))+' in '+str(arr))
ele = 18
print(str(ele)+' occurs at index '+str(get_index(arr, ele))+' in '+str(arr))
ele = 10
print(str(ele)+' occurs at index '+str(get_index(arr, ele))+' in '+str(arr))
ele = 2
print(str(ele)+' occurs at index '+str(get_index(arr, ele))+' in '+str(arr))This problem was asked by Amazon.
Given a string s and an integer k, break up the string into multiple lines such that each line has a length of k or less. You must break it up so that words don't break across lines. Each line has to have the maximum possible amount of words. If there's no way to break the text up, then return null.
You can assume that there are no spaces at the ends of the string and that there is exactly one space between each word.
For example, given the string "the quick brown fox jumps over the lazy dog" and k = 10, you should return: ["the quick", "brown fox", "jumps over", "the lazy", "dog"]. No string in the list has a length of more than 10.
def break_text(line, k):
words = line.split()
split_lines = []
cur_line = ''
for word in words:
word_len = len(word)
if word_len > k:
return 'null'
elif len(cur_line)+word_len+1 > k:
split_lines.append(cur_line)
cur_line = word
else:
cur_line += ' '+word
cur_line = cur_line.strip()
split_lines.append(cur_line)
return split_lines
if __name__ == '__main__':
in_str = 'the quick brown fox jumps over the lazy dog'
k = 10
print(in_str, break_text(in_str, k))This problem was asked by Google.
Given an undirected graph represented as an adjacency matrix and an integer k, write a function to determine whether each vertex in the graph can be colored such that no two adjacent vertices share the same color using at most k colors.
class Graph:
def __init__(self, n_vertices, graph):
self.n_vertices = n_vertices
self.graph = graph
def check_color(self, v_id, colors, cur_color):
for i in range(self.n_vertices):
if self.graph[v_id][i] == 1 and colors[i] == cur_color:
return False
return True
def fill_util(self, colors, k, n_filled):
# All filled!
if n_filled == self.n_vertices:
return True
for col in range(1, k+1):
if not self.check_color(n_filled, colors, col):
continue
colors[n_filled] = col
if self.fill_util(colors, k, n_filled+1):
return True
colors[n_filled] = 0
return False
def fill_color(self, k):
colors = [0 for i in range(self.n_vertices)]
for vert in range(self.n_vertices):
if not self.fill_util(colors, k, vert):
return False
return colors
if __name__ == '__main__':
matrix = [[0, 1, 1, 1],
[1, 0, 1, 0],
[1, 1, 0, 0],
[1, 0, 0, 0]]
graph = Graph(4, matrix)
print(matrix)
k = 3
print(k, graph.fill_color(k))
k = 2
print(k, graph.fill_color(k))This problem was asked by Microsoft.
Implement a URL shortener with the following methods:
shorten(url), which shortens the url into a six-character alphanumeric string, such as zLg6wl.
restore(short), which expands the shortened string into the original url. If no such shortened string exists, return null.
Hint: What if we enter the same URL twice?
# Imports
import random
class Bitly:
def __init__(self):
self.map = {}
self.rev_map = {}
self.lexicon = [str(num) for num in range(10)] + \
[chr(i) for i in range(65,91)] + \
[chr(i) for i in range(97,123)]
self.short_len = 6
def shorten(self, url):
if url not in self.map:
# TODO : This would cause lots of collisions after a while! And how to handle a distributed system?
short = ''.join(random.sample(self.lexicon, self.short_len))
while short in self.rev_map:
short = ''.join(random.sample(self.lexicon, self.short_len))
self.map[url] = short
self.rev_map[short] = url
return self.map[url]
def restore(self, short):
if short in self.rev_map:
return self.rev_map[short]
else:
return 'null'
if __name__ == '__main__':
shortner = Bitly()
urls = ['https://mail.google.com/mail/u/1/#inbox',
'https://mail.google.com/mail/u/0/#label/Daily+Coding+Problem/FMfcgxwCgLnwDmvSLmHJklKFKLssFHxC',
'https://music.youtube.com/watch?v=HWpZ_rOe_f0&list=RDMMKf781IbMQXk',
'https://github.com/NightFury13/Daily_Coding_Problem',
'https://github.com/NightFury13/Daily_Coding_Problem']
for url in urls:
shortner.shorten(url)
print(shortner.map)
print(shortner.rev_map)This problem was asked by Dropbox.
Sudoku is a puzzle where you're given a partially-filled 9 by 9 grid with digits. The objective is to fill the grid with the constraint that every row, column, and box (3 by 3 subgrid) must contain all of the digits from 1 to 9.
Implement an efficient sudoku solver.
# Imports
import numpy as np
def check(game, col, row, val):
# Check row-col
for i in range(9):
if game[i][row] == val:
return False
if game[col][i] == val:
return False
# Check 3x3 block
for i in range(3):
for j in range(3):
if game[3*(col/3)+i][3*(row/3)+j] == val:
return False
return True
def get_empty_cell(game):
for i in range(9):
for j in range(9):
if game[i][j] == 0:
return (i,j)
return False
def sudoku_solver(game):
empty_cell = get_empty_cell(game)
# Game solved!
if not empty_cell:
return True
col, row = empty_cell
for val in range(1, 10):
if check(game, col, row, val):
game[col][row] = val
if sudoku_solver(game):
return True
game[col][row] = 0
return False
if __name__ == '__main__':
game = np.zeros((9,9))
game[0][0] = 9
game[3][2] = 8
game[8][1] = 7
game[1][3] = 6
game[4][4] = 5
game[8][5] = 4
game[0][8] = 3
game[4][7] = 2
game[8][6] = 1
print(game)
print(sudoku_solver(game))
print(game)This problem was asked by Apple.
Implement a queue using two stacks. Recall that a queue is a FIFO (first-in, first-out) data structure with the following methods: enqueue, which inserts an element into the queue, and dequeue, which removes it.
class Stack:
def __init__(self, stack_list=None):
# IMPORTANT! Default instantiation for mutables workaround below
self.stack = stack_list if stack_list is not None else []
def push(self, ele):
self.stack.append(ele)
return
def pop(self):
if self.stack:
return self.stack.pop(-1)
return None
class StackQueue:
def __init__(self, stackA=Stack(), stackB=Stack()):
self.stackA = stackA
self.stackB = stackB
def enqueue(self, ele):
val = self.stackA.pop()
while val is not None:
self.stackB.push(val)
val = self.stackA.pop()
self.stackA.push(ele)
val = self.stackB.pop()
while val is not None:
self.stackA.push(val)
val = self.stackB.pop()
return
def dequeue(self):
return self.stackA.pop()
if __name__ == '__main__':
st_queue = StackQueue()
vals = [1,5,2,7,3,9,4]
for i, val in enumerate(vals):
st_queue.enqueue(val)
print('Enqueue', val, st_queue.stackA.stack)
if i%3==0:
pop_val = st_queue.dequeue()
print('Dequeue', pop_val, st_queue.stackA.stack)This problem was asked by Google.
Implement an LRU (Least Recently Used) cache. It should be able to be initialized with a cache size n, and contain the following methods:
set(key, value): sets key to value. If there are already n items in the cache and we are adding a new item, then it should also remove the least recently used item. get(key): gets the value at key. If no such key exists, return null. Each operation should run in O(1) time.
class DLL:
def __init__(self, root):
self.root = root
self.tail = root
self.size = 0 # Since the first node is a Null-Node
def push(self, node):
self.tail.child = node
node.parent = self.tail
self.tail = node
self.size += 1
return
def pop(self, node):
self.size -= 1
if self.root.child == node:
self.root.child = node.child
self.root.parent = None
return
if self.tail == node:
self.tail = node.parent
self.tail.child = None
return
node.parent.child = node.child
node.child.parent = node.parent
return
class LLNode:
def __init__(self, val, parent=None, child=None):
self.val = val
self.parent = parent
self.child = child
class LRU:
def __init__(self, n):
self.cache = {}
self.queue = DLL(LLNode('Null'))
self.allowed_max_size = n
def set(self, key, val):
if key in self.cache:
self.queue.pop(self.cache[key]['node'])
key_node = LLNode(key)
self.queue.push(key_node)
if self.queue.size > self.allowed_max_size:
self.cache.pop(self.queue.root.child.val) # Root is Null-Node
self.queue.pop(self.queue.root.child)
self.cache[key] = {'val':val, 'node':key_node}
return
def get(self, key):
if key not in self.cache:
return 'null'
self.queue.pop(self.cache[key]['node'])
key_node = LLNode(key)
self.queue.push(key_node)
self.cache[key]['node'] = key_node
return self.cache[key]['val']
def queue_to_list(queue):
out_arr = []
node = queue.root
while node.child:
out_arr.append(node.val)
node = node.child
out_arr.append(node.val)
return out_arr
if __name__ == '__main__':
lru_cache = LRU(3)
print(5, lru_cache.get(5))
lru_cache.set(1, 'A')
lru_cache.set(2, 'B')
lru_cache.set(3, 'C')
print(2, lru_cache.get(2))
lru_cache.set(1, 'A-new')
lru_cache.set(4, 'D')
print(3, lru_cache.get(3))
print(1, lru_cache.get(1))
print(2, lru_cache.get(2))
print(4, lru_cache.get(4))This problem was asked by Facebook.
Given a function that generates perfectly random numbers between 1 and k (inclusive), where k is an input, write a function that shuffles a deck of cards represented as an array using only swaps.
It should run in O(N) time.
Hint: Make sure each one of the 52! permutations of the deck is equally likely.
# Fisher-Yates Shuffle Algo : https://www.geeksforgeeks.org/shuffle-a-given-array-using-fisher-yates-shuffle-algorithm/
import random
def randk(k):
return random.randint(1, k)
def shuffle(cards):
for i in range(52):
# Get index between [i, 51] (inclusive)
swap_with = i+randk(52-i)-1
cards[i], cards[swap_with] = cards[swap_with], cards[i]
return cards
if __name__ == '__main__':
cards = range(52)
print(cards)
print(shuffle(cards))This problem was asked by Microsoft.
Suppose an arithmetic expression is given as a binary tree. Each leaf is an integer and each internal node is one of '+', '-', '*', or '/'.
Given the root to such a tree, write a function to evaluate it.
For example, given the following tree:
*
/ \
/ \ /
3 2 4 5
You should return 45, as it is (3 + 2) * (4 + 5).
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
def eval_tree(tree):
if tree:
if tree.val is '+':
return eval_tree(tree.left) + eval_tree(tree.right)
elif tree.val is '-':
return eval_tree(tree.left) - eval_tree(tree.right)
elif tree.val is '*':
return eval_tree(tree.left) * eval_tree(tree.right)
elif tree.val is '/':
# Handle division by zero if required
return eval_tree(tree.left) / eval_tree(tree.right)
else:
return tree.val
if __name__ == '__main__':
tree = Node('*', Node('+', Node(3), Node(2)), Node('+', Node(4), Node(5)))
print(eval_tree(tree))This problem was asked by Amazon.
Given an array of numbers, find the maximum sum of any contiguous subarray of the array.
For example, given the array [34, -50, 42, 14, -5, 86], the maximum sum would be 137, since we would take elements 42, 14, -5, and 86.
Given the array [-5, -1, -8, -9], the maximum sum would be 0, since we would not take any elements.
Do this in O(N) time.
def max_subarr_sum(arr):
max_sum = 0
cur_sum = 0
for ele in arr:
if cur_sum + ele > 0:
cur_sum += ele
if cur_sum > max_sum:
max_sum = cur_sum
else:
cur_sum = 0
return max_sum
if __name__ == '__main__':
in_arrs = [[34, -50, 42, 14, -5, 86], [-5, -1, -8, -9]]
for in_arr in in_arrs:
print(in_arr, max_subarr_sum(in_arr))This problem was asked by Google.
Given pre-order and in-order traversals of a binary tree, write a function to reconstruct the tree.
For example, given the following preorder traversal:
[a, b, d, e, c, f, g]
And the following inorder traversal:
[d, b, e, a, f, c, g]
You should return the following tree:
a
/ \
b c
/ \ / \
d e f g
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
def create_tree(preorder, inorder):
if not preorder:
return None
root_val = preorder[0]
root_inorder_idx = inorder.index(root_val)
left_inorder = inorder[:root_inorder_idx]
right_inorder = inorder[root_inorder_idx+1:]
left_preorder = [val for val in left_inorder]
left_preorder.sort(key=lambda x: preorder.index(x))
right_preorder = [val for val in right_inorder]
right_preorder.sort(key=lambda x: preorder.index(x))
tree = Node(root_val, \
create_tree(left_preorder, left_inorder), \
create_tree(right_preorder, right_inorder))
return tree
def print_postorder(tree):
if tree:
print_postorder(tree.left)
print_postorder(tree.right)
print(tree.val)
if __name__ == '__main__':
preorder = ['a', 'b', 'd', 'e', 'c', 'f', 'g']
inorder = ['d', 'b', 'e', 'a', 'f', 'c', 'g']
tree = create_tree(preorder, inorder)
print(preorder, inorder)
print('postorder :')
print_postorder(tree)This problem was asked by Facebook.
Given a array of numbers representing the stock prices of a company in chronological order, write a function that calculates the maximum profit you could have made from buying and selling that stock once. You must buy before you can sell it.
For example, given [9, 11, 8, 5, 7, 10], you should return 5, since you could buy the stock at 5 dollars and sell it at 10 dollars.
def max_profit(stocks):
min_price = stocks[0]
max_profit = 0
for price in stocks:
if price < min_price:
min_price = price
profit = price-min_price
if profit > max_profit:
max_profit = profit
return max_profit
if __name__ == '__main__':
stocks = [9, 11, 8, 5, 7, 10]
print(stocks, max_profit(stocks))This problem was asked by Amazon.
Given a string, find the longest palindromic contiguous substring. If there are more than one with the maximum length, return any one.
For example, the longest palindromic substring of "aabcdcb" is "bcdcb". The longest palindromic substring of "bananas" is "anana".
def longest_palin(in_str):
str_len = len(in_str)
palin_dp = [[False for i in range(str_len)] for j in range(str_len)]
longest = ''
# DP init
for i in range(str_len):
palin_dp[i][i] = True
if i < str_len-1 and in_str[i] == in_str[i+1]:
palin_dp[i][i+1] = True
# DP expand
for palin_len in range(3, str_len+1):
for i in range(str_len-palin_len+1):
j = i+palin_len-1
if palin_dp[i+1][j-1] and in_str[i] == in_str[j]:
palin_dp[i][j] = True
if len(in_str[i:j+1]) > len(longest):
longest = in_str[i:j+1]
return longest
if __name__ == '__main__':
in_strs = ['aabcdcb', 'bananas', 'palinilap']
for in_str in in_strs:
print(in_str, longest_palin(in_str))This problem was asked by Two Sigma.
Using a function rand5() that returns an integer from 1 to 5 (inclusive) with uniform probability, implement a function rand7() that returns an integer from 1 to 7 (inclusive).
Upgrade to premium and get in-depth solutions to every problem.
# Source : https://www.geeksforgeeks.org/generate-integer-from-1-to-7-with-equal-probability/
import random
def rand5():
return random.randint(1, 5)
def rand7():
val = 5*rand5() + rand5() - 5
if val < 22:
return val%7 + 1
return rand7()
if __name__ == '__main__':
ctr = {}
for i in range(10000):
val = rand7()
if val not in ctr:
ctr[val] = 1
else:
ctr[val] += 1
print("After 10k iterations")
print(ctr)This problem was asked by Google.
We can determine how "out of order" an array A is by counting the number of inversions it has. Two elements A[i] and
A[j] form an inversion if A[i] > A[j] but i < j. That is, a smaller element appears after a larger element.
Given an array, count the number of inversions it has. Do this faster than O(N^2) time.
You may assume each element in the array is distinct.
For example, a sorted list has zero inversions. The array [2, 4, 1, 3, 5] has three inversions: (2, 1), (4, 1), and (4, 3). The array [5, 4, 3, 2, 1] has ten inversions: every distinct pair forms an inversion.
def get_min_idx(arr):
min_idx = 0
for i in range(1, len(arr)):
if arr[i] < arr[min_idx]:
min_idx = i
return min_idx
def get_inversions(arr):
n_inv = 0
l_arr = len(arr)
for i in range(l_arr):
# Note this is min_idx of arr[i:] not arr[:]
min_idx = get_min_idx(arr[i:])
n_inv += min_idx
val = arr.pop(i+min_idx)
arr = [val]+arr
return n_inv
if __name__ == '__main__':
arrs = [[2, 4, 1, 3, 5], [5, 4, 3, 2, 1], [1, 2, 3, 4, 5]]
for arr in arrs:
print(arr)
print('Inversions : '+str(get_inversions(arr)))This problem was asked by Amazon.
Implement a stack that has the following methods:
push(val), which pushes an element onto the stack
pop(), which pops off and returns the topmost element of the stack. If there are no elements in the stack, then it
should throw an error or return null.
max(), which returns the maximum value in the stack currently. If there are no elements in the stack, then it should
throw an error or return null.
Each method should run in constant time.
class MaxStack:
def __init__(self):
self.stack = []
self.max = None
def push(self, val):
if not self.max:
self.stack.append(val)
self.max = val
elif val < self.max:
self.stack.append(val)
else:
self.stack.append(val + self.max)
self.max = val
def pop(self):
if not self.stack:
return 'null'
top_val = self.stack.pop()
if top_val < self.max:
return top_val
else:
self.max = top_val - self.max
return top_val - self.max
def getmax(self):
return self.max
if __name__ == '__main__':
arr = [1, 2, 5, 3, 3, 4]
m_stack = MaxStack()
for ele in arr:
m_stack.push(ele)
print(ele, m_stack.getmax(), m_stack.stack)
arr.reverse()
for ele in arr:
print('Pop : '+str(m_stack.pop()))
print(ele, m_stack.getmax(), m_stack.stack)This problem was asked by Google.
Given a list of integers S and a target number k, write a function that returns a subset of S that adds up to k. If such a subset cannot be made, then return null.
Integers can appear more than once in the list. You may assume all numbers in the list are positive.
For example, given S = [12, 1, 61, 5, 9, 2] and k = 24, return [12, 9, 2, 1] since it sums up to 24.
def sub_sum_k(arr, k):
if len(arr) == 0:
if k == 0:
return True
else:
return False
return sub_sum_k(arr[:-1], k-arr[-1]) or sub_sum_k(arr[:-1], k)
if __name__ == '__main__':
arr = [12, 1, 61, 5, 9, 2]
print(arr)
k = 24
print(k, sub_sum_k(arr, k))
k = 4
print(k, sub_sum_k(arr, k))This problem was asked by Facebook.
Given an unordered list of flights taken by someone, each represented as (origin, destination) pairs, and a starting airport, compute the person's itinerary. If no such itinerary exists, return null. If there are multiple possible itineraries, return the lexicographically smallest one. All flights must be used in the itinerary.
For example, given the list of flights [('SFO', 'HKO'), ('YYZ', 'SFO'), ('YUL', 'YYZ'), ('HKO', 'ORD')] and starting airport 'YUL', you should return the list ['YUL', 'YYZ', 'SFO', 'HKO', 'ORD'].
Given the list of flights [('SFO', 'COM'), ('COM', 'YYZ')] and starting airport 'COM', you should return null.
Given the list of flights [('A', 'B'), ('A', 'C'), ('B', 'C'), ('C', 'A')] and starting airport 'A', you should return the list ['A', 'B', 'C', 'A', 'C'] even though ['A', 'C', 'A', 'B', 'C'] is also a valid itinerary. However, the first one is lexicographically smaller.
def valid_iten(flights, iten):
if not flights:
return iten
cur_airport = iten[-1]
outbound_flights = [(idx, flt) for idx, flt in enumerate(flights) if flt[0] is cur_airport]
for idx, flt in sorted(outbound_flights, key=lambda x:x[1]):
iten.append(flt[1])
remain_flights = flights[:idx]+flights[idx+1:]
if valid_iten(remain_flights, iten) != 'null':
return iten
iten.pop()
return 'null'
if __name__ == '__main__':
flights = [[('SFO', 'HKO'), ('YYZ', 'SFO'), ('YUL', 'YYZ'), ('HKO', 'ORD')],
[('SFO', 'COM'), ('COM', 'YYZ')],
[('A', 'B'), ('A', 'C'), ('B', 'C'), ('C', 'A')]
]
origins = ['YUL', 'COM', 'A']
for flts, orig in zip(flights, origins):
print(flts)
print('Origin : '+orig)
iten = valid_iten(flts, [orig])
if iten != 'null':
print('Itinerary : '+'-->'.join(iten))
else:
print('Itinerary : null')This problem was asked by Google.
Given an array of integers where every integer occurs three times except for one integer, which only occurs once, find and return the non-duplicated integer.
For example, given [6, 1, 3, 3, 3, 6, 6], return 1. Given [13, 19, 13, 13], return 19.
Do this in O(N) time and O(1) space.
import numpy as np
# O(N) time and O(N) space
def non_triplet(arr):
set_sum = np.sum(list(set(arr)))
return (3*set_sum - np.sum(arr))/2
if __name__ == '__main__':
inp_arr = [6, 1, 3, 3, 3, 6, 6]
print(inp_arr, non_triplet(inp_arr))
inp_arr = [13, 19, 13, 13]
print(inp_arr, non_triplet(inp_arr))This problem was asked by Google.
The power set of a set is the set of all its subsets. Write a function that, given a set, generates its power set.
For example, given the set {1, 2, 3}, it should return {{}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}.
You may also use a list or array to represent a set.
def make_powerset(in_set):
powset_len = pow(2, len(in_set))
powerset = []
for i in range(powset_len):
subset = []
for ch_id, char in enumerate(format(i, 'b').zfill(3)):
if char is '1':
subset.append(in_set[ch_id])
powerset.append(subset)
return powerset
if __name__ == '__main__':
in_set = [1, 2, 3]
print('Input : '+str(in_set))
print('Powerset : '+str(make_powerset(in_set)))This problem was asked by Dropbox.
Given the root to a binary search tree, find the second largest node in the tree.
# Naive Solution (MaxHeap)
# >heapq._heapify_max(bst)
# >n=2
# >for i in range(n):
# > ans = heapq.heappop(bst)
# > return ans
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
def anti_in_order(root, order_arr, k):
if root:
anti_in_order(root.right, order_arr, k)
if len(order_arr) < k:
order_arr.append(root.val)
else:
return
anti_in_order(root.left, order_arr, k)
def k_max(bst, k):
max_arr = []
anti_in_order(bst, max_arr, k)
return max_arr[-1]
if __name__ == '__main__':
bst = Node(10, Node(5, Node(3), Node(6)), Node(12, None, Node(14, Node(13))))
print(k_max(bst, k=2))This problem was asked by Google.
Given an array of strictly the characters 'R', 'G', and 'B', segregate the values of the array so that all the Rs come first, the Gs come second, and the Bs come last. You can only swap elements of the array.
Do this in linear time and in-place.
For example, given the array ['G', 'B', 'R', 'R', 'B', 'R', 'G'], it should become ['R', 'R', 'R', 'G', 'G', 'B', 'B'].
def swap_sort(arr):
st_ptr = 0
end_ptr = len(arr)-1
while True:
while arr[st_ptr] is 'R' and st_ptr < end_ptr:
st_ptr += 1
while arr[end_ptr] is not 'R' and st_ptr <= end_ptr:
end_ptr -= 1
if st_ptr > end_ptr:
break
arr[st_ptr], arr[end_ptr] = arr[end_ptr], arr[st_ptr]
end_ptr = len(arr)-1
while True:
while arr[st_ptr] is not 'B' and st_ptr < end_ptr:
st_ptr += 1
while arr[end_ptr] is 'B' and st_ptr <= end_ptr:
end_ptr -= 1
if st_ptr > end_ptr:
break
arr[st_ptr], arr[end_ptr] = arr[end_ptr], arr[st_ptr]
return arr
if __name__ == '__main__':
in_arr = ['G', 'B', 'R', 'R', 'B', 'R', 'G']
print(in_arr)
print(swap_sort(in_arr))This problem was asked by Microsoft.
Compute the running median of a sequence of numbers. That is, given a stream of numbers, print out the median of the list so far on each new element.
Recall that the median of an even-numbered list is the average of the two middle numbers.
For example, given the sequence [2, 1, 5, 7, 2, 0, 5], your algorithm should print out:
2
1.5
2
3.5
2
2
2
def insert_into(ele, sorted_arr):
lo = 0
hi = len(sorted_arr)
while lo < hi:
mid = (hi+lo)/2
if ele < sorted_arr[mid]:
hi = mid
else:
lo = mid+1
return sorted_arr[:lo]+[ele]+sorted_arr[lo:]
def run_median(arr):
sorted_arr = []
for i in range(len(arr)):
sorted_arr = insert_into(arr[i], sorted_arr)
# Odd sized list
if i%2==0:
print(float(sorted_arr[i/2]))
else:
print(float(sorted_arr[i/2]+sorted_arr[(i/2)+1])/2)
return
if __name__ == '__main__':
arr = [2,1,5,7,2,0,5]
print(arr)
run_median(arr)This problem was asked by Jane Street.
Suppose you are given a table of currency exchange rates, represented as a 2D array. Determine whether there is a possible arbitrage: that is, whether there is some sequence of trades you can make, starting with some amount A of any currency, so that you can end up with some amount greater than A of that currency.
There are no transaction costs and you can trade fractional quantities.
# SIDE NOTE : This is basically Bellman-Ford, we are trying to see if there are any negative cycles.
# https://www.dailycodingproblem.com/blog/how-to-find-arbitrage-opportunities-in-python/
from math import log
def arbitrage(n_cur, exchange):
src = 0
dist = [float('inf')] * n_cur
n_edg = len(exchange)
dist[src] = 0
log_exchange = [(u, v, log(wt)) for u,v,wt in exchange]
for i in range(n_cur - 1):
for u, v, wt in log_exchange:
if dist[v] > dist[u] + wt:
dist[v] = dist[u] + wt
for u, v, wt in log_exchange:
if dist[v] > dist[u] + wt:
print(u, v, wt, dist[u], dist[v])
return True
return False
if __name__ == '__main__':
n_cur = 4
# The inputs should be create a cycle to test this. Ideally, they should be complete.
exchange = [(0, 1, 70.0), (1, 0, 1.0/70),
(1, 2, 10.0), (2, 1, 1.0/10),
(2, 3, 40.0), (3, 2, 1.0/40),
(0, 3, 100.0), (3, 0, 1.0/100)]
print(arbitrage(n_cur, exchange))
exchange = [(0, 1, 70.0), (1, 0, 1.0/70),
(1, 2, 10.0), (2, 1, 1.0/10),
(2, 3, 40.0), (3, 2, 1.0/40),
(0, 3, 28000.0), (3, 0, 1.0/28000)]
print(arbitrage(n_cur, exchange))This problem was asked by Google.
The edit distance between two strings refers to the minimum number of character insertions, deletions, and substitutions required to change one string to the other. For example, the edit distance between "kitten" and "sitting" is three: substitute the "k" for "s", substitute the "e" for "i", and append a "g".
Given two strings, compute the edit distance between them.
def editdistance(str1, str2, l1, l2, DP):
for i in range(l1+1):
for j in range(l2+1):
if i == 0:
DP[i][j] = j
elif j == 0:
DP[i][j] = i
elif str1[i-1] == str2[j-1]:
DP[i][j] = DP[i-1][j-1]
else:
DP[i][j] = 1 + min(DP[i-1][j-1], DP[i][j-1], DP[i-1][j])
return DP[l1][l2]
if __name__ == '__main__':
for str1, str2 in (('kitten', 'sitting'), ('mohit', 'rohit'), ('teapot', 'eats')):
l1 = len(str1)
l2 = len(str2)
DP = [[0 for j in range(l2+1)] for i in range(l1+1)]
print(str1, str2, editdistance(str1, str2, l1, l2, DP))This problem was asked by Facebook.
You are given an array of non-negative integers that represents a two-dimensional elevation map where each element is unit-width wall and the integer is the height. Suppose it will rain and all spots between two walls get filled up.
Compute how many units of water remain trapped on the map in O(N) time and O(1) space.
For example, given the input [2, 1, 2], we can hold 1 unit of water in the middle.
Given the input [3, 0, 1, 3, 0, 5], we can hold 3 units in the first index, 2 in the second, and 3 in the fourth index (we cannot hold 5 since it would run off to the left), so we can trap 8 units of water.
def max_fill(wall):
water_fill = 0
max_left, max_right = 0, 0
left_pt, right_pt = 0, len(wall)-1
while left_pt < right_pt:
if wall[left_pt] < wall[right_pt]:
if wall[left_pt] > max_left:
max_left = wall[left_pt]
water_fill += max_left - wall[left_pt]
left_pt += 1
else:
if wall[right_pt] > max_right:
max_right = wall[right_pt]
water_fill += max_right - wall[right_pt]
right_pt -= 1
return water_fill
if __name__ == '__main__':
walls = [[2, 1, 2], [3, 0, 1, 3, 0, 5], [6, 3, 0, 1, 3, 0, 5]]
for wall in walls:
print(wall, 'Max Fill : ', max_fill(wall))This problem was asked by Amazon.
Run-length encoding is a fast and simple method of encoding strings. The basic idea is to represent repeated successive characters as a single count and character. For example, the string "AAAABBBCCDAA" would be encoded as "4A3B2C1D2A".
Implement run-length encoding and decoding. You can assume the string to be encoded have no digits and consists solely of alphabetic characters. You can assume the string to be decoded is valid.
DIGITS = [str(i) for i in range(10)]
def encode(in_str):
enc = ''
ctr = 1
for i in range(1, len(in_str)):
if in_str[i] == in_str[i-1]:
ctr+=1
else:
enc += str(ctr)+in_str[i-1]
ctr = 1
enc+= str(ctr)+in_str[-1]
return enc
def decode(in_str):
dec = ''
idx = 0
repeat = ''
while idx < len(in_str):
if in_str[idx] in DIGITS:
repeat += in_str[idx]
else:
dec+=(in_str[idx]*int(repeat))
repeat = ''
idx+=1
return dec
if __name__ == '__main__':
in_str = 'AAAABBBCCDAA'
print('In : '+in_str+' | Encoded : '+encode(in_str)+' | Decoded : '+decode(encode(in_str))+' | Correct? : '+str(in_str==decode(encode(in_str))))This problem was asked by Palantir.
Write an algorithm to justify text. Given a sequence of words and an integer line length k, return a list of strings which represents each line, fully justified.
More specifically, you should have as many words as possible in each line. There should be at least one space between each word. Pad extra spaces when necessary so that each line has exactly length k. Spaces should be distributed as equally as possible, with the extra spaces, if any, distributed starting from the left.
If you can only fit one word on a line, then you should pad the right-hand side with spaces.
Each word is guaranteed not to be longer than k.
For example, given the list of words ["the", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog"] and k = 16, you should return the following:
["the quick brown", # 1 extra space on the left
"fox jumps over", # 2 extra spaces distributed evenly
"the lazy dog"] # 4 extra spaces distributed evenly
def fix_line(word_list, k):
n_words = len(word_list)
if n_words < 2:
return word_list
cur_len = len(''.join(word_list))
repeat = (k-cur_len)/(n_words-1)
word_list = [word_list[0]] + [(' '*repeat)+sub_word for sub_word in word_list[1:]]
extra = k - len(''.join(word_list))
for i in range(extra):
word_list[i] = word_list[i]+' '
return word_list
def justify(words, k):
text = []
line_len = len(words[0])
text.append([words[0]])
for idx, word in enumerate(words[1:]):
if line_len+len(' '+word) <= k:
text[-1].append(' '+word)
line_len += len(' '+word)
# Last word reached
if idx == len(words)-2:
text[-1] = fix_line(text[-1], k)
else:
# Fix last complete line
text[-1] = fix_line(text[-1], k)
# Place word in new line
text.append([word])
line_len = len(word)
text = [''.join(line) for line in text]
return text
if __name__ == '__main__':
all_words = (["the", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog"],
["my", "name", "is", "mohit", "jain.", "What", "is", "your", "name?"])
all_k = (16, 12)
for words, k in zip(all_words, all_k):
print(words, k)
text = justify(words, k)
for line in text:
print(line)This problem was asked by Facebook.
Given a string of round, curly, and square open and closing brackets, return whether the brackets are balanced (well-formed).
For example, given the string "([])", you should return true.
Given the string "([)]" or "((()", you should return false.
def well_formed(in_str):
stack = []
b_map = {')':'(', '}':'{', ']':'['}
for bracket in in_str:
if bracket not in b_map:
stack.append(bracket)
elif stack[-1] == b_map[bracket]:
stack.pop(-1)
else:
return False
if stack:
return False
return True
if __name__ == '__main__':
in_strs = ['([])[]({})', '([)]', '((()']
for in_str in in_strs:
print(in_str, well_formed(in_str))This problem was asked by Google.
Given a singly linked list and an integer k, remove the kth last element from the list. k is guaranteed to be smaller than the length of the list.
The list is very long, so making more than one pass is prohibitively expensive.
Do this in constant space and in one pass.
class Node:
def __init__(self, val, next_node=None):
self.val = val
self.next = next_node
def remove_k_last(ll_head, k):
front_ptr = ll_head
for i in range(k):
front_ptr = front_ptr.next
if not front_ptr:
ll_head = ll_head.next
else:
shadow_ptr = ll_head
while front_ptr.next:
front_ptr = front_ptr.next
shadow_ptr = shadow_ptr.next
shadow_ptr.next = shadow_ptr.next.next
return ll_head
def print_list(ll_head):
out_str = [str(ll_head.val)]
while ll_head.next:
ll_head = ll_head.next
out_str.append(str(ll_head.val))
print('('+') -> ('.join(out_str)+')')
if __name__ == '__main__':
ll_head = Node(1, Node(2, Node(3, Node(4, Node(5)))))
print_list(ll_head)
k = 2
print('k = '+str(k))
ll_head = remove_k_last(ll_head, k)
print_list(ll_head)
k = 3
print('k = '+str(k))
ll_head = remove_k_last(ll_head, k)
print_list(ll_head)
k = 3
print('k = '+str(k))
ll_head = remove_k_last(ll_head, k)
print_list(ll_head)This problem was asked by Facebook.
Implement regular expression matching with the following special characters:
. (period) which matches any single character
* (asterisk) which matches zero or more of the preceding element
That is, implement a function that takes in a string and a valid regular expression and returns whether or not the string matches the regular expression.
For example, given the regular expression "ra." and the string "ray", your function should return true. The same regular expression on the string "raymond" should return false.
Given the regular expression ".*at" and the string "chat", your function should return true. The same regular expression on the string "chats" should return false.
def re(in_str, re_exp):
if not re_exp:
return not in_str
prefix_match = bool(in_str) and re_exp[0] in (in_str[0], '.')
if len(re_exp) > 1 and re_exp[1] == '*':
return prefix_match and (re(in_str[1:], re_exp[2:]) or re(in_str[1:], re_exp))
else:
return prefix_match and re(in_str[1:], re_exp[1:])
if __name__ == '__main__':
in_strs = ['ray', 'raymond', 'chat', 'chats']
re_exps = ['ra.', 'ra.', '.*at', '.*at']
for in_str, re_exp in zip(in_strs, re_exps):
print(re_exp, in_str, re(in_str, re_exp))
print('~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~')This problem was asked by Google.
Implement locking in a binary tree. A binary tree node can be locked or unlocked only if all of its descendants or ancestors are not locked.
Design a binary tree node class with the following methods:
is_locked, which returns whether the node is locked
lock, which attempts to lock the node. If it cannot be locked, then it should return false. Otherwise, it should
lock it and return true.
unlock, which unlocks the node. If it cannot be unlocked, then it should return false. Otherwise, it should unlock
it and return true.
You may augment the node to add parent pointers or any other property you would like. You may assume the class is used in a single-threaded program, so there is no need for actual locks or mutexes. Each method should run in O(h), where h is the height of the tree.
class LockNode:
def __init__(self, val, lock_state=False, left=None, right=None, parent=None):
self.val = val
self.lock = lock_state
self.left = left
self.right = right
self.parent = parent
self.ct_locked_child = 0
def is_locked(self):
return self.lock
def _lockable_or_unlockable(self):
if self.ct_locked_child > 0:
return False
cur_node = self.parent
while cur_node:
if cur_node.lock:
return False
cur_node = cur_node.parent
return True
def lock(self):
if self._lockable_or_unlockable():
self.lock = True
cur_node = self.parent
while cur_node:
cur_node.ct_locked_child += 1
cur_node = cur_node.parent
return True
else:
return False
def unlock(self):
if self._lockable_or_unlockable():
self.lock = False
cur_node = self.parent
while cur_node:
cur_node.ct_locked_child -= 1
cur_node = cur_node.parent
return True
else:
return FalseThis problem was asked by Google.
You are given an M by N matrix consisting of booleans that represents a board. Each True boolean represents a wall. Each False boolean represents a tile you can walk on.
Given this matrix, a start coordinate, and an end coordinate, return the minimum number of steps required to reach the end coordinate from the start. If there is no possible path, then return null. You can move up, left, down, and right. You cannot move through walls. You cannot wrap around the edges of the board.
For example, given the following board:
[[f, f, f, f],
[t, t, f, t],
[f, f, f, f],
[f, f, f, f]]
and start = (3, 0) (bottom left) and end = (0, 0) (top left), the minimum number of steps required to reach the end is 7, since we would need to go through (1, 2) because there is a wall everywhere else on the second row
import numpy as np
def is_valid(x, y, board, visited):
b_xmax, b_ymax = len(board), len(board[0])
if x < b_xmax and \
y < b_ymax and \
x >= 0 and \
y >= 0 and \
visited[x][y] < 0 and \
not board[x][y]:
return True
return False
def min_steps(board, start, end):
# Start or End itself is a wall
if board[start[0]][start[1]] or board[end[0]][end[1]]:
return -1
visited = [ [-1 for i in range(len(board[0]))] for j in range(len(board)) ]
queue = [start]
visited[start[0]][start[1]] = 0
while queue:
node_x, node_y = queue.pop(0)
cur_depth = visited[node_x][node_y]
if (node_x, node_y) == end:
print(np.array(visited))
return cur_depth
if is_valid(node_x+1, node_y, board, visited):
visited[node_x+1][node_y] = cur_depth+1
queue.append((node_x+1, node_y))
if is_valid(node_x, node_y+1, board, visited):
visited[node_x][node_y+1] = cur_depth+1
queue.append((node_x, node_y+1))
if is_valid(node_x-1, node_y, board, visited):
visited[node_x-1][node_y] = cur_depth+1
queue.append((node_x-1, node_y))
if is_valid(node_x, node_y-1, board, visited):
visited[node_x][node_y-1] = cur_depth+1
queue.append((node_x, node_y-1))
return -1
if __name__ == '__main__':
board = [[False, False, False, False],
[True, True, False, True],
[False, False, False, False],
[False, False, False, False]]
start = (3,0)
end = (0,0)
print(min_steps(board, start, end))This problem was asked by Microsoft.
Given a dictionary of words and a string made up of those words (no spaces), return the original sentence in a list. If there is more than one possible reconstruction, return any of them. If there is no possible reconstruction, then return null.
For example, given the set of words 'quick', 'brown', 'the', 'fox', and the string "thequickbrownfox", you should return ['the', 'quick', 'brown', 'fox'].
Given the set of words 'bed', 'bath', 'bedbath', 'and', 'beyond', and the string "bedbathandbeyond", return either ['bed', 'bath', 'and', 'beyond] or ['bedbath', 'and', 'beyond'].
def breakdown(st, vocab, broken=[[]]):
for i in range(len(st)):
if st[:i+1] in vocab:
broken[-1].append(st[:i+1])
if i+1 == len(st):
broken.append([])
return broken
broken = breakdown(st[i+1:], vocab, broken)
return broken
if __name__ == '__main__':
vocab = ['bed', 'bath', 'bedbath', 'and', 'beyond']
st = 'bedbathandbeyond'
print(st, vocab)
broken = breakdown(st, vocab, [[]])
print(broken)
vocab = ['quick', 'brown', 'the', 'fox']
st = 'thequickbrownfoxx'
print(st, vocab)
broken = breakdown(st, vocab, [[]])
print(broken)This problem was asked by Snapchat.
Given an array of time intervals (start, end) for classroom lectures (possibly overlapping), find the minimum number of rooms required.
For example, given [(30, 75), (0, 50), (60, 150)], you should return 2.
def schedule(intervals):
sorted_intv = sorted(intervals, key=lambda x: x[1])
rooms = [[sorted_intv[0]]]
for interval in sorted_intv[1:]:
placed = False
for room in rooms:
if room[-1][1] < interval[0]:
room.append(interval)
placed = True
break
if not placed:
rooms.append([interval])
print(rooms)
return len(rooms)
if __name__ == '__main__':
intervals = [(30, 75), (0, 50), (60, 150)]
print(intervals)
print("Rooms : " +str(schedule(intervals)))This problem was asked by Google.
Given two singly linked lists that intersect at some point, find the intersecting node. The lists are non-cyclical.
For example, given A = 3 -> 7 -> 8 -> 10 and B = 99 -> 1 -> 8 -> 10, return the node with value 8.
In this example, assume nodes with the same value are the exact same node objects.
Do this in O(M + N) time (where M and N are the lengths of the lists) and constant space.
def intersection(A, B):
if len(A) > len(B):
diff = len(A) - len(B)
longer_list = A
shorter_list = B
else:
diff = len(B) - len(A)
longer_list = B
shorter_list = A
pt_1, pt_2 = longer_list[diff], shorter_list[0]
for i in range(1, len(shorter_list)):
if pt_1 == pt_2:
return pt_1
pt_1 = longer_list[diff+i]
pt_2 = shorter_list[i]
return None
if __name__ == '__main__':
A = [3, 7, 8, 10]
B = [99, 1, 8, 10]
print(intersection(A,B))This problem was asked by Facebook.
A builder is looking to build a row of N houses that can be of K different colors. He has a goal of minimizing cost while ensuring that no two neighboring houses are of the same color.
Given an N by K matrix where the nth row and kth column represents the cost to build the nth house with kth color, return the minimum cost which achieves this goal.
import numpy as np
def min_cost(nk_mat):
n_row, n_col = nk_mat.shape
costs = np.zeros((n_row, n_col))
costs[0] = nk_mat[0]
for row in range(1, n_row):
for col in range(n_col):
costs[row][col] = nk_mat[row][col] + np.min(np.concatenate([costs[row-1][:col], costs[row-1][col+1:]]))
return np.min(costs[-1])
if __name__ == '__main__':
arr = [[1,2,3], [4,5,6], [7,8,9]]
print(np.array(arr))
print(min_cost(np.array(arr)))
arr = [[1,4,7], [2,5,8], [3,6,9]]
print(np.array(arr))
print(min_cost(np.array(arr)))This problem was asked by Google.
Given an array of integers and a number k, where 1 <= k <= length of the array, compute the maximum values of each subarray of length k.
For example, given array = [10, 5, 2, 7, 8, 7] and k = 3, we should get: [10, 7, 8, 8], since:
10 = max(10, 5, 2)
7 = max(5, 2, 7)
8 = max(2, 7, 8)
8 = max(7, 8, 7)
Do this in O(n) time and O(k) space. You can modify the input array in-place and you do not need to store the results. You can simply print them out as you compute them.
# Not O(n)
def max_val(arr, k):
max_val = max(arr[:k])
print(max_val)
for i in range(len(arr)-k):
if arr[i+k] > max_val:
max_val = arr[i+k]
print(max_val)
elif arr[i] != max_val:
print(max_val)
else:
max_val = max(arr[i+1:i+1+k])
print(max_val)
if __name__ == '__main__':
arr = [10, 5, 2, 7, 8, 7]
k = 3
max_val(arr, k)This problem was asked by Google.
Suppose we represent our file system by a string in the following manner:
The string "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext" represents:
dir subdir1 subdir2 file.ext The directory dir contains an empty sub-directory subdir1 and a sub-directory subdir2 containing a file file.ext.
The string "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext" represents:
dir subdir1 file1.ext subsubdir1 subdir2 subsubdir2 file2.ext The directory dir contains two sub-directories subdir1 and subdir2. subdir1 contains a file file1.ext and an empty second-level sub-directory subsubdir1. subdir2 contains a second-level sub-directory subsubdir2 containing a file file2.ext.
We are interested in finding the longest (number of characters) absolute path to a file within our file system. For example, in the second example above, the longest absolute path is "dir/subdir2/subsubdir2/file2.ext", and its length is 32 (not including the double quotes).
Given a string representing the file system in the above format, return the length of the longest absolute path to a file in the abstracted file system. If there is no file in the system, return 0.
Note:
The name of a file contains at least a period and an extension.
The name of a directory or sub-directory will not contain a period.
def longest_abs_path(in_path):
chunks = in_path.split('\n')
cur_path_prefix_lens = []
cur_path_prefix = []
longest_abs_path = []
longest_abs_path_len = -1
for f_name in chunks:
depth = 0
while f_name[depth] == '\t':
depth += 1
if depth < len(cur_path_prefix):
cur_path_prefix = cur_path_prefix[:depth]
cur_path_prefix_lens = cur_path_prefix_lens[:depth]
cur_path_prefix.append(f_name.strip())
cur_path_prefix_lens.append(len(f_name.strip()) + 1)
if '.' in f_name:
if sum(cur_path_prefix_lens) > longest_abs_path_len:
longest_abs_path = '/'.join(cur_path_prefix)
longest_abs_path_len = sum(cur_path_prefix_lens)
return longest_abs_path
if __name__ == '__main__':
in_str = 'dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext'
print(longest_abs_path(in_str))This problem was asked by Twitter.
You run an e-commerce website and want to record the last N order ids in a log. Implement a data structure to accomplish this, with the following API:
record(order_id): adds the order_id to the log get_last(i): gets the ith last element from the log. i is guaranteed to be smaller than or equal to N. You should be as efficient with time and space as possible.
class API:
def __init__(self, N):
self.log = [None] * N
def record(self, order_id):
self.log = [order_id]+self.log[:-1]
def get_last(self, i):
return self.log[-i]
if __name__ == '__main__':
api = API(4)
for i in range(10):
api.record(i)
if i % 3:
log = []
for j in range(1,5):
log.append(api.get_last(j))
print(log)This problem was asked by Facebook.
Given a stream of elements too large to store in memory, pick a random element from the stream with uniform probability.
SOLUTION : This genre of problems is known as Reservoir Sampling (randomly select k elements from a 'large' set of n elements) readup : https://www.geeksforgeeks.org/select-a-random-number-from-stream-with-o1-space/ readup : https://www.geeksforgeeks.org/reservoir-sampling/
This problem was asked by Google.
The area of a circle is defined as (pi)r^2. Estimate (pi) to 3 decimal places using a Monte Carlo method.
Hint: The basic equation of a circle is x2 + y2 = r2.
# READUP : https://www.geeksforgeeks.org/estimating-value-pi-using-monte-carlo/
import random
def pi_estimate(n_iter):
circle_pts = 0
square_pts = 0
iter_size = 1000000
for epoch in range(n_iter):
for sample in range(iter_size):
rand_x = random.random()
rand_y = random.random()
if (rand_x*rand_x)+(rand_y*rand_y) <= 1:
circle_pts += 1
square_pts += 1
pi = 4*(float(circle_pts)/square_pts)
return pi
if __name__ == '__main__':
for i in range(1, 10):
print('Iter '+str(i), pi_estimate(i))This problem was asked by Amazon.
Given an integer k and a string s, find the length of the longest substring that contains at most k distinct characters.
For example, given s = "abcba" and k = 2, the longest substring with k distinct characters is "bcb".
def longest_subs(s, k):
chars_map = {}
cur_st = 0
longest_st = 0
longest_len = 0
for i, char in enumerate(s):
if char in chars_map:
chars_map[char] += 1
else:
chars_map[char] = 1
if len(chars_map) <= k:
if i-cur_st+1 > longest_len:
longest_len = i-cur_st+1
longest_st = cur_st
else:
if chars_map[s[cur_st]] == 1:
chars_map.pop(s[cur_st])
else:
chars_map[s[cur_st]] -= 1
cur_st += 1
return (s[longest_st:longest_st+longest_len])
if __name__ == '__main__':
s = 'abcba'
k = 2
print(s, k, '->', longest_subs(s, k))
s = 'abcadcacacaca'
k = 3
print(s, k, '->', longest_subs(s, k))This problem was asked by Amazon.
There exists a staircase with N steps, and you can climb up either 1 or 2 steps at a time. Given N, write a function that returns the number of unique ways you can climb the staircase. The order of the steps matters.
For example, if N is 4, then there are 5 unique ways:
1, 1, 1, 1
2, 1, 1
1, 2, 1
1, 1, 2
2, 2
EXTRA : What if, instead of being able to climb 1 or 2 steps at a time, you could climb any number from a set of positive integers X? For example, if X = {1, 3, 5}, you could climb 1, 3, or 5 steps at a time.
def climb_ways(n):
if n == 1:
return [[1]]
if n == 2:
return [[1,1], [2]]
return [[1]+way for way in climb_ways(n-1)] + [[2]+way for way in climb_ways(n-2)]
if __name__ == '__main__':
n = 4
print(n)
ways = climb_ways(n)
for way in ways:
print way
n = 7
print(n)
ways = climb_ways(n)
for way in ways:
print wayThis problem was asked by Twitter.
Implement an autocomplete system. That is, given a query string s and a set of all possible query strings, return all strings in the set that have s as a prefix.
For example, given the query string de and the set of strings [dog, deer, deal], return [deer, deal].
Hint: Try preprocessing the dictionary into a more efficient data structure to speed up queries.
class Node:
def __init__(self, val=None, children=[], c_vals=[]):
self.val = val
self.child = children
self.c_vals = c_vals
def create_tree(head, all_s):
for word in all_s:
ptr = Node(word[0])
head.child.append(ptr)
head.c_vals.append(word[0])
for char in word[1:]:
if char not in ptr.c_vals:
ptr.c_vals.append(char)
c_node = Node(char)
ptr.child.append(c_node)
ptr = [c_node for c_node in ptr.child if c_node.val is char][0]
# Add EOW char ' '
ptr.c_vals.append(' ')
ptr.child.append(Node(' '))
return head
def subtree_walk(ptr, prefix):
suggestions = []
node_stack = ptr.child
word_stack = [prefix+node.val for node in node_stack]
while node_stack:
pop_node = node_stack.pop(0)
pop_word = word_stack.pop(0)
if pop_node.val == ' ':
suggestions.append(pop_word.strip())
print(suggestions)
else:
child_nodes = pop_node.child
add_word_stack = [pop_word+node.val for node in child_nodes]
node_stack += child_nodes
word_stack += add_word_stack
return suggestions
def autocomplete(s, s_tree_head):
prefix = ''
ptr = s_tree_head
for char in s:
if char in ptr.c_vals:
prefix += char
ptr = [node for node in ptr.child if node.val is char][0]
else:
return []
return subtree_walk(ptr, prefix)
if __name__ == '__main__':
s = 'de'
all_s = ['dog', 'deer', 'deal']
print('All Strings : '+ str(all_s))
print('Query : '+ s)
head = Node()
s_tree_head = create_tree(head, all_s)
print(s_tree_head.child)
vals = autocomplete(s, s_tree_head)
print('AutoComplete : '+ str(vals))This problem was asked by Apple.
Implement a job scheduler which takes in a function f and an integer n, and calls f after n milliseconds.
import sched
import time
def printer():
print("Executed at : "+ str(time.time()))
def job_sched(scheduler, f, n):
scheduler.enter(float(n)/100, 1, f, ())
scheduler.run()
if __name__ == '__main__':
scheduler = sched.scheduler(time.time, time.sleep)
print("Start at : "+ str(time.time()))
job_sched(scheduler, printer, 10)This problem was asked by Airbnb.
Given a list of integers, write a function that returns the largest sum of non-adjacent numbers. Numbers can be 0 or negative.
For example, [2, 4, 6, 2, 5] should return 13, since we pick 2, 6, and 5. [5, 1, 1, 5] should return 10, since we pick 5 and 5.
Follow-up: Can you do this in O(N) time and constant space?
def maxsum(arr):
if not arr:
return 0
elif len(arr) == 1:
return arr[0]
sum = [0] * (len(arr)+1)
sum[0] = 0
sum[1] = arr[0]
sum[2] = max(arr[0], arr[1])
for i in range(3, len(arr)+1):
sum[i] = max(sum[i-2] + arr[i-1], sum[i-1])
return sum[-1]
if __name__ == '__main__':
arr = [2,4,6,2,5]
print(arr, maxsum(arr))
arr = [5,1,1,5]
print(arr, maxsum(arr))This problem was asked by Google.
A unival tree (which stands for "universal value") is a tree where all nodes under it have the same value.
Given the root to a binary tree, count the number of unival subtrees.
For example, the following tree has 5 unival subtrees:
0
/ \
1 0
/ \
1 0
/ \
1 1
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
self.is_unival = False
def unival_ct(node):
node_ct = 0
# Leaf nodes are unival
if not node.left or not node.right:
node.is_unival = True
node_ct = 1
return node_ct
# Recurse
left_ct = unival_ct(node.left)
right_ct = unival_ct(node.right)
if node.left.is_unival and node.right.is_unival and \
node.left.val == node.right.val:
node.is_unival = True
node_ct = 1
return node_ct + left_ct + right_ct
if __name__ == '__main__':
head = Node(0)
head.left = Node(1)
#head.left.left = Node(1)
#head.left.right = Node(1)
head.right = Node(0)
head.right.left = Node(1)
head.right.left.left = Node(1)
head.right.left.right = Node(1)
head.right.right = Node(0)
print(unival_ct(head))This problem was asked by Facebook.
Given the mapping a = 1, b = 2, ... z = 26, and an encoded message, count the number of ways it can be decoded.
For example, the message '111' would give 3, since it could be decoded as 'aaa', 'ka', and 'ak'.
You can assume that the messages are decodable. For example, '001' is not allowed.
def count_decode(message):
arr = [0] * (len(message)+1)
arr[0] = 1
arr[1] = 1
for i in range(2, len(message)+1):
arr[i] = 0
if message[i-1] > '0':
arr[i] = arr[i-1]
if message[i-2] == '1' or (message[i-2] == '2' and message[i-1] < '7'):
arr[i] += arr[i-2]
return arr[-1]
if __name__ == '__main__':
mes = '111'
print(mes, count_decode(mes))
mes = '131'
print(mes, count_decode(mes))
mes = '1312'
print(mes, count_decode(mes))
mes = '1013'
print(mes, count_decode(mes))This problem was asked by Google.
An XOR linked list is a more memory efficient doubly linked list. Instead of each node holding next and prev fields, it holds a field named both, which is an XOR of the next node and the previous node. Implement an XOR linked list; it has an add(element) which adds the element to the end, and a get(index) which returns the node at index.
If using a language that has no pointers (such as Python), you can assume you have access to get_pointer and dereference_pointer functions that converts between nodes and memory addresses.
https://www.geeksforgeeks.org/xor-linked-list-a-memory-efficient-doubly-linked-list-set-1/
cons(a, b) constructs a pair, and car(pair) and cdr(pair) returns the first and last element of that pair. For example, car(cons(3, 4)) returns 3, and cdr(cons(3, 4)) returns 4.
Given this implementation of cons:
def cons(a, b):
def pair(f):
return f(a, b)
return pairImplement car and cdr.
def cons(a, b):
def pair(f):
return f(a, b)
return pair
def car(f):
def left(a, b):
return a
return f(left)
def cdr(f):
def right(a, b):
return b
return f(right)
if __name__ == '__main__':
print(car(cons(3, 4)))
print(cdr(cons(3, 4)))Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well.
For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3.
You can modify the input array in-place.
SOLUTIONS : Brute force - search for all numbers n+1 if array of size n [ O(n^2)] Sort - then simple linear search [ O(logn + n) ] Hash - keep a dict of all +ve int, then do another linear iteration of N+1 elements in hash [ O(n) + O(n) space ]
Best -
- Segregate positive numbers from others i.e., move all non-positive numbers to left side. In the following code, segregate() function does this part.
- Now we can ignore non-positive elements and consider only the part of array which contains all positive elements. We traverse the array containing all positive numbers and to mark presence of an element x, we change the sign of value at index x to negative. We traverse the array again and print the first index which has positive value.
def segregate(arr):
pos_idx = 0
for i in range(len(arr)):
if arr[i] < 1:
arr[i], arr[pos_idx] = arr[pos_idx], arr[i]
pos_idx = pos_idx+1
return pos_idx
# Making general solution for finding Kth missing int
def missing_int(arr, k):
pos_idx = segregate(arr)
arr = arr[pos_idx:]
l_arr = len(arr)
for i in range(l_arr):
if arr[i] - 1 < l_arr:
arr[arr[i]-1] = -1 * abs(arr[arr[i]-1])
miss_ctr = 0
for i in range(l_arr):
if arr[i] > 0:
miss_ctr += 1
if miss_ctr is k:
return i+1
if __name__ == '__main__':
for k, arr in ((1, [3,4,-1,1]), (2, [3,4,-1,1]), (3, [1,2,0]), (1, [3,5,-1,5,0,2,7,1])):
print(arr, k, missing_int(arr, k))Given the root to a binary tree, implement serialize(root), which serializes the tree into a string, and deserialize(s), which deserializes the string back into the tree.
For example, given the following Node class
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = rightThe following test should pass:
node = Node('root', Node('left', Node('left.left')), Node('right'))
assert deserialize(serialize(node)).left.left.val == 'left.left'
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
def serialize(node):
if not node:
return '()'
if not node.left and not node.right:
return '("'+node.val+'")'
return '("'+node.val+'", '+serialize(node.left)+', '+serialize(node.right)+')'
def deserialize(tree_str):
code_string = 'tree = '+ tree_str.replace(', ()','').replace('(','Node(')
print("Deserialized : "+code_string)
exec code_string
return tree
if __name__ == '__main__':
node = Node('root', Node('left', Node('left.left')), Node('right'))
print("Serialized : "+serialize(node))
assert deserialize(serialize(node)).left.left.val == 'left.left'Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i.
For example, if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], the expected output would be [2, 3, 6]
import numpy as np
def not_prod(arr):
tot_prod = np.prod(arr)
prod_arr = [tot_prod/num for num in arr]
return prod_arr
def not_prod_nodiv(arr):
l_arr = len(arr)
left, right = np.ones(l_arr), np.ones(l_arr)
for i in range(1, l_arr):
left[i] = left[i-1]*arr[i-1]
right[l_arr-1-i] = right[l_arr-i]*arr[l_arr-i]
prod_arr = [left[i]*right[i] for i in range(l_arr)]
return prod_arrGiven a list of numbers and a number k, return whether any two numbers from the list add up to k.
For example, given [10, 15, 3, 7] and k of 17, return true since 10 + 7 is 17.
NOTE : If negative numbers are also to be handled, just add the smallest number of the set to all elements and carry on with below algos
# Brute Force
def sum_true_bf(ele_list, k):
for i in range(len(ele_list)):
for j in range(i+1, len(ele_list)):
if ele_list[i]+ele_list[j] == k:
return True
return False
# Single Pass
def sum_true_f(ele_list, k):
diff_list = []
for i in range(len(ele_list)):
diff = k - ele_list[i]
if diff in diff_list:
return True
diff_list.append(ele_list[i])
return False