1277

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/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// 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;
    }
};