How do you approach Matrix Diagonal Sum?

For row i, the primary diagonal element is mat[i][i]. Which column index holds the secondary diagonal element?

What is the time complexity of Matrix Diagonal Sum?

Time complexity: O(n) - one element pair per row. Space complexity: O(1).

What are common mistakes on Matrix Diagonal Sum?

Transposing row and column in the index. It reads fine and fails on any non-square input. Overwriting cells a later iteration still needs to read - most in-place matrix problems want a marker pass first. Bounds checks that cover only some of the four edges.

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Matrix Diagonal Sum

EasyAlgorithms

Description

Given a square matrix mat, return the sum of the matrix diagonals.

Include the sum of all the elements on the primary diagonal (top-left to bottom-right) and all the elements on the secondary diagonal (top-right to bottom-left) that are not part of the primary diagonal.

Example 1

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

Output: 25

Explanation: Primary diagonal: 1 + 5 + 9 = 15. Secondary diagonal: 3 + 7 = 10 (the center 5 is counted only once). Total 25.

Example 2

Input: mat = [[1,1,1,1],[1,1,1,1],[1,1,1,1],[1,1,1,1]]

Output: 8

Example 3

Input: mat = [[5]]

Output: 5

Constraints

- n == mat.length == mat[i].length - 1 <= n <= 100 - 1 <= mat[i][j] <= 100

Matrix Diagonal Sum: approach, complexity and pitfalls

An easy matrix problem, graded against 6 test cases (3 of them hidden).

The pattern: Matrix

2D grid traversal, rotation, and in-place manipulation.

Reach for it when you see: A grid input, spiral or diagonal traversal, or an in-place transformation constraint.

If you are stuck

Show the 3-step hint ladder

For row i, the primary diagonal element is mat[i][i]. Which column index holds the secondary diagonal element?
The secondary diagonal in row i is mat[i][n - 1 - i]. One pass over the rows covers both diagonals.
When n is odd, i == n - 1 - i at the center row - make sure you don't add that element twice.

The optimal approach

Show the approach: Single Pass over Rows

In row `i` the primary diagonal contributes `mat[i][i]` and the secondary contributes `mat[i][n - 1 - i]`. Sum both per row, then handle the only possible overlap: when `n` is odd the two indices coincide at the center, so skip (or subtract) that one duplicate.

Time: O(n) - one element pair per row
Space: O(1)

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

What the tests check

Read off this problem's own test suite, so these are the cases a submission actually has to survive.

Single element. A one-element case is included. Two-pointer and window solutions frequently return the wrong thing when the two indices start out equal.
Duplicate values. Repeated values appear in the input, so any logic that assumes distinct elements - or that skips over equal neighbours - is exercised.

Common Matrix mistakes

These apply to the pattern as a whole, not just this problem.

Transposing row and column in the index. It reads fine and fails on any non-square input.
Overwriting cells a later iteration still needs to read - most in-place matrix problems want a marker pass first.
Bounds checks that cover only some of the four edges.