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package g1501_1600.s1504_count_submatrices_with_all_ones;

// #Medium #Array #Dynamic_Programming #Matrix #Stack #Monotonic_Stack
// #2022_04_07_Time_9_ms_(85.86%)_Space_43.1_MB_(87.32%)

/**
 * 1504 - Count Submatrices With All Ones\.
 *
 * Medium
 *
 * Given an `m x n` binary matrix `mat`, _return the number of **submatrices** that have all ones_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/10/27/ones1-grid.jpg)
 *
 * **Input:** mat = \[\[1,0,1],[1,1,0],[1,1,0]]
 *
 * **Output:** 13
 *
 * **Explanation:** 
 *
 * There are 6 rectangles of side 1x1.
 *
 * There are 2 rectangles of side 1x2. 
 *
 * There are 3 rectangles of side 2x1.
 *
 * There is 1 rectangle of side 2x2. 
 *
 * There is 1 rectangle of side 3x1. 
 *
 * Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/10/27/ones2-grid.jpg)
 *
 * **Input:** mat = \[\[0,1,1,0],[0,1,1,1],[1,1,1,0]]
 *
 * **Output:** 24
 *
 * **Explanation:**
 *
 * There are 8 rectangles of side 1x1.
 *
 * There are 5 rectangles of side 1x2.
 *
 * There are 2 rectangles of side 1x3.
 *
 * There are 4 rectangles of side 2x1.
 *
 * There are 2 rectangles of side 2x2.
 *
 * There are 2 rectangles of side 3x1.
 *
 * There is 1 rectangle of side 3x2. 
 *
 * Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.
 *
 * **Constraints:**
 *
 * *   `1 <= m, n <= 150`
 * *   `mat[i][j]` is either `0` or `1`.
**/
public class Solution {
    public int numSubmat(int[][] mat) {
        int[][] dp = new int[mat.length][mat[0].length];
        for (int i = 0; i < mat.length; i++) {
            int c = 0;
            for (int j = mat[0].length - 1; j >= 0; j--) {
                if (mat[i][j] == 1) {
                    c++;
                } else {
                    c = 0;
                }
                dp[i][j] = c;
            }
        }
        int ans = 0;
        for (int i = 0; i < mat.length; i++) {
            for (int j = 0; j < mat[0].length; j++) {
                int x = Integer.MAX_VALUE;
                for (int k = i; k < mat.length; k++) {
                    x = Math.min(x, dp[k][j]);
                    ans += x;
                }
            }
        }
        return ans;
    }
}