1170EasyArrayMatrix

Adding Up Both Diagonals of a Panel

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A square panel panel has as many rows as columns. Add up the tiles along the diagonal running from the top-left corner to the bottom-right one, and the tiles along the diagonal running from the top-right corner to the bottom-left one. A tile lying on both diagonals is counted only once.

Return that total.

Examples

Example 1

Input: panel = [[1, 2], [3, 4]]
Output: 10

On a two by two panel the two diagonals between them cover all four tiles.

Example 2

Input: panel = [[3, 1, 1], [1, 3, 1], [1, 1, 3]]
Output: 11

The first diagonal holds the three 3 tiles for 9. The other adds the two corner 1 tiles, while the middle tile lies on both and is counted once.

Example 3

Input: panel = [[100, 1], [1, 100]]
Output: 202

The two 100 tiles sit on the first diagonal and the two 1 tiles on the other.

Constraints

1 <= panel.length <= 100
panel[i].length == panel.length
1 <= panel[i][j] <= 100

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def diagonal_sum(panel: list[list[int]]) -> int:

Java

public int diagonalSum(int[][] panel)

The intended approach

Time O(n) over the rows.Space O(1).

Each row holds exactly one tile of each diagonal: in row `i` the first diagonal sits at column `i` and the second at column `n - 1 - i`. So one pass down the rows reaches every tile that counts, with no need to walk the whole panel. The two columns coincide only in the middle row of a panel with an odd side, where `i` equals `n - 1 - i`. Checking for that before adding the second tile is the whole of the special handling, and it is exactly what keeps the shared tile from being counted twice.

Pythonentry point: diagonal_sum

def diagonal_sum(panel):
    n = len(panel)
    total = 0
    for i in range(n):
        total += panel[i][i]
        if i != n - 1 - i:
            total += panel[i][n - 1 - i]
    return total

The obvious version

Gather the positions of both diagonals into a set of row and column pairs, which quietly drops any position named twice, then add up the tiles at those positions. Costs an extra pass and a set rather than one walk down the rows.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_diagonal_sum(panel):
    n = len(panel)
    picked = set()
    for i in range(n):
        picked.add((i, i))
        picked.add((i, n - 1 - i))
    total = 0
    for spot in picked:
        total += panel[spot[0]][spot[1]]
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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