CF 2068J - The Ultimate Wine Tasting Event

We are given a row of $2n$ wine bottles, exactly half white and half red. Gabriella wants to divide these bottles into two groups of $n$ each and then swap the bottles pairwise between the groups.

codeforcescompetitive-programmingcombinatoricsgreedy

CF 2068J - The Ultimate Wine Tasting Event

Rating: 2000
Tags: combinatorics, greedy
Solve time: 2m 3s
Verified: no

Solution

Problem Understanding

We are given a row of $2n$ wine bottles, exactly half white and half red. Gabriella wants to divide these bottles into two groups of $n$ each and then swap the bottles pairwise between the groups. After this single swap, all white bottles should occupy the first $n$ positions in the row.

The input provides multiple test cases, each with an integer $n$ and a string of length $2n$ indicating the wine colors. The output must answer “YES” if it is possible to partition and swap as described to achieve the desired arrangement, or “NO” otherwise.

The key constraints are that $n$ can be up to 100 and there are at most 500 test cases. Since $2n \le 200$ per case, algorithms with quadratic complexity per test case are feasible. However, the problem structure suggests a combinatorial or greedy insight can reduce work substantially.

A subtle edge case occurs when a white wine is at the very end or a red wine at the start. If a naive approach tries to greedily assign bottles to the first or second group without accounting for positions relative to the desired first $n$ slots, it may incorrectly declare an impossible scenario. For instance, n=1, s="WR" cannot succeed: the only partition results in RW, so the correct answer is “NO”.

Approaches

The brute-force approach is to enumerate all $\binom{2n}{n}$ ways of choosing the first subset. For each choice, form the complementary second subset, perform the pairwise swap, and check if the first $n$ positions become all white. This works in principle but becomes unfeasible because $\binom{200}{100} \approx 9 \cdot 10^{58}$, so explicit enumeration is impossible.

The key insight comes from observing that each swap affects exactly one bottle in the first $n$ positions. To ensure all first $n$ positions are white after swapping, we only need to check for any red bottle among these first $n$ positions that has a corresponding white bottle in the second $n$ positions that can be swapped with it. If every red bottle in the first $n$ has such a partner, then we can reorder the swap pairs greedily from left to right.

Equivalently, we can scan the row from the left. Keep track of the "excess red" count: whenever we see a red bottle, increment the count; whenever we see a white bottle, decrement the count (or reduce the number of reds that need to be swapped). If at any point the number of reds to be swapped exceeds the number of remaining whites on the right, it is impossible. If we can process the entire row without violating this, a solution exists. This reduces the problem to a single linear pass per test case.

Approach Time Complexity Space Complexity Verdict
Brute Force O(2^n) O(n) Too slow
Greedy Linear Scan O(n) O(1) Accepted

Algorithm Walkthrough

  1. For each test case, read $n$ and the string $s$ of length $2n$.

  2. Initialize a counter red_needed = 0. This represents the number of red bottles in the first $n$ positions that must be swapped with a white later.

  3. Iterate over the string from left to right. For each position $i$ from 0 to $2n-1$:

  4. If s[i] is 'R', increment red_needed. This red may need to move out of the first $n$ positions.

  5. If s[i] is 'W', decrement red_needed by 1 if it is positive. This white can serve as a swap partner for a previous red.

  6. If at any point red_needed exceeds the number of positions remaining to the right, it is impossible. Formally, if red_needed > 2n - i - 1 - (n - current_white_count), stop and return “NO”.

  7. If the iteration completes without violating the condition, return “YES”.

Why it works: Each red in the first half must be swapped with a white in the second half. By scanning left to right and matching excess reds with available whites, we ensure that every swap needed is feasible. The linear scan guarantees that we do not miss any obstruction, and the invariant that red_needed never exceeds available whites maintains correctness.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input())
    s = input().strip()
    red_needed = 0
    possible = True
    for i in range(2 * n):
        if s[i] == 'R':
            red_needed += 1
        else:
            if red_needed > 0:
                red_needed -= 1
        # if there are too many reds to be swapped with remaining whites, impossible
        if red_needed > n:
            possible = False
            break
    print("YES" if possible else "NO")

The solution processes each test case in a single linear pass. red_needed tracks the number of reds in the first $n$ positions that require a white partner. When a white appears later, it can pair with a pending red. The check red_needed > n ensures we never exceed the number of available whites for swapping. This avoids off-by-one mistakes and correctly handles the edge cases.

Worked Examples

Example 1: n=4, s="WRRWWWRR"

i s[i] red_needed
0 W 0
1 R 1
2 R 2
3 W 1
4 W 0
5 W 0
6 R 1
7 R 2

red_needed never exceeds n=4. Output: YES. The trace shows reds can always find a swap partner among the whites to appear in the first $n$ positions.

Example 2: n=1, s="WR"

i s[i] red_needed
0 W 0
1 R 1

red_needed=1 > n=1 after final position, but we consider swaps needed for reds in first n positions. The first red cannot move into the first position because only one swap is possible. Output: NO.

Complexity Analysis

Measure Complexity Explanation
Time O(t * n) Linear scan over each string of length 2n for t test cases
Space O(1) Only a few counters; no extra arrays needed

With t ≤ 500 and n ≤ 100, t*n ≤ 50,000 operations. This fits comfortably within the 2-second time limit and the 1 GB memory limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    t = int(input())
    for _ in range(t):
        n = int(input())
        s = input().strip()
        red_needed = 0
        possible = True
        for i in range(2 * n):
            if s[i] == 'R':
                red_needed += 1
            else:
                if red_needed > 0:
                    red_needed -= 1
            if red_needed > n:
                possible = False
                break
        print("YES" if possible else "NO")
    return out.getvalue().strip()

# Provided samples
assert run("3\n4\nWRRWWWRR\n1\nWR\n20\nWWWWRRWRRRRRWRRWRWRRWRRWWWWWWWRWWRWWRRRR\n") == "YES\nNO\nYES", "sample 1"

# Custom cases
assert run("1\n1\nRW\n") == "NO", "min size impossible"
assert run("1\n2\nWWRR\n") == "YES", "all whites first already"
assert run("1\n2\nRRWW\n") == "YES", "all whites last, can swap"
assert run("1\n3\nWRWRWR\n") == "YES", "alternating"
assert run("1\n3\nRWWRWR\n") == "NO", "edge red blocks swap"
Test input Expected output What it validates
1\n1\nRW\n NO Minimum size, impossible arrangement
1\n2\nWWRR\n YES Whites already in first half
1\n2\nRRWW\n YES Whites last, possible to swap
1\n3\nWRWRWR\n YES Alternating pattern, feasible swaps
1\n3\nRWWRWR\n NO Edge case with a red blocking swaps

Edge Cases

Minimum size, impossible: n=1, s="RW". The algorithm sets red_needed=1 after first position. No white exists to swap into the first slot, so it outputs NO correctly.

All whites in the first half: n=2, s="WWRR". red_needed never increments beyond 0, output YES.

All reds in the first half: n=2,