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Weekly Contest 165 Q3
Array
Dynamic Programming
Matrix

1277. Count Square Submatrices with All Ones

中文文档

Description

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

Solutions

Solution 1

Python3

class Solution:
    def countSquares(self, matrix: List[List[int]]) -> int:
        m, n = len(matrix), len(matrix[0])
        f = [[0] * n for _ in range(m)]
        ans = 0
        for i, row in enumerate(matrix):
            for j, v in enumerate(row):
                if v == 0:
                    continue
                if i == 0 or j == 0:
                    f[i][j] = 1
                else:
                    f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1
                ans += f[i][j]
        return ans

Java

class Solution {
    public int countSquares(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int[][] f = new int[m][n];
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) {
                    continue;
                }
                if (i == 0 || j == 0) {
                    f[i][j] = 1;
                } else {
                    f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i - 1][j], f[i][j - 1])) + 1;
                }
                ans += f[i][j];
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countSquares(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int ans = 0;
        vector<vector<int>> f(m, vector<int>(n));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) continue;
                if (i == 0 || j == 0)
                    f[i][j] = 1;
                else
                    f[i][j] = min(f[i - 1][j - 1], min(f[i - 1][j], f[i][j - 1])) + 1;
                ans += f[i][j];
            }
        }
        return ans;
    }
};

Go

func countSquares(matrix [][]int) int {
	m, n, ans := len(matrix), len(matrix[0]), 0
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, n)
	}
	for i, row := range matrix {
		for j, v := range row {
			if v == 0 {
				continue
			}
			if i == 0 || j == 0 {
				f[i][j] = 1
			} else {
				f[i][j] = min(f[i-1][j-1], min(f[i-1][j], f[i][j-1])) + 1
			}
			ans += f[i][j]
		}
	}
	return ans
}

TypeScript

function countSquares(matrix: number[][]): number {
    const [m, n] = [matrix.length, matrix[0].length];
    const f = Array.from({ length: m }, () => Array(n));
    const dfs = (i: number, j: number): number => {
        if (i === m || j === n || !matrix[i][j]) return 0;
        f[i][j] ??= 1 + Math.min(dfs(i + 1, j), dfs(i, j + 1), dfs(i + 1, j + 1));
        return f[i][j];
    };
    let ans = 0;

    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            ans += dfs(i, j);
        }
    }

    return ans;
}

JavaScript

function countSquares(matrix) {
    const [m, n] = [matrix.length, matrix[0].length];
    const f = Array.from({ length: m }, () => Array(n));
    const dfs = (i, j) => {
        if (i === m || j === n || !matrix[i][j]) return 0;
        f[i][j] ??= 1 + Math.min(dfs(i + 1, j), dfs(i, j + 1), dfs(i + 1, j + 1));
        return f[i][j];
    };
    let ans = 0;

    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            ans += dfs(i, j);
        }
    }

    return ans;
}