Leetcode #1996: The Number of Weak Characters in the Game
In this guide, we solve Leetcode #1996 The Number of Weak Characters in the Game in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Greedy, Array, Sorting, Monotonic Stack
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: properties = [[5,5],[6,3],[3,6]]
Output: 0
Explanation: No character has strictly greater attack and defense than the other.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.