1996. The Number of Weak Characters in the Game

1996. The Number of Weak Characters in the Game (Medium)

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

Similar Questions:

• Russian Doll Envelopes (Hard)
• Maximum Height by Stacking Cuboids (Hard)

Solution 1. Sorting

// OJ: https://leetcode.com/problems/the-number-of-weak-characters-in-the-game/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int numberOfWeakCharacters(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](auto &a, auto &b) { return a[0] < b[0]; });
        multiset<int> s;
        for (auto &a : A) s.insert(a[1]);
        int N = A.size(), ans = 0;
        for (int i = 0; i < N; ) {
            vector<int> ds;
            int at = A[i][0];
            while (i < N && A[i][0] == at) {
                ds.push_back(A[i][1]);
                s.erase(s.find(A[i][1]));
                ++i;
            }
            for (int d : ds) {
                ans += s.upper_bound(d) != s.end();
            }
        }
        return ans;
    }
};

Solution 2. LIS

Similar to 354. Russian Doll Envelopes (Hard)

// OJ: https://leetcode.com/problems/the-number-of-weak-characters-in-the-game/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int numberOfWeakCharacters(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](auto &a, auto &b) { return a[0] != b[0] ? a[0] > b[0] : a[1] < b[1]; });
        vector<int> dp;
        int ans = 0;
        for (auto &c : A) {
            int i = lower_bound(begin(dp), end(dp), c[1], greater<>()) - begin(dp);
            if (i == dp.size()) dp.emplace_back();
            dp[i] = c[1];
            ans += i > 0;
        }
        return ans;
    }
};

TODO

Use Bucket Sort. Work on O(N) solution.