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Formatted question description: https://leetcode.ca/all/1996.html

# 1996. The Number of Weak Characters in the Game

Medium

## 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] = [attack_i, defense_i] represents the properties of the i-th 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 attack_j > attack_i and defense_j > defense_i.

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 <= 10^5
• properties[i].length == 2
• 1 <= attack_i, defense_i <= 10^5

## Solution

Sort properties according to attack in ascending order and then according to defense in ascending order. Loop over properties backwards. When a character is visited, the characters with greater attack values have already been visited. If there exists at least one character among these characters that has a greater defense value, then the current character is weak.

• class Solution {
public int numberOfWeakCharacters(int[][] properties) {
Arrays.sort(properties, new Comparator<int[]>() {
public int compare(int[] property1, int[] property2) {
if (property1[0] != property2[0])
return property1[0] - property2[0];
else
return property1[1] - property2[1];
}
});
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int[] property : properties) {
int attack = property[0], defense = property[1];
if (defense > map.getOrDefault(attack, 0))
map.put(attack, defense);
}
int count = 0;
int length = properties.length;
int index = length - 2;
while (index >= 0 && properties[index][0] == properties[index + 1][0])
index--;
if (index < 0)
return 0;
int maxDefense = properties[length - 1][1];
for (int i = index; i >= 0; i--) {
int attack = properties[i][0], defense = properties[i][1];
if (defense < maxDefense)
count++;
if (i > 0 && attack > properties[i - 1][0]) {
int curMaxDefense = map.get(attack);
maxDefense = Math.max(maxDefense, curMaxDefense);
}
}
return count;
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/the-number-of-weak-characters-in-the-game/
// 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;
}
};

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

############

# 1996. The Number of Weak Characters in the Game
# https://leetcode.com/problems/the-number-of-weak-characters-in-the-game

class Solution:
def numberOfWeakCharacters(self, A: List[List[int]]) -> int:
A.sort(key = lambda x:(x[0], -x[1]))
stack = []
res = 0

for a,b in A:
while stack and a > stack[-1][0] and b > stack[-1][1]:
stack.pop()
res += 1

stack.append((a, b))

return res


• 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;
};