We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Greedy
  4. Luck Balance
  5. Discussions

Luck Balance

  • tao_zhang
     + 18 comments

    This is the same problem I have done in Week Of Code 21. I copied my solution in Java:

    import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        int total = 0;
        List<Integer> importantContests = new ArrayList<>();
        for (int i=0; i<n; i++){
            int luck = scanner.nextInt();
            int importance = scanner.nextInt();
            total += luck;
            if (importance == 1) {
                importantContests.add(luck);
            } 
        }
        Collections.sort(importantContests);
        int luckToFlip = 0;
        int mustWinImprCount = importantContests.size() - k;
        for (int i=0; i<mustWinImprCount; i++){
            luckToFlip += importantContests.get(i);
        }
        int result = total - 2*luckToFlip;
        System.out.println(result);
    }
}