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LeetCode 1996: The Number of Weak Characters in the Game Solution

Master LeetCode problem 1996 (The Number of Weak Characters in the Game), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1996. The Number of Weak Characters in the Game

Medium
ArrayStackGreedySortingMonotonic Stack

Problem Explanation

Explanation

To solve this problem, we need to iterate through the characters' properties and determine if there exists another character with strictly greater attack and defense levels for each character. We can achieve this by sorting the characters based on decreasing attack and defense levels. Then, we iterate through the sorted characters and check if there is any character with both attack and defense levels strictly greater than the current character.

Algorithm

  1. Sort the characters in descending order of attack levels and, for characters with the same attack level, in descending order of defense levels.
  2. Initialize a count variable to track the number of weak characters.
  3. Iterate through the sorted characters and compare the current character's attack and defense levels with the maximum attack and defense levels encountered so far. If there exists a character with both attack and defense levels greater than the current character, increment the count.
  4. Return the count as the number of weak characters.

Time Complexity

The time complexity of this algorithm is O(n log n), where n is the number of characters.

Space Complexity

The space complexity of this algorithm is O(n) to store the sorted characters.

Solution Code

import java.util.*;

class Solution {
    public int numberOfWeakCharacters(int[][] properties) {
        Arrays.sort(properties, (a, b) -> {
            if (a[0] == b[0]) {
                return b[1] - a[1];
            }
            return b[0] - a[0];
        });
        
        int count = 0;
        int maxDefense = 0;
        
        for (int[] character : properties) {
            if (character[1] < maxDefense) {
                count++;
            }
            maxDefense = Math.max(maxDefense, character[1]);
        }
        
        return count;
    }
}

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