Sum of Both Diagonals of a Square Matrix

Easy10–15 min

Topics

ArrayMatrix

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Given a square matrix mat of size n × n, return the sum of the elements on both diagonals — the primary diagonal (top-left to bottom-right) and the secondary diagonal (top-right to bottom-left). If the matrix has an odd number of rows and columns, the very center cell sits on both diagonals at once — count it only once. Checking every cell is correct but wasteful: the diagonal positions are known in advance by a simple formula, so a single pass that visits only those 2n (or 2n-1) cells finds the answer in O(n) instead of O(n²).

Test Case 1:

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

Output:25

Explanation:Primary diagonal 1+5+9=15, secondary diagonal 3+5+7=15 — the shared center (5) is only counted once, so 15+15-5=25.

Test Case 2:

Input:mat = [[5]]

Output:5

Explanation:A single cell sits on both diagonals at once — it's still just counted once.

Test Case 3:

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

Output:10

Explanation:An even-sized matrix has no shared center: primary 1+4=5, secondary 2+3=5, total 10.

Constraints

◆1 ≤ n ≤ 100
◆1 ≤ mat[i][j] ≤ 100
◆mat is a square matrix (n × n)

🚀

Try the Dry Run

Don't just read the solution — watch it execute, one step at a time.

🧪Try your own test case

mat (rows separated by ;) — 2-6 rows, square (n×n)

1	class Solution {
2	public int diagonalSum(int[][] mat) {
3	int n = mat.length;
4	int sum = 0;
5	for (int i = 0; i < n; i++) {
6	sum += mat[i][i];
7	if (i != n - 1 - i) sum += mat[i][n - 1 - i];
8	}
9	return sum;
10	}
11	}
12

Variables

n3

sum0

INITIALIZE

Start sum at 0. Walk i from 0 to 2, adding mat[i][i] and mat[i][n-1-i] at each step — no need to scan the whole matrix.

Step 1 / 8

Approach & Solutions

Brute Force

Brute

Check every cell of the matrix, and add it to the running sum if it sits on either diagonal — the primary diagonal (i === j) or the secondary diagonal (i + j === n - 1). Correct, but visiting every cell revisits n² - 2n+1 cells that were never going to be added.

TimeO(n²)

SpaceO(1)

Java

Try it out

class Solution {
    public int diagonalSum(int[][] mat) {
        int n = mat.length;
        int sum = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == j || i + j == n - 1) sum += mat[i][j];
            }
        }
        return sum;
    }
}

Optimal — Single Pass

Optimal

There's no need to check every cell — the diagonal positions are already known by formula. Walk i from 0 to n-1 once, adding mat[i][i] (primary) and mat[i][n-1-i] (secondary) at each step, skipping the secondary add when i equals n-1-i to avoid double-counting the shared center of an odd-sized matrix.

TimeO(n)