# LeetCode – Maximal Square (Java)

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.

For example, given the following matrix:

```1101
1101
1111
```

Return 4.

Analysis

This problem can be solved by dynamic programming. The changing condition is:
t[i][j] = min(t[i][j-1], t[i-1][j], t[i-1][j-1]) + 1. It means the square formed before this point.

Java Solution

```public int maximalSquare(char[][] matrix) { if(matrix==null||matrix.length==0){ return 0; }   int result=0; int[][] dp = new int[matrix.length][matrix[0].length];   for(int i=0; i<matrix.length; i++){ dp[i][0]=matrix[i][0]-'0'; result=Math.max(result, dp[i][0]); }   for(int j=0; j<matrix[0].length; j++){ dp[0][j]=matrix[0][j]-'0'; result=Math.max(result, dp[0][j]); }   for(int i=1; i<matrix.length; i++){ for(int j=1; j<matrix[0].length; j++){ if(matrix[i][j]=='1'){ int min = Math.min(dp[i-1][j], dp[i][j-1]); min = Math.min(min, dp[i-1][j-1]); dp[i][j]=min+1;   result = Math.max(result, min+1); }else{ dp[i][j]=0; } } }   return result*result; }```
Category >> Algorithms >> Interview
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• alexwest11

other solutions ( worse) would be

ver1 force , for each sub matrix, check all 1
ver2 force +hashing, hash each row/column , exist 0 or not
then, for each submatrix(x,y, size m*n) check exist hashed 0 or not per row , column

• Israel Lopez

I Think it is a better solution.

private static void process (int[][] arrray){

int max=0, maxAux=0, maxRow=0, maxAuxRow=0;
HashSet complete = new HashSet();

for (int rows= 0; rows<arrray.length;rows++){

max=0; maxAux=0;
for (int cols = 0; colsmaxAux){
maxAux = max;
max = 0;
}else if (maxAux>max){
max = maxAux;
maxAux = 0;
}
}

maxRow =Math.max(max, maxAux);

}

if (maxRow>maxAuxRow){
maxAuxRow = maxRow;
maxRow = 0;
}else if (maxAuxRow>maxRow){
maxRow = maxAuxRow;
maxAuxRow = 0;
}

}

System.out.println(“max “+Math.max(maxRow, maxAuxRow));

}

• Alina Rozenbaum

Ok, this may be stupidly redundant, but with a lot of the matrices problems, why not just pass as int[][], why do char[][] and then spend your time converting back in the method?

• Guest

Why can’t we keep track of max and update it inside ‘if (matrix[i][j] == ‘1’)’ with

max = Math.min(min, max)