# Maximal Square in C++

Suppose we have a 2D binary matrix filled with 0's and 1's. We have to find the largest square containing only 1's and return its area. So if the matrix is like −

 1 0 1 0 0 1 0 1 1 0 1 1 1 1 1 1 0 0 1 0

Then the output will be 4

To solve this, we will follow these steps −

• ans := 0, n := number of rows, c := number of rows

• if n is 0, then return 0

• Create another matrix of order (n x c)

• for i in range 0 to n – 1

• for j in range 0 to c – 1

• m[i, j] := matrix[i, j]

• ans := maximum of m[i, j] and ans

• for j in range 0 to c – 1

• if m[i, j] is not 0, then

• m[i, j] := 1 + minimum of m[i + 1, j], m[i, j-1], m[i + 1, j-1],

• ans := maximum of m[i, j] and ans

• return ans*ans

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
int ans = 0;
int n = matrix.size();
if(!n)return 0;
int c = matrix.size();
vector<vector<int>> m(n, vector <int> (c));
for(int i =0;i<n;i++){
for(int j = 0; j<c;j++){
m[i][j] = matrix[i][j] - '0';
ans = max(m[i][j],ans);
}
}
for(int i =n-2;i>=0;i--){
for(int j =1;j<c;j++){
if(m[i][j]){
m[i][j] = 1 + min({m[i+1][j],m[i][j-1],m[i+1][j-1]});
}
ans = max(ans,m[i][j]);
}
}
return ans*ans;
}
};
main(){
vector<vector<char>> v = {{'1','0','1','0','0'},{'1','0','1','1','1'},{'1','1','1','1','1'},         {'1','0','0','1','0'}};
Solution ob;
cout << ((ob.maximalSquare(v)));
}

## Input

[["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]

## Output

4