Overview

This problem asks you to determine the maximum number of times your supported player \(K\) can win when you are free to decide the tournament matchups. Since the player with the higher rating always wins, player \(K\) can only beat players with a lower rating.

Analysis

Key Insight 1: Match outcomes are completely determined

All players have distinct ratings, and the player with the higher rating always wins, so the result of any match between two players is known in advance. Player \(K\) will always beat players with a lower rating and always lose to players with a higher rating.

Key Insight 2: The matchup order can be freely chosen

In each round, you can choose any two remaining players to compete, so you have complete control over who player \(K\) faces. This differs from a standard tournament where the bracket is fixed.

Key Insight 3: As long as weaker players remain, K can keep winning

As long as there are players weaker than player \(K\) (players with lower ratings) still remaining, Takahashi can match player \(K\) against one of those weaker players. Player \(K\) will definitely win and advance to the next round.

• If player \(K\) is the strongest (has the highest rating): they can beat everyone, so the answer is \(N - 1\).
• Otherwise: they can defeat weaker players one by one. Once all weaker players are eliminated, only stronger players remain, so player \(K\) will definitely lose in the next match.

Concrete Example

For \(N = 5, K = 3, S = [10, 30, 20, 50, 40]\):

• Player \(3\)’s rating is \(20\)
• Weaker players: Player \(1\) (rating \(10\)) → 1 player
• Stronger players: Players \(2, 4, 5\) (ratings \(30, 50, 40\)) → 3 players

Takahashi matches player \(3\) against player \(1\), and player \(3\) wins. After that, only stronger players remain, so player \(3\) loses. The answer is 1.

Algorithm

1. Get player \(K\)’s rating \(S_K\).
2. Count the number of players with a rating lower than \(S_K\) (\(\text{num\_weaker}\)).
3. Count the number of players with a rating higher than \(S_K\) (\(\text{num\_stronger}\)).
4. If \(\text{num\_stronger} = 0\) (player \(K\) is the strongest), the answer is \(N - 1\).
5. Otherwise, the answer is \(\text{num\_weaker}\).

Complexity

• Time complexity: \(O(N)\) (just a single scan through all players’ ratings to count)
• Space complexity: \(O(N)\) (for storing the input)

Implementation Notes

• Player numbers are \(1\)-indexed, so you need to use S[K-1] when accessing the array.

• Don’t forget to handle the case where player \(K\) is the strongest. In that case, they can win all \(N-1\) matches.

• Sorting is not essential to the solution; it suffices to simply count the number of elements smaller than \(S_K\).

  Source Code

import sys

def solve():
    input_data = sys.stdin.read().split()
    N = int(input_data[0])
    K = int(input_data[1])
    S = list(map(int, input_data[2:2+N]))
    
    sk = S[K-1]
    
    # Count how many players have rating less than sk
    # Player K can beat all players with rating < sk
    # The question is: how many of them can K actually face in an optimal tournament?
    
    # Players weaker than K (K can beat them)
    weaker = sorted([s for s in S if s < sk])
    # Players stronger than K (K loses to them)
    stronger = sorted([s for s in S if s > sk])
    
    num_weaker = len(weaker)
    num_stronger = len(stronger)
    
    # Strategy: We want K to fight as many weaker players as possible before
    # eventually facing a stronger player (if any).
    # 
    # But we also need to "eliminate" stronger players by having them fight each other
    # or fight weaker players, so that K doesn't have to face them.
    # Wait, K will eventually lose to a stronger player (or win the tournament if K is strongest).
    
    # If K is the strongest player, K can win N-1 times (fight everyone one by one).
    if num_stronger == 0:
        print(N - 1)
        return
    
    # K can beat all weaker players. The question is how many weaker players
    # can K face before being forced to face a stronger player.
    #
    # At each round, we choose any two remaining players to fight.
    # We want to maximize K's wins.
    #
    # Strategy: Have K fight weaker players one by one. Meanwhile, we need
    # to also eliminate other players (stronger ones can fight each other or 
    # fight weaker ones) but we need to keep enough weaker players for K to fight.
    #
    # Total N players, N-1 matches total. K will play some matches, winning against
    # weaker opponents, and eventually lose to a stronger opponent (or never face one
    # if we can eliminate all stronger players - but we can't since K can't beat them
    # and stronger players can only be eliminated by even stronger players).
    #
    # Actually, we need all stronger players to be eliminated by other stronger players
    # or... no. The tournament ends when 1 player remains. K will be eliminated when
    # K faces a stronger player. We want to delay this as long as possible.
    #
    # We can have K fight one weaker player each round while having stronger players
    # fight each other (or fight weaker players) in other matches.
    #
    # Each round we can have multiple matches? No - "each round" we pick 2 players
    # for 1 match. So rounds are sequential, one match per round.
    #
    # So in each round, we pick 2 players. We can alternate: have K fight a weaker
    # player, then have two other players fight, then K fights another weaker player, etc.
    #
    # K needs to fight weaker players. Each time K fights, that uses one round and 
    # eliminates one weaker player. We can also use rounds to have other players 
    # fight each other (not involving K) to reduce the total count.
    #
    # K will eventually need to face a stronger player (since stronger players exist
    # and the tournament must end with 1 player). K loses that match.
    #
    # We want to maximize the number of K's wins. K can fight a weaker player 
    # whenever there's at least one weaker player remaining. K loses when forced
    # to face a stronger player.
    #
    # But K is never "forced" to face a stronger player as long as weaker players exist.
    # So K can keep fighting weaker players until no weaker players remain.
    # Then K must face a stronger player and lose.
    #
    # So the answer is simply: number of weaker players = num_weaker.
    # 
    # Wait, but we need other matches to happen too to keep the tournament going.
    # Actually no - we just freely choose who fights whom each round. As long as
    # there are weaker players, we can always have K fight one of them. 
    # K wins, the weaker player is eliminated, and we continue.
    # When no weaker players remain, K faces a stronger player and loses.
    
    print(num_weaker)

solve()

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