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

Solution: DP

dp[i][j] := edge of largest square with bottom right corner at (i, j)
dp[i][j] = min(dp[i – 1][j], dp[i – 1][j – 1], dp[i][j – 1]) if m[i][j] == 1 else 0
ans = sum(dp)

Time complexity: O(n*m)
Space complexity: O(n*m)

// Author: Huahua
class Solution {
public:
  int countSquares(vector<vector<int>>& matrix) {
    const int n = matrix.size();
    const int m = matrix[0].size();
    // dp[i][j] := edge of largest square with right bottom corner at (i, j)
    vector<vector<int>> dp(n, vector<int>(m));
    int ans = 0;
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) {
        dp[i][j] = matrix[i][j];
        if (i && j && dp[i][j])
          dp[i][j] = min({dp[i - 1][j - 1], dp[i][j - 1], dp[i - 1][j]}) + 1;
        ans += dp[i][j];
      }
    return ans;
  }
};

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