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2020-05-21

## 221. Maximal Square

### Question:

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

#### Example:

``````Input:

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

Output: 4
``````

### Solution:

Using DP to solve the problem. The DP function is `dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1])`. The DP means that the side length of the square with right bottom point ends at `matrix[i][j]`

The following solution using O(1) space, but require cast from integer to character and from character to integer.

``````class Solution {
public int maximalSquare(char[][] matrix) {
int max = 0;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] == '1') {
if (i == 0 || j == 0) {
matrix[i][j] = (char)(1 + '0');
} else {
matrix[i][j] = (char)(1 + Math.min(Math.min((int)matrix[i][j-1], (int)matrix[i-1][j]), (int)matrix[i-1][j-1]));
}
max = Math.max((int)(matrix[i][j] - '0'), max);
}
}
}
return max * max;
}
}
``````