Solution Explanation:

This problem asks us to find the number of "weak" characters in a list of characters, where each character has an attack and defense value. A character is considered weak if another character exists with strictly greater attack and defense values.

The most efficient approach uses sorting and a single traversal. The core idea is to cleverly order the characters to make the weakness check straightforward.

1. Sorting:

We sort the properties array using a custom comparator. The sorting criteria is as follows:

• Primary Key: Characters are primarily sorted in descending order of their attack value (-x[0] in Python, b[0] - a[0] in Java/C++/etc.). This means characters with higher attack values appear earlier in the sorted array.
• Secondary Key (Tiebreaker): If two characters have the same attack value, they are then sorted in ascending order of their defense value (x[1] in Python, a[1] - b[1] in Java/C++/etc.).

2. Traversal and Weakness Check:

After sorting, we iterate through the sorted array. We maintain a variable mx (maximum defense encountered so far). For each character:

• Weakness Check: If the current character's defense value (x[1]) is less than mx, it means a character with both higher attack (because of the sorting) and higher defense has already been encountered. Therefore, the current character is weak, and we increment the ans (answer) counter.
• Update mx: Regardless of whether the character is weak or not, we update mx to be the maximum of its current value and the current character's defense value.

Time Complexity: The dominant operation is the sorting, which takes O(n log n) time, where n is the number of characters. The traversal takes O(n) time. Therefore, the overall time complexity is O(n log n).

Space Complexity: The space complexity is determined by the sorting algorithm. Most efficient sorting algorithms (like merge sort or quicksort) use O(log n) space in the average case due to the recursive calls. In some implementations, in-place sorting might be used, leading to O(1) space complexity.

Code Examples (Multiple Languages):

The code examples below demonstrate the solution in several programming languages. They all follow the same core logic: sorting and traversal.

Python3

class Solution:
    def numberOfWeakCharacters(self, properties: List[List[int]]) -> int:
        properties.sort(key=lambda x: (-x[0], x[1])) # Sort by attack (descending), then defense (ascending)
        ans = mx = 0
        for _, x in properties: # Iterate through defense values
            if x < mx:
                ans += 1
            else:
                mx = x
        return ans

Java

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]); //Custom comparator for sorting
        int ans = 0, mx = 0;
        for (int[] p : properties) {
            if (p[1] < mx) ans++;
            else mx = p[1];
        }
        return ans;
    }
}

C++

class Solution {
public:
    int numberOfWeakCharacters(vector<vector<int>>& properties) {
        sort(properties.begin(), properties.end(), [](const auto& a, const auto& b) {
            return a[0] == b[0] ? a[1] < b[1] : a[0] > b[0]; //Custom lambda for sorting
        });
        int ans = 0, mx = 0;
        for (const auto& p : properties) {
            if (p[1] < mx) ans++;
            else mx = p[1];
        }
        return ans;
    }
};

JavaScript

var numberOfWeakCharacters = function(properties) {
    properties.sort((a, b) => (b[0] - a[0]) || (a[1] - b[1])); // Sort using a comparator
    let ans = 0;
    let mx = 0;
    for (let i = 0; i < properties.length; i++) {
        if (properties[i][1] < mx) ans++;
        else mx = properties[i][1];
    }
    return ans;
};

The other languages (Go, TypeScript) would have similar structures, prioritizing efficient sorting and then a linear scan to count weak characters. The key is the careful sorting to make the weakness check simple and efficient during the traversal.