LeetCode 1996. The Number of Weak Characters in the Game Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Sorting + Traversal

We can sort all characters in descending order of attack power and ascending order of defense power.

Then, traverse all characters. For the current character, if its defense power is less than the previous maximum defense power, it is a weak character, and we increment the answer by one. Otherwise, update the maximum defense power.

After the traversal, we get the answer.

The time complexity is O(n × log n), and the space complexity is O(log n). Here, n is the number of characters.

PythonJavaC++GoTypeScriptJavaScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

/** * @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; };(code-box)