CF 401A - Vanya and Cards

Rating: 800
Tags: implementation, math
Solve time: 4m 10s
Verified: no

Solution

Problem Understanding

Vanya has discovered a subset of cards from his original infinite collection, where each card carries an integer between -x and x. His goal is to balance the sum of these found cards to zero by potentially adding additional cards from the infinite supply. The task is to determine the minimum number of extra cards required to achieve a total sum of zero.

The input gives the number of found cards n, the maximum absolute value x, and the list of integers on the found cards. The output is a single integer representing the minimum number of extra cards Vanya must pick.

The constraints are modest: n can be up to 1000 and x up to 1000, so an O(n) or O(1) solution is acceptable. There is no need for advanced data structures or optimizations beyond simple arithmetic.

A subtle edge case occurs when the sum of the found cards is already zero. In that case, no additional cards are needed. Another edge case arises when the absolute value of the sum is smaller than or equal to x; only one additional card suffices. A careless implementation might attempt to add multiple small cards unnecessarily instead of using a single card of magnitude equal to the sum.

Approaches

The brute-force approach would be to try every combination of extra cards until the sum becomes zero. This works because you can choose any integer between -x and x, but it becomes inefficient since the number of combinations grows exponentially with the number of extra cards, which is unacceptable even for n=1000.

The key observation is that the sum of found cards, call it s, determines the problem completely. The smallest number of extra cards needed is simply the smallest number of integers with absolute value at most x that sum up to -s. The optimal strategy is always to use as few cards as possible with the maximum magnitude x. This reduces the problem to a simple arithmetic formula: divide the absolute value of s by x and round up. This works because using numbers smaller than x never reduces the count of required cards.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^k)	O(k)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

Compute the sum s of all found cards. This captures the net imbalance that we need to correct.
Take the absolute value of s. The sign is irrelevant because we can choose positive or negative cards.
Divide the absolute sum by x using integer division to compute how many full x-magnitude cards are needed.
If the absolute sum is not divisible by x, increment the count by one to account for the remainder. This ensures we reach exactly zero without overshooting.
Output the final count.

Why it works: the invariant is that using the maximum magnitude card minimizes the number of cards required. Any other distribution would either require more cards of smaller magnitude or fail to reach zero exactly. The algorithm always produces the smallest possible number of additional cards.

Python Solution

import sys
input = sys.stdin.readline

n, x = map(int, input().split())
cards = list(map(int, input().split()))

total = sum(cards)
abs_total = abs(total)

# number of extra cards needed
count = abs_total // x
if abs_total % x != 0:
    count += 1

print(count)

The code first reads the input efficiently. We calculate the sum and take its absolute value to simplify subsequent arithmetic. Using integer division captures the number of full x-magnitude cards, and the modulo check ensures any leftover imbalance is corrected by adding one more card. There is no need for loops or conditional logic beyond this simple arithmetic.

Worked Examples

Sample 1:

Input: 3 2, -1 1 2

Step	total	abs_total	count
sum cards	2	2	1
divide by x	2 // 2	-	1
modulo check	2 % 2 = 0	-	count unchanged

The sum is 2, which can be balanced with one card of value -2.

Sample 2:

Input: 2 3, 1 1

Step	total	abs_total	count
sum cards	2	2	0
divide by x	2 // 3	-	0
modulo check	2 % 3 = 2	-	count += 1 → 1

The sum is 2, less than x=3, so one card of value -2 suffices.

These traces confirm the correctness of computing the absolute sum and dividing by the maximum allowed card value.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Computing the sum of n cards
Space	O(n)	Storing the list of cards

Given n ≤ 1000, this is efficient. Memory usage is small, and arithmetic operations are trivial, so the solution easily fits within the 1-second limit and 256 MB memory constraint.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, x = map(int, input().split())
    cards = list(map(int, input().split()))
    total = sum(cards)
    abs_total = abs(total)
    count = abs_total // x
    if abs_total % x != 0:
        count += 1
    return str(count)

# Provided samples
assert run("3 2\n-1 1 2\n") == "1", "sample 1"
assert run("2 3\n2 2\n") == "2", "sample 2"

# Custom cases
assert run("1 5\n0\n") == "0", "sum zero requires no cards"
assert run("5 3\n3 3 3 3 3\n") == "5", "sum 15, x=3, need 5 cards"
assert run("4 10\n-5 5 -10 10\n") == "0", "sum zero with mix of positives and negatives"
assert run("3 2\n1 1 1\n") == "2", "sum 3, max card 2, need 2 cards"
assert run("1 1\n1\n") == "1", "sum 1, max card 1, need 1 card"

Test input	Expected output	What it validates
`1 5\n0`	0	Zero sum requires no extra cards
`5 3\n3 3 3 3 3`	5	Multiple cards needed when sum divisible by x
`4 10\n-5 5 -10 10`	0	Mix of positives and negatives summing to zero
`3 2\n1 1 1`	2	Sum exceeds max card, need multiple smaller cards
`1 1\n1`	1	Minimal input with sum equal to max card

Edge Cases

When the sum is already zero, as in 4 10\n-5 5 -10 10, the algorithm immediately computes abs_total = 0 and count = 0. The modulo check does not increment the count, correctly producing zero extra cards.

When the sum is smaller than x, for example 2 3\n1 1, abs_total = 2, which is less than x=3. The integer division yields 0, but the modulo check detects a remainder and increments the count to 1, correctly picking a single card of value -2.