CF 2010A - Alternating Sum of Numbers

CF 2010A - Alternating Sum of Numbers

Rating: 800
Tags: brute force, implementation, math
Solve time: 1m 27s
Verified: yes

Solution

Problem Understanding

The task is to compute an alternating sum over a sequence of integers. Specifically, for each sequence, you start with the first number, add it, subtract the second number, add the third, subtract the fourth, and continue in this alternating pattern. The input provides multiple sequences, each with its length and the sequence itself, and the output must produce one integer per sequence representing this alternating sum.

The constraints are small: each sequence has at most 50 elements, and there can be up to 1000 sequences. Each integer is at most 100. This means that a simple linear pass over each sequence will work without performance concerns, because in the worst case we would process 50,000 numbers, which is trivial for a modern CPU. Integer overflow is not a concern in Python because integers are arbitrary precision.

Edge cases to consider include sequences with only one element, sequences with all identical numbers, and sequences where all numbers are even or odd. For example, if the sequence is just [100], the alternating sum is 100. If the sequence is [100, 100], the alternating sum is 0. A naive implementation could accidentally start with subtraction or misalign the signs, leading to wrong results.

Approaches

The straightforward approach is a direct simulation. Iterate through each element of the sequence, maintaining a running sum. For each element, decide whether to add or subtract it based on its position in the sequence: add the first, subtract the second, and so on. This method works because the sequence lengths are small and the operation count is linear in the size of the sequence.

The insight that can slightly simplify the implementation is to realize that alternating signs correspond to multiplying each element by +1 or -1 depending on its index. If indices start at zero, the sign is 1 for even indices and -1 for odd indices. This removes conditional logic and reduces the problem to a simple sum of products.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n) per test case	O(1)	Accepted
Sign-based Multiplication	O(n) per test case	O(1)	Accepted

Both approaches are fast enough given the constraints.

Algorithm Walkthrough

1. Read the number of test cases t. This tells us how many sequences we will process.

2. Loop over each test case. For each test case:

3. Read the length of the sequence n.

4. Read the n integers of the sequence into a list a.

5. Initialize a variable total to zero. This will store the running alternating sum.

6. Loop over each element a[i] in the sequence:

• If the index i is even, add a[i] to total.
• If the index i is odd, subtract a[i] from total.

5. After processing the sequence, print total.
6. Repeat until all test cases are processed.

Why it works: By alternating the sign based on the index, we guarantee that the first element is added, the second is subtracted, and so on. This exactly matches the definition of the alternating sum. Since each number is processed exactly once, no values are missed, and the invariant that total always holds the correct alternating sum up to the current index is maintained.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    total = 0
    for i in range(n):
        if i % 2 == 0:
            total += a[i]
        else:
            total -= a[i]
    print(total)

The code first reads t and then iterates over each test case. For each sequence, it reads n and the integers. The variable total tracks the alternating sum. The loop adds the element if its index is even and subtracts if the index is odd. Using i % 2 guarantees that the first element is always added, preserving the alternating pattern.

Worked Examples

Example 1

Input sequence: [1, 2, 3, 17]

i	a[i]	Operation	total
0	1	+1	1
1	2	-2	-1
2	3	+3	2
3	17	-17	-15

The output is -15, which matches the expected result.

Example 2

Input sequence: [3, 1, 4, 1, 5]

i	a[i]	Operation	total
0	3	+3	3
1	1	-1	2
2	4	+4	6
3	1	-1	5
4	5	+5	10

The output is 10, confirming the algorithm correctly handles sequences of odd length.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * t)	Each sequence of length n is processed in a single loop; t sequences total.
Space	O(n)	Space is needed to store each sequence temporarily.

Given the constraints t <= 1000 and n <= 50, the maximum number of operations is 50,000, well within the 2-second limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))
        total = 0
        for i in range(n):
            if i % 2 == 0:
                total += a[i]
            else:
                total -= a[i]
        print(total)
    return output.getvalue().strip()

# Provided samples
assert run("4\n4\n1 2 3 17\n1\n100\n2\n100 100\n5\n3 1 4 1 5\n") == "-15\n100\n0\n10", "sample 1"

# Custom cases
assert run("1\n1\n50\n") == "50", "single element"
assert run("1\n2\n99 1\n") == "98", "two elements"
assert run("1\n5\n5 5 5 5 5\n") == "5", "all equal"
assert run("1\n50\n" + " ".join(["1"]*50) + "\n") == "0", "maximum size, alternating sum zero"
assert run("1\n3\n1 2 3\n") == "2", "small odd sequence"

Test input	Expected output	What it validates
1\n1\n50\n	50	Single-element sequence
1\n2\n99 1\n	98	Two-element sequence, subtraction correct
1\n5\n5 5 5 5 5\n	5	All elements equal, checks correct alternating sum
1\n50\n1 ... 1	0	Maximum-size sequence, ensures loop handles long inputs
1\n3\n1 2 3\n	2	Small odd-length sequence

Edge Cases

For a single-element sequence like [100], the algorithm adds the first element and never subtracts, resulting in 100, which is correct. For a two-element sequence like [100, 100], the algorithm adds the first element and subtracts the second, giving 0. In the maximum-size sequence of 50 elements all equal to 1, the alternating sum is zero because additions and subtractions cancel exactly. The indexing by i % 2 handles both odd and even sequence lengths correctly without special cases.