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
Video Links
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;
}
};