LeetCode Link: https://leetcode.com/problems/count-square-submatrices-with-all-ones/
Language: C#
Problem Statement
Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.
Examples
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
public class Solution
{
public int CountSquares(int[][] matrix) => GetNoOfSquares(matrix);
private int GetNoOfSquares(int[][] matrix)
{
int noOfSquares = 0;
var x = matrix.Length;
var y = matrix[0].Length;
for (int i=0; i<x; i++)
{
for (int j=0; j<y; j++)
{
if (matrix[i][j] == 1)
{
if (i!=0 && j!=0 && matrix[i-1][j] > 0 && matrix[i][j-1] > 0 && matrix[i-1][j-1] > 0)
{
matrix[i][j] = Min(matrix[i-1][j-1], matrix[i-1][j], matrix[i][j-1]) + 1;
}
noOfSquares += matrix[i][j];
}
}
}
return noOfSquares;
}
private int Min(params int[] values)
{
int min = int.MaxValue;
foreach (var item in values)
{
if (item < min)
{
min = item;
}
}
return min;
}
}
Complexities
Time Complexity: O(m*n)
Space Complexity: O(1)