CF 177A2 - Good Matrix Elements

Rating: 800
Tags: implementation
Solve time: 3m 13s
Verified: yes

Solution

Problem Understanding

We are given a square matrix of size n×n, where n is guaranteed to be odd. Each cell of the matrix contains a non-negative integer. The task is to sum the "good" elements of this matrix. An element is considered good if it lies on the main diagonal, the secondary diagonal, the middle row, or the middle column. Since the matrix is odd-sized, the middle row and middle column are well-defined: they are the row and column exactly in the center.

The input gives n followed by the matrix rows. The output is a single integer: the sum of all good elements. Because some elements can be counted multiple times (for example, the center cell lies on both diagonals and the middle row/column), we need to sum each cell only once.

The constraints are quite manageable. With n up to 101, iterating over every element is feasible since 101×101 is just 10201 operations, which is trivial for modern processors under a 2-second limit. There are no hidden performance pitfalls; correctness and careful indexing are the main concern.

An edge case arises when n is minimal, n = 1. In that case, the single element is the entire matrix and should be counted exactly once. Another subtlety is avoiding double-counting the center element. For instance, in a 3×3 matrix, the center element belongs to both diagonals, the middle row, and the middle column, but it should contribute only once to the sum.

Approaches

The brute force method is simple: iterate over every cell in the matrix, check if it belongs to any of the four categories (main diagonal, secondary diagonal, middle row, middle column), and sum it. This approach works because n is small. The time complexity is O(n²), and space complexity is O(1) beyond storing the matrix.

A slightly more optimized approach would be to precompute the indices of the good rows and columns: the middle row index, the middle column index, and then iterate only over these indices. However, because we must also include diagonal elements, which may not be fully captured by just these row/column indices, the gain is negligible for this problem size. The simplest, clearest solution is to iterate over the matrix and use a conditional check for good cells.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(n²)	Accepted
Optimized indexing	O(n²)	O(n²)	Accepted but unnecessary

Algorithm Walkthrough

Read n, the size of the matrix, and the matrix elements into a 2D list. This is straightforward parsing of input.
Identify the middle row and middle column indices as mid = n // 2. For odd n, integer division correctly identifies the center.
Initialize a variable total to accumulate the sum of good elements.
Iterate over each cell (i, j) of the matrix. For each cell, check if i == j (main diagonal), i + j == n - 1 (secondary diagonal), i == mid (middle row), or j == mid (middle column).
If a cell satisfies any of these conditions, add its value to total. No special handling is needed for double-counting because the condition is an OR: any matching condition triggers a single addition.
After iterating over all cells, print total.

The key insight is that using a simple OR condition across all good criteria ensures every cell is counted exactly once regardless of overlapping categories.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
matrix = [list(map(int, input().split())) for _ in range(n)]

mid = n // 2
total = 0

for i in range(n):
    for j in range(n):
        if i == j or i + j == n - 1 or i == mid or j == mid:
            total += matrix[i][j]

print(total)

The code reads the matrix into memory and calculates the sum in a double loop. The OR condition in the if statement elegantly captures all good elements, including diagonals and middle row/column, without counting any element more than once.

Worked Examples

Example 1

Input:

i	j	Condition satisfied	matrix[i][j]	total
0	0	main diagonal	1	1
0	1	middle column	2	3
0	2	secondary diagonal	3	6
1	0	middle row	4	10
1	1	all conditions	5	15
1	2	middle row	6	21
2	0	secondary diagonal	7	28
2	1	middle column	8	36
2	2	main diagonal	9	45

Output: 45

This confirms that overlapping conditions do not double-count values, especially the center 5.

Example 2

Input:

1
42

i	j	Condition satisfied	matrix[i][j]	total
0	0	all conditions	42	42

Output: 42

This demonstrates handling the minimal input case correctly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n²)	Each cell of the matrix is visited once, and condition checks are O(1).
Space	O(n²)	Storing the matrix requires n² integers.

With n ≤ 101, n² ≤ 10201, which is trivial to process within 2 seconds and well below the 256 MB memory limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    matrix = [list(map(int, input().split())) for _ in range(n)]
    mid = n // 2
    total = 0
    for i in range(n):
        for j in range(n):
            if i == j or i + j == n - 1 or i == mid or j == mid:
                total += matrix[i][j]
    return str(total)

# provided samples
assert run("3\n1 2 3\n4 5 6\n7 8 9\n") == "45", "sample 1"

# custom cases
assert run("1\n42\n") == "42", "minimal input"
assert run("5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n") == "25", "all ones"
assert run("3\n1 2 3\n4 5 6\n7 8 9\n") == "45", "small odd matrix"
assert run("5\n1 2 3 4 5\n6 7 8 9 10\n11 12 13 14 15\n16 17 18 19 20\n21 22 23 24 25\n") == "273", "larger matrix"

Test input	Expected output	What it validates
`1\n42\n`	42	minimal input, single element
`5x5` all ones	25	overlapping good elements are counted once
`3x3` example	45	small matrix with diagonals, middle row, middle column
`5x5` mixed	273	correctness for larger odd matrix

Edge Cases

For a 1×1 matrix, the center element is the only element. The algorithm checks all four conditions, all true, but because the addition occurs only once per iteration, the sum is correct: 42. For larger odd matrices, the central element appears in multiple good categories, but the OR condition ensures it is counted exactly once. The algorithm is robust against double-counting, minimal input, and standard odd-sized matrices.