CF 177A1 - Good Matrix Elements

CF 177A1 - Good Matrix Elements

Rating: 800
Tags: implementation
Solve time: 1m 37s
Verified: no

Solution

Problem Understanding

We are given a square matrix of size n × n, where n is guaranteed to be an odd number. Each element of the matrix is a non-negative integer. The task is to compute the sum of "good" elements, which are defined as elements that belong to at least one of these four categories: the main diagonal, the secondary diagonal, the middle row, or the middle column. The middle row and column are uniquely defined because n is odd. The output is a single integer representing the sum of all these elements, counting each element only once even if it belongs to multiple categories.

Given that n can go up to 101, the total number of matrix elements is at most 10201. Any algorithm with complexity O(n²) or less will execute comfortably under the 2-second time limit. This rules out the need for advanced optimization techniques; the problem is primarily about careful indexing to avoid double-counting elements that belong to multiple "good" sets.

Non-obvious edge cases include the smallest matrix (1×1) where every element is trivially good, and larger matrices where the center element lies on both diagonals and in both the middle row and column. For example, in a 3×3 matrix:

1 2 3
4 5 6
7 8 9

All elements are good, including the center element 5, which belongs to all four categories. A naive implementation might accidentally sum it multiple times if it treats each category independently without tracking overlaps.

Approaches

The brute-force approach is straightforward: iterate over each element of the matrix, check if it belongs to any of the four categories, and sum it if it does. This is correct but requires careful handling to avoid double-counting, particularly at the center of the matrix.

The key observation for an optimal solution is that because n is odd, we can directly identify indices that belong to each category. The middle row and column both have index n // 2 using zero-based indexing. The main diagonal has indices where the row index equals the column index. The secondary diagonal has indices where the row index plus the column index equals n - 1. We can iterate over the entire matrix and add an element if it satisfies any of these four conditions. Because the set of conditions covers all possible good elements and the size of the matrix is small, this is efficient and avoids complex data structures.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(1)	Accepted
Optimal	O(n²)	O(1)	Accepted

Algorithm Walkthrough

1. Read the integer n from input. This defines the size of the square matrix.
2. Read the next n lines to construct the matrix. Each line contains n integers.
3. Compute the middle index mid = n // 2. This index identifies both the middle row and middle column.
4. Initialize a variable total = 0 to accumulate the sum of good elements.
5. Iterate over each element of the matrix using two nested loops: the outer loop for rows i and the inner loop for columns j.
6. For each element at position (i, j), check if it is good. The element is good if any of these conditions hold: i == j (main diagonal), i + j == n - 1 (secondary diagonal), i == mid (middle row), j == mid (middle column).
7. If the element is good, add its value to total.
8. After iterating through all elements, print total.

The invariant is that we only add each element once. Since we only check the conditions and add once per element, even if an element satisfies multiple conditions, it is summed exactly once.

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 first reads the matrix using list comprehension and maps each line into integers. It computes the middle index once, then iterates over all elements with nested loops. The condition for good elements directly implements the definition, ensuring no element is added twice. Using i == j and i + j == n - 1 correctly identifies both diagonals, and i == mid and j == mid capture the central row and column.

Worked Examples

Example 1:

Input:

3
1 2 3
4 5 6
7 8 9

i	j	matrix[i][j]	Good?	total
0	0	1	Yes	1
0	1	2	Yes	3
0	2	3	Yes	6
1	0	4	Yes	10
1	1	5	Yes	15
1	2	6	Yes	21
2	0	7	Yes	28
2	1	8	Yes	36
2	2	9	Yes	45

All elements are good because the matrix is small and the middle row and column overlap with the diagonals. The sum is 45.

Example 2:

Input:

5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

The middle index is 2 (0-based). The good elements include all diagonal elements, row 2, and column 2. The total sum is 273.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n²)	We iterate over every element in the n × n matrix once.
Space	O(n²)	We store the entire matrix.

With n ≤ 101, the total number of operations is at most 10201, which executes comfortably under 2 seconds. Memory usage is within the 256 MB 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"
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", "sample 2"

# Minimum-size input
assert run("1\n42\n") == "42", "1x1 matrix"

# All equal values
assert run("3\n5 5 5\n5 5 5\n5 5 5\n") == "45", "all elements equal"

# Maximum-size input (n=101)
import random
values = "\n".join(" ".join(str(random.randint(0,100)) for _ in range(101)) for _ in range(101))
# Not asserting exact value; just testing execution
run(f"101\n{values}\n")

# Boundary diagonals and middle overlap
assert run("3\n1 2 3\n4 5 6\n7 8 9\n") == "45", "overlapping center"

Test input	Expected output	What it validates
1\n42\n	42	Single element matrix
3\n5 5 5\n5 5 5\n5 5 5\n	45	All elements are equal, sum correctness
3\n1 2 3\n4 5 6\n7 8 9\n	45	Center element overlaps all categories
5x5 random	varies	Execution and correctness on larger inputs

Edge Cases

For a 1×1 matrix:

1
99

The middle row and column are the same as the only element, which is also both diagonals. Our algorithm identifies it as good, adding it once. The sum is 99.

For an odd matrix like 5×5, the center element lies at (2,2). This element satisfies i == j, i + j == n - 1, i == mid, and j == mid. The algorithm adds it exactly once because we only add an element