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1996. The Number of Weak Characters in the Game

Description

You are playing a game that contains multiple characters, and each of the characters has two main properties: attack and defense. You are given a 2D integer array properties where properties[i] = [attacki, defensei] represents the properties of the ith character in the game.

A character is said to be weak if any other character has both attack and defense levels strictly greater than this character's attack and defense levels. More formally, a character i is said to be weak if there exists another character j where attackj > attacki and defensej > defensei.

Return the number of weak characters.

 

Example 1:

Input: properties = [[5,5],[6,3],[3,6]]
Output: 0
Explanation: No character has strictly greater attack and defense than the other.

Example 2:

Input: properties = [[2,2],[3,3]]
Output: 1
Explanation: The first character is weak because the second character has a strictly greater attack and defense.

Example 3:

Input: properties = [[1,5],[10,4],[4,3]]
Output: 1
Explanation: The third character is weak because the second character has a strictly greater attack and defense.

 

Constraints:

• 2 <= properties.length <= 105
• properties[i].length == 2
• 1 <= attacki, defensei <= 105

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• class Solution {
      public int numberOfWeakCharacters(int[][] properties) {
          Arrays.sort(properties, (a, b) -> b[0] - a[0] == 0 ? a[1] - b[1] : b[0] - a[0]);
          int ans = 0, mx = 0;
          for (var x : properties) {
              if (x[1] < mx) {
                  ++ans;
              }
              mx = Math.max(mx, x[1]);
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int numberOfWeakCharacters(vector<vector<int>>& properties) {
          sort(properties.begin(), properties.end(), [&](auto& a, auto& b) { return a[0] == b[0] ? a[1] < b[1] : a[0] > b[0]; });
          int ans = 0, mx = 0;
          for (auto& x : properties) {
              ans += x[1] < mx;
              mx = max(mx, x[1]);
          }
          return ans;
      }
  };
  

• class Solution:
      def numberOfWeakCharacters(self, properties: List[List[int]]) -> int:
          properties.sort(key=lambda x: (-x[0], x[1]))
          ans = mx = 0
          for _, x in properties:
              ans += x < mx
              mx = max(mx, x)
          return ans
  
  

• func numberOfWeakCharacters(properties [][]int) (ans int) {
  	sort.Slice(properties, func(i, j int) bool {
  		a, b := properties[i], properties[j]
  		if a[0] == b[0] {
  			return a[1] < b[1]
  		}
  		return a[0] > b[0]
  	})
  	mx := 0
  	for _, x := range properties {
  		if x[1] < mx {
  			ans++
  		} else {
  			mx = x[1]
  		}
  	}
  	return
  }
  

• function numberOfWeakCharacters(properties: number[][]): number {
      properties.sort((a, b) => (a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]));
      let ans = 0;
      let mx = 0;
      for (const [, x] of properties) {
          if (x < mx) {
              ans++;
          } else {
              mx = x;
          }
      }
      return ans;
  }
  
  

• /**
   * @param {number[][]} properties
   * @return {number}
   */
  var numberOfWeakCharacters = function (properties) {
      properties.sort((a, b) => (a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]));
      let ans = 0;
      let mx = 0;
      for (const [, x] of properties) {
          if (x < mx) {
              ans++;
          } else {
              mx = x;
          }
      }
      return ans;
  };
  
  

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