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304.RangeSumQuery2D-Immutable(bruteforce2).py
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304.RangeSumQuery2D-Immutable(bruteforce2).py
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"""
Given a 2D matrix matrix, find the sum of the elements inside the rectangle
defined by its upper left corner (row1, col1) and lower right corner
(row2, col2).
Range Sum Query 2D
[
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, [2, 0, 1,] 5],
[4, [1, 0, 1,] 7],
[1, [0, 3, 0,] 5]
]
The above rectangle (with the red border) is defined by
(row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.
Example:
Given matrix = [
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, 2, 0, 1, 5],
[4, 1, 0, 1, 7],
[1, 0, 3, 0, 5]
]
sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12
Note:
1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.
"""
#Difficulty: Medium
#12 / 12 test cases passed.
#Runtime: 1032 ms
#Memory Usage: 16.3 MB
#Runtime: 1032 ms, faster than 20.82% of Python3 online submissions for Range Sum Query 2D - Immutable.
#Memory Usage: 16.3 MB, less than 6.03% of Python3 online submissions for Range Sum Query 2D - Immutable.
class NumMatrix:
def __init__(self, matrix: List[List[int]]):
self.matrix = matrix
self.sum = 0
for row in matrix:
self.sum += sum(row)
def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
matrix_sum = self.sum
for i in range(len(self.matrix)):
if i not in range(row1, row2+1):
matrix_sum -= sum(self.matrix[i])
else:
matrix_sum -= sum(self.matrix[i][:col1])
matrix_sum -= sum(self.matrix[i][col2+1:])
return matrix_sum
# Your NumMatrix object will be instantiated and called as such:
# obj = NumMatrix(matrix)
# param_1 = obj.sumRegion(row1,col1,row2,col2)