Files Committed

coding_freshmen/C++/Ashutosh_Panda/Readme.md

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+# Problem Title: Print Matrix in Spiral order
+
+## Problem Explanation
+
+Ex:- Given Matrix :
+                    1  2  3
+                    4  5  6
+                    7  8  9
+Output- 1 2 3 6 9 8 7 4 5
+
+## Logic and Intuition
+
+* An optimized approach will be to create 4 variables that store the upper ,lower ,left and right bound of the Matrix which decreases after each iteration.
+* We will iterate through every row and column in a spiral manner until the lower and right bound are greater than upper and left bound respectively.
+
+## Time Complexity and Space Complexity
+
+* Time Complexity: O(m*n)
+* Space Complexity: O(1)
+
+
+# Problem Title: Sieve of Eratosthenes
+
+To find all the prime numbers upto a given number.
+
+## Logic & Intuition
+
+* We will create a bool array which is used to check if a number is prime or not (initially all are set to true).
+* We will iterate the array $\sqrt{n}$ times and in each iteration we will mark all the multiples of i as false.
+* At the end of the loop the remaining array will contain only the prime numbers.
+
+## Time Complexity and Space Complexity
+
+* Time Complexity: O(n * log(log(n)) )
+* Space Complexity: O(n)

coding_freshmen/C++/Ashutosh_Panda/SoErato.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+void SioErato(int n) // Sieve of Eratosthenes
+{
+    int i;
+    bool t[n + 1];
+    fill(t, t + n + 1, true);
+    t[0] = false, t[1] = false;
+    for (i = 2; i * i <= n; i++)
+        if(t[j])
+            for (int j = 2 * i; j <= n; j += i)
+                if (j % i == 0)
+                    t[j] = false;
+
+    for (i = 1; i <= n; i++)
+        if (t[i] == true)
+            cout << i << " ";
+}
+int main()
+{
+    int n;
+    cout<<"Enter upper limit: ";
+    // cin>>n;
+    SioErato(1000);
+    return 0;
+}

coding_freshmen/C++/Ashutosh_Panda/spiral.cpp

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+#include<bits/stdc++.h>
+using namespace std;
+void spiOrd(vector<vector<int>> &m)
+{
+    int i, l = 0, u = 0, d = m.size(), r = m[0].size();
+    while (l < r and u < d)
+    {
+        for (i = l; i < r; i++)
+            cout<<m[u][i]<<" ";
+        u++;
+        for (i = u; i < d; i++)
+            cout<<m[i][r - 1]<<" ";
+        r--;
+        if (u < d)
+        {
+            for (i = r - 1; i >= l; i--)
+                cout<<m[d - 1][i]<<" ";
+            d--;
+        }
+        if (l < r)
+        {
+            for (i = d - 1; i >= u; i--)
+                cout<<m[i][l]<<" ";
+            l++;
+        }
+    }
+}
+
+int main()
+{
+    int n,r,c,i,j;
+    cout<<"enter no. of rows and columns: ";
+    cin>>r>>c;
+    vector<vector<int>>v(r,vector<int>(c,0));
+    for(i=0;i<r;i++)
+        for(j=0;j<c;j++)
+            cin>>v[i][j];
+    spiOrd(v);
+    return 0;
+}