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package g0201_0300.s0221_maximal_square;

// #Medium #Top_100_Liked_Questions #Array #Dynamic_Programming #Matrix
// #Dynamic_Programming_I_Day_16 #Big_O_Time_O(m*n)_Space_O(m*n)
// #2022_07_04_Time_7_ms_(72.35%)_Space_59.5_MB_(10.55%)

/**
 * 221 - Maximal Square\.
 *
 * Medium
 *
 * Given an `m x n` binary `matrix` filled with `0`'s and `1`'s, _find the largest square containing only_ `1`'s _and return its area_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/26/max1grid.jpg)
 *
 * **Input:** matrix = \[\["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
 *
 * **Output:** 4 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/26/max2grid.jpg)
 *
 * **Input:** matrix = \[\["0","1"],["1","0"]]
 *
 * **Output:** 1 
 *
 * **Example 3:**
 *
 * **Input:** matrix = \[\["0"]]
 *
 * **Output:** 0 
 *
 * **Constraints:**
 *
 * *   `m == matrix.length`
 * *   `n == matrix[i].length`
 * *   `1 <= m, n <= 300`
 * *   `matrix[i][j]` is `'0'` or `'1'`.
**/
public class Solution {
    public int maximalSquare(char[][] matrix) {
        int m = matrix.length;
        if (m == 0) {
            return 0;
        }
        int n = matrix[0].length;
        if (n == 0) {
            return 0;
        }
        int[][] dp = new int[m + 1][n + 1];
        int max = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '1') {
                    // 1 + minimum from cell above, cell to the left, cell diagonal upper-left
                    int next = 1 + Math.min(dp[i][j], Math.min(dp[i + 1][j], dp[i][j + 1]));
                    // keep track of the maximum value seen
                    if (next > max) {
                        max = next;
                    }
                    dp[i + 1][j + 1] = next;
                }
            }
        }
        return max * max;
    }
}




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