The Number of Weak Characters in the Game

// Accepted solution for LeetCode #1996: The Number of Weak Characters in the Game
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;
    }
}

// Accepted solution for LeetCode #1996: The Number of Weak Characters in the Game
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
}

# Accepted solution for LeetCode #1996: The Number of Weak Characters in the Game
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

// Accepted solution for LeetCode #1996: The Number of Weak Characters in the Game
/**
 * [1996] The Number of Weak Characters in the Game
 *
 * 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 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 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 <= 10^5
 * 	properties[i].length == 2
 * 	1 <= attacki, defensei <= 10^5
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/the-number-of-weak-characters-in-the-game/
// discuss: https://leetcode.com/problems/the-number-of-weak-characters-in-the-game/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn number_of_weak_characters(properties: Vec<Vec<i32>>) -> i32 {
        let mut properties = properties;
        properties.sort_unstable_by_key(|p| (-p[0], p[1]));
        properties
            .iter()
            .fold((0, -1), |(count, max_def), prop| {
                if max_def > prop[1] {
                    (count + 1, max_def)
                } else {
                    (count, prop[1])
                }
            })
            .0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1996_example_1() {
        let properties = vec![vec![5, 5], vec![6, 3], vec![3, 6]];

        let result = 0;

        assert_eq!(Solution::number_of_weak_characters(properties), result);
    }

    #[test]
    fn test_1996_example_2() {
        let properties = vec![vec![2, 2], vec![3, 3]];

        let result = 1;

        assert_eq!(Solution::number_of_weak_characters(properties), result);
    }

    #[test]
    fn test_1996_example_3() {
        let properties = vec![vec![1, 5], vec![10, 4], vec![4, 3]];

        let result = 1;

        assert_eq!(Solution::number_of_weak_characters(properties), result);
    }
}

// Accepted solution for LeetCode #1996: The Number of Weak Characters in the Game
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;
}