Python Brute Force Solution

For the left-to-right diagonal: Loop through all elements, sum elements where row index == column index.

For the right-to-left diagonal: Loop through all elements, sum elements where row index + column index == n-1.

def diagonalDifference(arr):
    n = len(arr)
    left_diag = 0
    right_diag = 0
    for i in range(n):
        for j in range(n):
            if i == j:
                left_diag += arr[i][j]
            if i + j == n-1:
                right_diag += arr[i][j]
    return abs(left_diag - right_diag)

Time Complexity: O(n²), two nested loops. Space Complexity: O(1)

Optimal Solution You don't need to traverse every element, each diagonal has only n elements (one per row/column).

Access the diagonals directly using index formulas in a single loop.

def diagonalDifference(arr):
    n = len(arr)
    left_diag = 0
    right_diag = 0
    for i in range(n):
        left_diag += arr[i][i]
        right_diag += arr[i][n-i-1]
    return abs(left_diag - right_diag)

Time Complexity: O(n), just one pass Space Complexity: O(1) >

.

public static int diagonalDifference(List> arr) { // Write your code here int sum=0; for(int i=0; i

public static int diagonalDifference(List> arr) { int n = arr.size(); /// tam de la matriz int diagnonalPrincipal = 0; int diagonalSecundaria = 0;

    for(int i = 0; i<n;i++){
        diagnonalPrincipal += arr.get(i).get(i);
        diagonalSecundaria += arr.get(i).get(n-1-i);
    }

    return Math.abs(diagnonalPrincipal - diagonalSecundaria);


}

This is my code using C: int diagonalDifference(int arr_rows, int arr_columns, int** arr) { int x = 0; int y = 0; for (int i = 0; i < arr_rows; i ++) { x += arr[i][i]; y += arr[arr_rows - i - 1][i]; }
return abs(x-y); }

Can some one explain why it is not possible to let int N = arr_rows?

O(n) typescript solution

function diagonalDifference(arr: number[][]): number {
    // Write your code here
    let primarySum:number = 0;
    let secoundarySum:number = 0;
    
    for(let i=0; i<arr.length; i++) {
        let lastIndex:number = arr[i].length - 1;
        primarySum += arr[i][i];
        secoundarySum += arr[i][lastIndex - i];
    }
    
    return Math.abs(primarySum - secoundarySum);
}