# 1277. Count Square Submatrices with All Ones

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 one 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`

Example:

[0,1,1,1],

[1,1,1,1],

[0,1,1,1]

- count single ones
- count 2×2
- count 3×3
- ……

Goal: reduce n x n to (n-1)x(n-1) … to 2×2

Now scan 2×2 block, for each position(i,j) scan (i+1,j) (i,j+1) (i+1, j+1)

[*0,1*,1,1],

[*1,1*,1,1],

[0,1,1,1]

if the 2×2 scanning block contains 0, update (i,j) to 0

[0,1,1,1],

[**0**,*1*,1,1],

[*0*,*1*,1,1]

then we can reduce it by removing (ignoring) the last row and column.

[0,1,1],

[0,1,1]

Then walk through with the 2×2 scanning block again.

the counter could be updated when the whole 2×2 block is 1, OR count number of ones in the next round.

Version A, update counter during 2×2 scan

class Solution { public int countSquares(int[][] matrix) { int c = 0; int m = matrix.length; int n = matrix[0].length; for (int i = 0; i < m; i ++){ for (int j = 0; j < n; j ++){ if (matrix[i][j] == 1){ c++; } } } while (m > 1 && n > 1){ for (int i = 0; i < m-1; i ++){ for (int j = 0; j < n-1; j ++){ if ( matrix[i][j] == 0 || matrix[i+1][j] == 0 || matrix[i][j+1] == 0 || matrix[i+1][j+1] == 0) { matrix[i][j] = 0; } else { c++; } } } // lower m and n (ignoring last row and column) m --; n --; } return c; } }

B, a more compact version, update counter next round, count number of ones.

slightly slower for more loops

class Solution { public int countSquares(int[][] matrix) { int c = 0; int m = matrix.length; int n = matrix[0].length; while (m > 0 && n > 0){ for (int i = 0; i < m; i ++){ for (int j = 0; j < n; j ++){ if (matrix[i][j] == 1){ c++; } } } for (int i = 0; i < m-1; i ++){ for (int j = 0; j < n-1; j ++){ if (matrix[i][j] == 0 || matrix[i+1][j] == 0 || matrix[i][j+1] == 0 || matrix[i+1][j+1] == 0){ matrix[i][j] = 0; } } } // lower m and n (ignoring last row and column) m --; n --; } return c; } }

A: 86-110 ms

B: 160-170 ms

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