LeetCode 3128: Right Triangles Solution

Master LeetCode problem 3128 (Right Triangles), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3128. Right Triangles

Medium

Array Hash Table Math Combinatorics Counting

Problem Explanation

Explanation

To solve this problem, we can iterate through each cell in the grid and treat it as the top vertex of a potential right triangle. We then check for other vertices in the same row and column to form a right triangle with the current vertex. If such vertices exist, we increment the count of right triangles.

Algorithm

Initialize a variable count to 0 to store the count of right triangles.
Iterate through each cell in the grid.
For each cell, check all other cells in the same row and column to find potential vertices for a right triangle with the current cell.
If a right triangle can be formed with the current cell as the top vertex, increment the count.
Return the final count of right triangles.

Time Complexity

The time complexity of this algorithm is O(n^2), where n is the size of the grid.

Space Complexity

The space complexity of this algorithm is O(1) as we are not using any additional data structures.

Solution Code

class Solution {
    public int countRightTriangles(int[][] grid) {
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    for (int k = 0; k < grid.length; k++) {
                        if (grid[k][j] == 1 && i != k) {
                            for (int l = 0; l < grid[0].length; l++) {
                                if (grid[i][l] == 1 && j != l) {
                                    count++;
                                }
                            }
                        }
                    }
                }
            }
        }
        return count;
    }
}

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