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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
#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[0].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
Updated on: 30-Apr-2020
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