# 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 || matrix[0].length == 0) return 0;   int m = matrix.length; int n = matrix[0].length;   int[][] t = new int[m][n];   //top row for (int i = 0; i < m; i++) { t[i][0] = Character.getNumericValue(matrix[i][0]); }   //left column for (int j = 0; j < n; j++) { t[0][j] = Character.getNumericValue(matrix[0][j]); }   //cells inside for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { if (matrix[i][j] == '1') { int min = Math.min(t[i - 1][j], t[i - 1][j - 1]); min = Math.min(min,t[i][j - 1]); t[i][j] = min + 1; } else { t[i][j] = 0; } } }   int max = 0; //get maximal length for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (t[i][j] > max) { max = t[i][j]; } } }   return max * max; }```
Category >> Algorithms >> Interview
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• 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)