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.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[i] = dp[i] = 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.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.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.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;
}
}