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Leetcode 3238. Find the Number of Winning Players

Problem Statement

In a game, each player has a unique skill level represented by an integer. You are given a list of unique integers skills where skills[i] is the skill level of the i-th player. There is a concept of winning and losing players. A player i is considered a winner if there are exactly k players with a skill level less than skills[i]. You need to write a function to find the number of winning players.

The function signature should be:

int findWinningPlayers(vector<int>& skills, int k);

Input

• skills: A list of unique integers representing the skill levels of the players. [skills[i]] (1 ≤ skills.length ≤ 1000, -10^9 ≤ skills[i] ≤ 10^9).
• k: An integer specifying the required count of players with lesser skill levels. (0 ≤ k < skills.length).

Output

• Return the number of winning players.

Example

vector<int> skills = {3, 5, 7, 2, 8};
int k = 2;
// Output: 2
// Explanation: The players with skills 5 and 7 are winning players since there are exactly 2 players with skill levels less than theirs.

Clarifying Questions

1. Will the skill levels always be unique?
   • Yes, each player’s skill level is unique.
2. What should be the behavior if no player meets the criteria for being a winner?
   • Simply return 0.
3. Is the order of skill levels in the input list significant?
   • No, the order is not significant; only the skill values matter.

Strategy

To solve the problem:

1. Sort the skills List: First, sort the list of skills.
2. Determine Winning Players: Iterate through the sorted list and check if the current index i matches the count k. If so, increment the counter for winning players.

Detailed Steps

1. Sort the skills list. Sorting simplifies the process of determining the number of players with lower skill levels.
2. Count winners: For each skill level in the sorted list, compare the current index with k. Count the players whose situation matches the condition.

Code

Below is the implementation in C++:

#include <vector>
#include <algorithm>
using namespace std;

int findWinningPlayers(vector<int>& skills, int k) {
    // Step 1: Sort the skills array
    sort(skills.begin(), skills.end());
    
    int count = 0;
    
    // Step 2: Iterate through the sorted array and count the winners
    for (int i = 0; i < skills.size(); ++i) {
        if (i == k) {
            count++;
        }
    }
    
    return count;
}

Time Complexity

• Sorting the Array: The time complexity for sorting the skills array is (O(n \log n)), where (n) is the number of players.
• Counting the Winners: This is a linear scan of the sorted array, so it takes (O(n)) time.

Thus, the overall time complexity is (O(n \log n)) due to the sorting step. The space complexity is (O(1)) if we don’t consider the input array itself.

This should efficiently handle the constraints given in the problem.

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