Leetcode/src/main/java/com/fishercoder/solutions/_1277.java at master · dodola/Leetcode · GitHub

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
package com.fishercoder.solutions;

/**
 * 1277. Count Square Submatrices with All Ones
 *
 * Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.
 *
 * Example 1:
 * Input: matrix =
 * [
 *   [0,1,1,1],
 *   [1,1,1,1],
 *   [0,1,1,1]
 * ]
 * Output: 15
 * Explanation:
 * There are 10 squares of side 1.
 * There are 4 squares of side 2.
 * There is  1 square of side 3.
 * Total number of squares = 10 + 4 + 1 = 15.
 *
 * Example 2:
 * Input: matrix =
 * [
 *   [1,0,1],
 *   [1,1,0],
 *   [1,1,0]
 * ]
 * Output: 7
 * Explanation:
 * There are 6 squares of side 1.
 * There is 1 square of side 2.
 * Total number of squares = 6 + 1 = 7.
 *
 * Constraints:
 * 1 <= arr.length <= 300
 * 1 <= arr[0].length <= 300
 * 0 <= arr[i][j] <= 1
 * */
public class _1277 {
    public static class Solution1 {
        /**
         * In-place solution.
         *
         * credit: https://leetcode.com/problems/count-square-submatrices-with-all-ones/discuss/441306/Python-DP-solution
         * */
        public int countSquares(int[][] matrix) {
            int m = matrix.length;
            int n = matrix[0].length;
            int count = 0;
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (matrix[i][j] > 0 && i > 0 && j > 0) {
                        matrix[i][j] = Math.min(matrix[i - 1][j - 1], Math.min(matrix[i - 1][j], matrix[i][j - 1])) + 1;
                    }
                    count += matrix[i][j];
                }
            }
            return count;
        }
    }

    public static class Solution2 {
        /**
         * Use m*n extra space solution.
         *
         * credit: https://leetcode.com/problems/count-square-submatrices-with-all-ones/discuss/441312/Java-Simple-DP-solution
         */
        public int countSquares(int[][] matrix) {
            int m = matrix.length;
            int n = matrix[0].length;
            int[][] dp = new int[m][n];
            for (int i = 0; i < m; i++) {
                if (matrix[i][0] == 1) {
                    dp[i][0] = 1;
                }
            }
            for (int j = 0; j < n; j++) {
                if (matrix[0][j] == 1) {
                    dp[0][j] = 1;
                }
            }
            for (int i = 1; i < m; i++) {
                for (int j = 1; j < n; j++) {
                    if (matrix[i][j] == 1) {
                        dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
                    }
                }
            }
            int count = 0;
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    count += dp[i][j];
                }
            }
            return count;
        }
    }
}