Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1277.html

1277. Count Square Submatrices with All Ones (Medium)

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

Related Topics:
Array, Dynamic Programming

Solution 1. DP

Let dp[i][j] be the side length of the max square with all ones.

dp[i + 1][j + 1] = min( dp[i][j], dp[i + 1][j], dp[i][j + 1] ) + 1
dp[0][i] = dp[i][0] = 0

The answer is the sum of all dp[i + 1][j + 1].

// OJ: https://leetcode.com/problems/count-square-submatrices-with-all-ones/
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int countSquares(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        vector<vector<int>> dp(M + 1, vector<int>(N + 1));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 0) continue;
                dp[i + 1][j + 1] = min({ dp[i][j], dp[i + 1][j], dp[i][j + 1] }) + 1;
                ans += dp[i + 1][j + 1];
            }
        }
        return ans;
    }
};

Solution 2. DP

// OJ: https://leetcode.com/problems/count-square-submatrices-with-all-ones/
// Time: O(MN)
// Space: O(N)
class Solution {
public:
    int countSquares(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        vector<int> dp(N + 1);
        for (int i = 0; i < M; ++i) {
            int prev = 0;
            for (int j = 0; j < N; ++j) {
                int cur = dp[j + 1];
                if (A[i][j] == 0) dp[j + 1] = 0;
                else dp[j + 1] = min({ prev, dp[j], dp[j + 1] }) + 1;
                ans += dp[j + 1];
                prev = cur;
            }
        }
        return ans;
    }
};

Java

  • class Solution {
    public int countSquares(int[][] matrix) {
        int rows = matrix.length, columns = matrix[0].length;
        boolean[][] flags = new boolean[rows][columns];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++)
                flags[i][j] = true;
        }
        int count = 0;
        int maxSide = Math.min(rows, columns);
        for (int side = 1; side <= maxSide; side++) {
            int rowEnd = rows - side, columnEnd = columns - side;
            for (int i = 0; i <= rowEnd; i++) {
                int fromRow = i, toRow = i + side - 1;
                for (int j = 0; j <= columnEnd; j++) {
                    if (!flags[i][j])
                        continue;
                    int fromColumn = j, toColumn = j + side - 1;
                    boolean isAllOne = true;
                    outer:
                    for (int row = fromRow; row <= toRow; row++) {
                        for (int column = fromColumn; column <= toColumn; column++) {
                            if (matrix[row][column] != 1) {
                                flags[i][j] = false;
                                isAllOne = false;
                                break outer;
                            }
                        }
                    }
                    if (isAllOne)
                        count++;
                }
            }
        }
        return count;
    }
}

  • // OJ: https://leetcode.com/problems/count-square-submatrices-with-all-ones/
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int countSquares(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        vector<vector<int>> dp(M + 1, vector<int>(N + 1));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 0) continue;
                dp[i + 1][j + 1] = min({ dp[i][j], dp[i + 1][j], dp[i][j + 1] }) + 1;
                ans += dp[i + 1][j + 1];
            }
        }
        return ans;
    }
};

  • 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

All Problems

All Solutions