updated structure

Signed-off-by: akshatnema <[email protected]>

GFG/Product array puzzle - GFG/README.md

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+# Product array puzzle
+## Easy 
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given an array <strong>nums[]</strong> of size <strong>n</strong>, construct a Product Array <strong>P</strong> (of same size n) such that<strong> P[i]</strong> is equal to the product of all the elements of <strong>nums</strong> except nums[i].</span></p>
+
+<p>&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Example 1:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:
+</strong>n = 5
+nums[] = {10, 3, 5, 6, 2}
+<strong>Output:
+</strong>180 600 360 300 900<strong>
+Explanation: </strong>
+For i=0, P[i] = 3*5*6*2 = 180.
+For i=1, P[i] = 10*5*6*2 = 600.
+For i=2, P[i] = 10*3*6*2 = 360.
+For i=3, P[i] = 10*3*5*2 = 300.
+For i=4, P[i] = 10*3*5*6 = 900.</span>
+</pre>
+
+<p><span style="font-size:18px"><strong>Example 2:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:
+</strong>n = 2
+nums[] = {12,0}
+<strong>Output:
+</strong>0 12</span>
+
+</pre>
+
+<p><span style="font-size:18px"><strong>Your Task:</strong><br>
+You do not have to read input. Your task is to complete the function&nbsp;<strong>productExceptSelf() </strong>that takes array nums[] and n&nbsp;as input parameters and returns a list of n&nbsp;integers denoting the product array P.&nbsp;If the array has only one element the returned list should&nbsp;should contains one value i.e {1}</span><br>
+<span style="font-size:18px"><strong>Note:&nbsp;</strong>Try to solve this problem without using the division operation.</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Expected Time Complexity:&nbsp;</strong>O(n)<br>
+<strong>Expected Auxiliary Space:&nbsp;</strong>O(n)</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Constraints:</strong><br>
+1 &lt;= n &lt;= 1000<br>
+0 &lt;= nums<sub>i</sub> &lt;= 200</span><br>
+<span style="font-size:18px">Array&nbsp;may contain duplicates.</span></p>
+ <p></p>
+            </div>

GFG/Product array puzzle - GFG/product-array-puzzle.cpp

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+// { Driver Code Starts
+//Initial template for C++
+
+#include<bits/stdc++.h>
+using namespace std;
+
+
+
+ // } Driver Code Ends
+//User function template for C++
+
+class Solution{
+  public:
+    // nums: given vector
+    // return the Product vector P that hold product except self at each index
+    vector<long long int> productExceptSelf(vector<long long int>& nums, int n) {
+       vector<long long int> pre1(n);
+       vector<long long int> pre2(n);
+       pre1[0]=nums[0];
+       for(int i=1;i<n;i++)
+        pre1[i]=pre1[i-1]*nums[i];
+       pre2[n-1]=nums[n-1];
+       for(int i=n-2;i>=0;i--){
+           pre2[i]=pre2[i+1]*nums[i];
+       }
+       vector<long long int> ans;
+       for(int i=0;i<n;i++){
+           long long int x=1,y=1;
+           if(i-1>=0) x=x*pre1[i-1];
+           if(i+1<n) y=y*pre2[i+1];
+           ans.push_back(x*y);
+       }
+       return ans;
+    }
+};
+
+
+// { Driver Code Starts.
+int main()
+ {
+    int t;  // number of test cases
+    cin>>t;
+    while(t--)
+    {
+        int n;  // size of the array
+        cin>>n;
+        vector<long long int> arr(n),vec(n);
+
+        for(int i=0;i<n;i++)    // input the array
+        {
+            cin>>arr[i];
+        }
+        Solution obj;
+        vec = obj.productExceptSelf(arr,n);   // function call
+
+        for(int i=0;i<n;i++)    // print the output
+        {
+            cout << vec[i] << " ";
+        }
+        cout<<endl;
+    }
+	return 0;
+}  // } Driver Code Ends

GFG/Replace O's with X's - GFG/README.md

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+# Replace O's with X's
+## Medium 
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given a matrix <strong>mat</strong> of size <strong>N x M</strong> where every element is either ‘O’ or ‘X’.<br>
+Replace all ‘O’ with ‘X’ that are surrounded by ‘X’.<br>
+A ‘O’ (or a set of ‘O’) is considered to be by surrounded by ‘X’ if there are ‘X’ at locations just below, just above, just left and just right of it.</span></p>
+
+<p><strong><span style="font-size:18px">Example 1:</span></strong></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong> n = 5, m = 4
+mat = {{'X', 'X', 'X', 'X'}, 
+&nbsp;      {'X', 'O', 'X', 'X'}, 
+&nbsp;      {'X', 'O', 'O', 'X'}, 
+&nbsp;      {'X', 'O', 'X', 'X'}, 
+&nbsp;      {'X', 'X', 'O', 'O'}}
+<strong>Output:</strong> ans = {{'X', 'X', 'X', 'X'}, 
+&nbsp;              {'X', 'X', 'X', 'X'}, 
+&nbsp;              {'X', 'X', 'X', 'X'}, 
+&nbsp;              {'X', 'X', 'X', 'X'}, 
+&nbsp;              {'X', 'X', 'O', 'O'}}
+<strong>Explanation:</strong> Following the rule the above 
+matrix is the resultant matrix. </span>
+</pre>
+
+<p><span style="font-size:18px"><strong>Your Task:</strong><br>
+You do not need to read input or print anything. Your task is to complete the function <strong>fill()</strong> which takes n, m and mat as input parameters ad returns a 2D array representing the resultant matrix.</span></p>
+
+<p><span style="font-size:18px"><strong>Expected Time Complexity:</strong> O(n*m)<br>
+<strong>Expected Auxiliary Space:</strong> O(n*m)</span></p>
+
+<p><span style="font-size:18px"><strong>Constraints:</strong><br>
+1 ≤ n, m ≤ 500</span></p>
+ <p></p>
+            </div>

GFG/Replace O's with X's - GFG/replace-os-with-xs.cpp

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+// { Driver Code Starts
+// Initial Template for C++
+
+#include <bits/stdc++.h>
+using namespace std;
+
+ // } Driver Code Ends
+// User function Template for C++
+
+class Solution{
+public:
+    void change(vector<vector<char>> &mat, int i, int j){
+        mat[i][j]='*';
+        int dx[]={0,0,-1,1};
+        int dy[]={-1,1,0,0};
+        for(int k=0;k<4;k++){
+           int l=i+dx[k];
+            int m=j+dy[k];
+            if(l>=0 && m>=0 && l<mat.size() && m<mat[0].size() && mat[l][m]=='O')
+             change(mat,l,m);
+        }
+        return ;
+    }
+
+    vector<vector<char>> fill(int n, int m, vector<vector<char>> mat)
+    {
+        for(int i=0;i<n;i++)
+          for(int j=0;j<m;j++){
+              if(i==0 || j==0 || i==n-1 || j==m-1){
+                  if(mat[i][j]=='O')
+                   change(mat,i,j);
+              }
+         }
+
+        for(int i=0;i<n;i++)
+          for(int j=0;j<m;j++){
+               if(mat[i][j]=='O')
+                 mat[i][j]='X';
+          } 
+
+        for(int i=0;i<n;i++)
+          for(int j=0;j<m;j++){
+               if(mat[i][j]=='*')
+                 mat[i][j]='O';
+          } 
+
+        return mat;
+    }
+};
+
+// { Driver Code Starts.
+
+int main(){
+    int t;
+    cin>>t;
+    while(t--){
+        int n, m;
+        cin>>n>>m;
+        vector<vector<char>> mat(n, vector<char>(m, '.'));
+        for(int i = 0;i < n;i++)
+            for(int j=0; j<m; j++)
+                cin>>mat[i][j];
+
+        Solution ob;
+        vector<vector<char>> ans = ob.fill(n, m, mat);
+        for(int i = 0;i < n;i++) {
+            for(int j = 0;j < m;j++) {
+                cout<<ans[i][j]<<" ";
+            }
+            cout<<"\n";
+        }
+    }
+    return 0;
+}  // } Driver Code Ends

GFG/Union of two arrays - GFG/README.md

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+# Union of two arrays
+## Easy
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given two arrays <strong>a[]</strong>&nbsp;and <strong>b[]</strong>&nbsp;of size <strong>n</strong>&nbsp;and <strong>m</strong> respectively. The task is to find union between these two arrays. </span></p>
+
+<p><span style="font-size:18px">Union of the two arrays can be defined as the set containing distinct elements from both the arrays. If there are repetitions, then only one occurrence of element should be printed in the union.</span></p>
+
+<p><span style="font-size:18px"><strong>Example 1:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:
+</strong>5 3
+1 2 3 4 5
+1 2 3
+<strong>Output: 
+</strong>5<strong>
+Explanation: 
+</strong>1, 2, 3, 4 and 5 are the
+elements which comes in the union set
+of both arrays. So count is 5.</span>
+</pre>
+
+<p><span style="font-size:18px"><strong>Example 2:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:
+</strong>6 2 
+85 25 1 32 54 6
+85 2 
+<strong>Output: 
+</strong>7<strong>
+Explanation: 
+</strong>85, 25, 1, 32, 54, 6, and
+2 are the elements which comes in the
+union set of both arrays. So count is 7.</span></pre>
+
+<p><strong><span style="font-size:18px">Your Task:</span></strong><br>
+<span style="font-size:18px">Complete <strong>doUnion </strong>funciton that takes<strong> a, n, b, m as parameters and returns</strong> the count of union elements of the&nbsp;two arrays. The <strong>printing </strong>is done by the <strong>driver </strong>code.</span></p>
+
+<p><span style="font-size:18px"><strong>Constraints:</strong></span><br>
+<span style="font-size:18px">1 ≤ n, m&nbsp;≤ 10<sup>5</sup><br>
+0 ≤ a[i], b[i] &lt;&nbsp;10<sup>5</sup></span></p>
+
+<p><span style="font-size:18px"><strong>Expected Time Complexity </strong>: O((n+m)log(n+m))<br>
+<strong>Expected Auxilliary Space</strong> : O(n+m)</span></p>
+ <p></p>
+            </div>

GFG/Union of two arrays - GFG/union-of-two-arrays.cpp

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+// { Driver Code Starts
+//Initial template for C++
+
+#include <bits/stdc++.h>
+using namespace std;
+
+ // } Driver Code Ends
+//User function template in C++
+
+class Solution{
+    public:
+    //Function to return the count of number of elements in union of two arrays.
+    int doUnion(int a[], int n, int b[], int m)  {
+        set<int> s;
+        for(int i=0;i<n;i++) s.insert(a[i]);
+        for(int i=0;i<m;i++) s.insert(b[i]);
+        return s.size();
+    }
+};
+
+// { Driver Code Starts.
+
+int main() {
+
+	int t;
+	cin >> t;
+
+	while(t--){
+
+	    int n, m;
+	    cin >> n >> m;
+	    int a[n], b[m];
+
+	    for(int i = 0;i<n;i++)
+	       cin >> a[i];
+
+	    for(int i = 0;i<m;i++)
+	       cin >> b[i];
+	    Solution ob;
+	    cout << ob.doUnion(a, n, b, m) << endl;
+
+	}
+
+	return 0;
+}  // } Driver Code Ends

README.md

@@ -1 +1,5 @@
-# DSA-Practice
+# DSA-Practice
+
+This repository contains the solution of the problems done on the Leetcode and the Geeksforgeeks. The sheet solved to practice the problems are as follows:
+
+[Coding Sheet](https://docs.google.com/spreadsheets/d/1NmGtqSNSeLmErfFZF90V77SSf-RdKiIMbufBqfeA4Ms/edit#gid=0)