1277. Count Square Submatrices with All Ones

Problem Statement

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

Intuition

Not Intuituive

Basically we Keep an element at bottom left and check for top, left, daigonal
Take min of all three +1

Take addition of all dp matrix

dp[i][j] = how many squares end at i,j

Links

https://leetcode.com/problems/count-square-submatrices-with-all-ones/description/

Video Links

https://www.youtube.com/watch?v=auS1fynpnjo&ab_channel=takeUforward

Approach 1:

Tabulation

C++

class Solution {
public:
    int countSquares(vector<vector<int>>& arr) {
        int m = arr.size();
        int n = arr[0].size();
        vector<vector<int>> dp(m, vector<int> (n,0));

        for(int i=0; i<m; i++){
            if(arr[i][0] == 1)
                dp[i][0] = 1;
        }

        for(int i=0; i<n; i++){
            if(arr[0][i] == 1)
                dp[0][i] = 1;
        }

        for(int i=1; i<m; i++){
            for(int j=1; j<n; j++){
                if(arr[i][j] == 1)
                    dp[i][j] = min( {dp[i-1][j], dp[i][j-1], dp[i-1][j-1]} ) + 1;
            }
        }

        int ans = 0;

        for(auto &it: dp)
            ans += accumulate(it.begin(), it.end(), 0);
        
        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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