Merge pull request #398 from AbhiC7721/devincept2

Added maximal square question

Author: smv1999

Dynamic Programming/Maximal_square.cpp

@@ -0,0 +1,87 @@
+/*
+	Question => https://leetcode.com/problems/maximal-square/
+	
+	Description:
+	Given an m x n binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
+
+	Eg: 
+    m = 4 and n = 5
+
+	1 0  1 0  0
+	1 0 |1 1| 1
+	1 1 |1 1| 1 
+	1 0  0 1  0
+    
+	The answer here is: 4 (2x2) shown here
+
+	Soltuion:
+
+	The approach is to  use dynamic programming.
+	We use another matrix to store the maximal square possible using the (i,j)-th element of matrix
+	as the bottom-left corner of the matrix.
+	Base cases :
+	a) Fill the left-most column with 1 and (0 where matrix is 0) dp[i][n-1]=1 or 0
+	b) Fill the bottom-most row with 1 and (0 where matrix is 0). dp[m-1][i]=1 or 0
+	Then for general case:
+	dp[i][j] = 0 => if matrix[i][j]=0
+	dp[i][j] = 1 + min(dp[i+1][j], dp[i+1][j+1], dp[i][j+1])
+	for matrix[i][j]=> dp[i][j] is 1 + minimum of cell on right, bottom and diagonally right
+
+	The answer is the max. value in the dp matrix.
+    
+*/
+
+// The code for the solution => 
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int maximalSquare(vector<vector<char>>& matrix) {
+    int n = matrix.size();
+    int m = matrix[0].size();
+    vector<vector<int>>dp(n, vector<int>(m, 0)); // dp matrix
+    dp[n-1][m-1] = matrix[n-1][m-1]-'0';  // dp[n-][m-1] is same as matrix[n-1][m-1]
+    int ans = dp[n-1][m-1];
+    //last column is 1 or 0
+    for(int i=0;i<n-1;i++){
+        dp[i][m-1]=matrix[i][m-1]-'0';
+        ans = max(ans, dp[i][m-1]);
+    } 
+    //last row is 1 or 0
+    for(int j=0;j<m-1;j++){
+        dp[n-1][j]=matrix[n-1][j]-'0';
+        ans = max(ans, dp[n-1][j]);
+    }    
+    for(int i=n-2;i>=0;i--)
+    {
+        for(int j=m-2;j>=0;j--)
+        {
+            // dp[i][j] is 0 when matrix[i][j] is 0
+            if(matrix[i][j]=='0')
+                dp[i][j]=0; 
+            //dp[i][j] is 1 + min(right_hand, diagonally_down, down)  
+            else
+                dp[i][j] = 1+min(dp[i][j+1],min(dp[i+1][j],dp[i+1][j+1]));
+            // ans is max of all values in dp
+            ans =  max(dp[i][j], ans);
+        }
+    }
+    return ans*ans;
+}
+
+int main()
+{
+    int m,n;
+    cin>>m>>n;
+    vector<vector<char>>matrix(m, vector<char>(n, 0));
+    for(int i=0; i<m; i++)
+    {
+        for(int j=0; j<n; j++)
+        {
+            cin>>matrix[i][j];
+        }
+    }
+    //max. area of 1s
+    int ans = maximalSquare(matrix);
+    cout<<ans<<"\n";
+}