How do you solve The Number of Weak Characters in the Game on LeetCode?

The Number of Weak Characters in the Game (LeetCode #1996) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting helps in efficiently grouping characters by attack values.

What is the time complexity of The Number of Weak Characters in the Game?

The optimal solution for The Number of Weak Characters in the Game runs in O(n log n) time and uses O(1) space. The sorting step dominates the time complexity, making it O(n log n). The subsequent single pass through the list is O(n).

What is the brute force solution for The Number of Weak Characters in the Game?

The brute force approach for The Number of Weak Characters in the Game is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each character against every other character to determine if it is weak. This means we compare the attack and defense values of all pairs of characters. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve The Number of Weak Characters in the Game?

The Number of Weak Characters in the Game uses the following concepts: Array, Stack, Greedy, Sorting, Monotonic Stack. The recommended patterns to study are: Sorting, Greedy Algorithms.

What are the interview tips for The Number of Weak Characters in the Game?

Always consider sorting as a first step in problems involving comparisons. Be mindful of edge cases, such as characters with the same attack but different defenses. Explain your thought process clearly while coding; it helps the interviewer understand your approach.

What are common mistakes when solving The Number of Weak Characters in the Game?

Not sorting correctly, which can lead to incorrect comparisons. Failing to handle cases where multiple characters have the same attack value.

#1996

The Number of Weak Characters in the Game

Medium

ArrayStackGreedySortingMonotonic StackSortingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

By sorting the characters based on their attack values and using a single pass to track the maximum defense seen so far, we can efficiently determine the number of weak characters.

⚙️

Algorithm

5 steps

1Step 1: Sort the characters by attack in ascending order. If attacks are equal, sort by defense in descending order.
2Step 2: Initialize a variable to keep track of the maximum defense encountered so far.
3Step 3: Iterate through the sorted list and for each character, check if its defense is less than the maximum defense seen so far.
4Step 4: If it is, increment the weak character count.
5Step 5: Update the maximum defense with the current character's defense if it's greater.

solution.py14 lines

1# Full working Python code
2properties = [[5,5],[6,3],[3,6]]
3
4def countWeakCharacters(properties):
5    properties.sort(key=lambda x: (x[0], -x[1]))
6    weak_count = 0
7    max_defense = 0
8    for attack, defense in properties:
9        if defense < max_defense:
10            weak_count += 1
11        max_defense = max(max_defense, defense)
12    return weak_count
13
14print(countWeakCharacters(properties))

ℹ

Complexity note: The sorting step dominates the time complexity, making it O(n log n). The subsequent single pass through the list is O(n).

1Sorting helps in efficiently grouping characters by attack values.
2Using a single pass after sorting allows us to leverage previously seen defenses.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.