Upper Triangular Matrix Conversion

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Given a 2D square matrix of integers where all elements are non-zero, convert it into an upper triangular matrix by setting all elements below the main diagonal to zero. The main diagonal consists of elements where the row index equals the column index (i.e., for element at position (i, j), if i > j, set it to zero).

Examples:

Input: [[1, 2], [3, 4]]

Output: [[1, 2], [0, 4]]

Input: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Output: [[1, 2, 3], [0, 5, 6], [0, 0, 9]]

Input: [[5]]

Output: [[5]]

Constraints

The matrix is square (n x n) with 1 ≤ n ≤ 100.
All elements in the input matrix are non-zero integers.
After conversion, elements below the main diagonal are set to zero; elements on or above the diagonal remain unchanged.

Editorial

Intuition:

To convert a square matrix to upper triangular, we zero out all elements where the row index is greater than the column index (i > j), as these lie below the main diagonal.

Approach:

Get the size n of the square matrix.
Iterate through each row i from 0 to n-1.
For each row, iterate through each column j from 0 to n-1.
If i > j, set matrix[i][j] = 0.

#include <vector>
using namespace std;

class Solution {
public:
    void convertToUpperTriangular(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i > j) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
};
class Solution {
    public void convertToUpperTriangular(int[][] matrix) {
        int n = matrix.length;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i > j) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
}
def convertToUpperTriangular(matrix):
    n = len(matrix)
    for i in range(n):
        for j in range(n):
            if i > j:
                matrix[i][j] = 0
function convertToUpperTriangular(matrix) {
    const n = matrix.length;
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            if (i > j) {
                matrix[i][j] = 0;
            }
        }
    }
}

Complexity Analysis:

Time Complexity: O(n²), where n is the matrix dimension. We traverse all n² elements once.

Space Complexity: O(1), as we modify the matrix in-place without extra space.

Notes

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