The exact problem description for LeetCode problem “3238. Find the Number of Winning Players” is not provided. However, the problem title suggests that it involves determining the count of players who meet certain criteria to be classified as “winning” from a given list of scores or similar measurements.
Typically, such problems are related to sports competitions, gaming leaderboards, or score lists where players’ performances are evaluated. A winning player might be defined as one whose score is higher than a certain threshold, in the top X percentage, or satisfying some custom criteria.
To better understand and solve the problem, the following clarifications are necessary:
Assuming a basic scenario where we have a list of player scores and a threshold score, and we need to count how many players have scores above that threshold:
For instance, if we are given the list of scores [10, 20, 30, 40, 50]
and the threshold is 25
, the winning players are those with scores 30
, 40
, and 50
. Hence, the result should be 3
.
def count_winning_players(scores, threshold):
"""
Count the number of players with scores above the threshold.
:param scores: List[int] - A list of player scores.
:param threshold: int - The threshold score to be considered a winning player.
:return: int - The count of winning players.
"""
# Initialize the count of winning players
winning_count = 0
# Iterate through the scores and count how many are above the threshold
for score in scores:
if score > threshold:
winning_count += 1
return winning_count
# Example usage:
scores = [10, 20, 30, 40, 50]
threshold = 25
print(count_winning_players(scores, threshold)) # Output: 3
The time complexity of this solution is (O(n)), where (n) is the number of scores in the list. This is because we need to iterate through the entire list once to count the scores above the threshold.
The space complexity is (O(1)) because we only use a constant amount of extra space for the counter.
This solution effectively counts the number of winning players based on a given threshold. Depending on further clarification about the problem’s constraints and requirements, the code and strategy can be adjusted accordingly.
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