algoadvance

Leetcode 3238. Find the Number of Winning Players

Problem Statement

LeetCode Problem 3238: Find the Number of Winning Players

Given an integer array scores that represents the scores of players in a competition, you are asked to determine the number of winning players.

A player is considered winning if their score is strictly greater than the average score of all players.

Return the count of winning players.

Example:

Input: scores = [1, 2, 3, 4, 5]
Output: 2
Explanation: The average score is 3, and the players with scores 4 and 5 are the winning players.

Clarifying Questions

  1. Q: What is the range of the integers in the scores array?
    • A: The integers are likely to be within a typical competitive range, possibly from 0 to 100.
  2. Q: Can the scores array be empty?
    • A: No, the constraints will clarify that the array always contains at least one element.
  3. Q: Can the scores be negative?
    • A: Unlikely, given the context of the problem is a competition. Assume scores are non-negative.
  4. Q: What is the expected time complexity?
    • A: Given that calculating the average and comparing values are the primary operations, a linear time complexity, O(n), is expected to be efficient and sufficient.

Strategy

The core steps to solving this problem are:

  1. Calculate the average score of all players.
  2. Iterate through the scores array and count the number of players whose scores are strictly greater than the calculated average.

Steps:

  1. Calculate the Average Score:
    • Sum all the elements in the scores array.
    • Divide the sum by the number of elements.
  2. Count the Winning Players:
    • Iterate through the scores array.
    • For each score, check if it is greater than the calculated average.
    • Increment a counter each time the condition is met.

Code

public class NumberOfWinningPlayers {
    public int findWinningPlayers(int[] scores) {
        if (scores == null || scores.length == 0) {
            return 0;
        }

        // Calculate the average score
        double sum = 0;
        for (int score : scores) {
            sum += score;
        }
        double average = sum / scores.length;

        // Count the number of players with scores greater than the average
        int winningPlayersCount = 0;
        for (int score : scores) {
            if (score > average) {
                winningPlayersCount++;
            }
        }

        return winningPlayersCount;
    }

    public static void main(String[] args) {
        NumberOfWinningPlayers solution = new NumberOfWinningPlayers();
        int[] scores = {1, 2, 3, 4, 5};
        System.out.println(solution.findWinningPlayers(scores)); // Output: 2
    }
}

Time Complexity

The time complexity of this solution is O(n), where n is the number of elements in the scores array. Here’s why:

  1. Sum Calculation: We iterate through the array once to calculate the sum.
  2. Count Calculation: We iterate through the array a second time to count the number of winning players. Thus, the linear time complexity, O(n), ensures efficiency even for large input sizes.

Feel free to ask if any more clarifying questions are needed or if additional explanations on any part of the problem or solution would be helpful!

Cut your prep time in half and DOMINATE your interview with AlgoAdvance AI