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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[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/
// 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/
// 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;
}
};

• class Solution {
public int countSquares(int[][] matrix) {
int rows = matrix.length, columns = matrix[0].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;
}
}

############

class Solution {
public int countSquares(int[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;
int[][] f = new int[m][n];
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (matrix[i][j] == 0) {
continue;
}
if (i == 0 || j == 0) {
f[i][j] = 1;
} else {
f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i - 1][j], f[i][j - 1])) + 1;
}
ans += f[i][j];
}
}
return ans;
}
}

• // 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[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;
}
};

• class Solution:
def countSquares(self, matrix: List[List[int]]) -> int:
m, n = len(matrix), len(matrix[0])
f = [[0] * n for _ in range(m)]
ans = 0
for i, row in enumerate(matrix):
for j, v in enumerate(row):
if v == 0:
continue
if i == 0 or j == 0:
f[i][j] = 1
else:
f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1
ans += f[i][j]
return ans


• func countSquares(matrix [][]int) int {
m, n, ans := len(matrix), len(matrix[0]), 0
f := make([][]int, m)
for i := range f {
f[i] = make([]int, n)
}
for i, row := range matrix {
for j, v := range row {
if v == 0 {
continue
}
if i == 0 || j == 0 {
f[i][j] = 1
} else {
f[i][j] = min(f[i-1][j-1], min(f[i-1][j], f[i][j-1])) + 1
}
ans += f[i][j]
}
}
return ans
}

func min(a, b int) int {
if a < b {
return a
}
return b
}