https://github.com/emahtab/count-square-submatrices-with-all-ones
Count square submatrices with all ones
https://github.com/emahtab/count-square-submatrices-with-all-ones
dynamic-programming leetcode problem-solving
Last synced: 3 months ago
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Count square submatrices with all ones
- Host: GitHub
- URL: https://github.com/emahtab/count-square-submatrices-with-all-ones
- Owner: eMahtab
- Created: 2020-06-08T14:55:13.000Z (about 5 years ago)
- Default Branch: master
- Last Pushed: 2020-06-08T15:14:17.000Z (about 5 years ago)
- Last Synced: 2025-02-02T03:26:11.758Z (5 months ago)
- Topics: dynamic-programming, leetcode, problem-solving
- Homepage:
- Size: 8.79 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# Count square submatrices with all ones
## https://leetcode.com/problems/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 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. 1 <= arr.length <= 300
2. 1 <= arr[0].length <= 300
3. 0 <= arr[i][j] <= 1
## Implementation 1 : Time : O(rows * columns) , Space : O(rows * columns)
```java
class Solution {
public int countSquares(int[][] matrix) {
if(matrix == null || matrix.length == 0)
return 0;
int rows = matrix.length;
int columns = matrix[0].length;
int total = 0;
int[][] dp = new int[rows+1][columns+1];
for(int i = 1; i <= rows; i++) {
for(int j = 1; j <= columns; j++) {
if(matrix[i-1][j-1] == 1) {
int min = Math.min(dp[i][j-1], dp[i-1][j]);
min = Math.min(min, dp[i-1][j-1]);
dp[i][j] = min + 1;
total += dp[i][j];
}
}
}
return total;
}
}
```
### Implementation 2 : Time : O(rows * columns), Space : O(1) , Mutating the input array
```java
class Solution {
public int countSquares(int[][] matrix) {
if(matrix == null || matrix.length == 0)
return 0;
int rows = matrix.length;
int columns = matrix[0].length;
int total = 0;
for(int i = 0; i < rows; i++) {
for(int j = 0; j < columns; j++) {
if(matrix[i][j] == 0) continue; // doesn't contribute to total
else if(i == 0 || j == 0) //means cell value is 1, its either on 1st row or 1st column
total++;
else {
int min = Math.min(matrix[i][j-1], matrix[i-1][j]);
min = Math.min(min, matrix[i-1][j-1]);
matrix[i][j] = min + 1;
total += matrix[i][j];
}
}
}
return total;
}
}
```
**Note : Mutating the method inputs is not considered a good practice, but if its OK to modify the method input, then the second implementation will save the extra space.**
# References :
https://www.youtube.com/watch?v=7xMVc2lPXhI