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

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve The Number of Weak Characters in the Game?

This problem uses Array, Stack, Greedy, Sorting, Monotonic Stack. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving The Number of Weak Characters in the Game?

Insight 1: Sorting the characters based on attack allows us to efficiently compare defenses, as we only need to consider characters that have a greater attack value. Insight 2: When attack values are equal, sorting defenses in ascending order ensures that characters with the same attack will not incorrectly be counted as weak against each other.

Which companies ask LeetCode 1996. The Number of Weak Characters in the Game in interviews?

This problem is frequently asked at Pinterest.

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The Number of Weak Characters in the Game - LeetCode 1996 Solution

The Number of Weak Characters in the Game - Complete Solution Guide

The Number of Weak Characters in the Game is LeetCode problem 1996, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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] = [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

Detailed Explanation

The problem asks us to find the number of "weak" characters in a game. A character is considered weak if there exists another character in the game with strictly greater attack and defense values. The input is a 2D array `properties` where each element `properties[i]` contains the attack and defense values of the i-th character. We need to return the total number of weak characters.

Solution Approach

The solution sorts the characters based on their attack values in descending order. If two characters have the same attack value, they are sorted based on their defense values in ascending order. After sorting, the code iterates through the sorted list of characters and keeps track of the maximum defense value encountered so far. If a character's defense is less than the current maximum defense, it's considered a weak character, and the counter is incremented. This works because, due to sorting, any character encountered before the current one will have either a higher attack or the same attack and lower or equal defense. Therefore, if its defense is less than the maximum defense seen so far, there must be at least one character before it with higher attack and defense.

Step-by-Step Algorithm

Step 1: Sort the `properties` array. Sort by attack in descending order. If attack values are equal, sort by defense in ascending order. This sorting strategy is crucial for efficient comparison.
Step 2: Initialize `weak_characters_count` to 0 and `max_defense` to 0. These variables keep track of the number of weak characters and the maximum defense value seen so far, respectively.
Step 3: Iterate through the sorted `properties` array.
Step 4: For each character, compare its defense with the `max_defense`. If the character's defense is less than `max_defense`, increment `weak_characters_count`.
Step 5: If the character's defense is greater than or equal to `max_defense`, update `max_defense` to the character's defense.
Step 6: After iterating through all the characters, return `weak_characters_count`.

Key Insights

Insight 1: Sorting the characters based on attack allows us to efficiently compare defenses, as we only need to consider characters that have a greater attack value.
Insight 2: When attack values are equal, sorting defenses in ascending order ensures that characters with the same attack will not incorrectly be counted as weak against each other.
Insight 3: Maintaining the maximum defense seen so far enables us to quickly determine if a character is weak by comparing its defense against this maximum value.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Stack, Greedy, Sorting, Monotonic Stack.

Companies

Asked at: Pinterest.

The Number of Weak Characters in the Game | LeetCode 1996

Medium

Array Stack Greedy Sorting Monotonic Stack

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int numberOfWeakCharacters(int[][] properties) {
        Arrays.sort(properties, (a, b) -> {
            if (a[0] != b[0]) {
                return b[0] - a[0];
            } else {
                return a[1] - b[1];
            }
        });

        int weak_characters_count = 0;
        int max_defense = 0;

        for (int[] property : properties) {
            int defense = property[1];
            if (defense < max_defense) {
                weak_characters_count++;
            } else {
                max_defense = defense;
            }
        }

        return weak_characters_count;
    }
}

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Problem Info

Description

Problem 1996: The Number of Weak Characters in the Game - Detailed explanation and solution approaches.

Topics

Array Stack Greedy Sorting Monotonic Stack

Companies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1996

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Sort the `properties` array. Sort by attack in descending order. If attack values are equal, sort by defense in ascending order. This sorting strategy is crucial for efficient comparison.
2
Step 2: Initialize `weak_characters_count` to 0 and `max_defense` to 0. These variables keep track of the number of weak characters and the maximum defense value seen so far, respectively.
3
Step 3: Iterate through the sorted `properties` array.
4
Step 4: For each character, compare its defense with the `max_defense`. If the character's defense is less than `max_defense`, increment `weak_characters_count`.
5
Step 5: If the character's defense is greater than or equal to `max_defense`, update `max_defense` to the character's defense.
6
Step 6: After iterating through all the characters, return `weak_characters_count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n log n) due to the sorting algorithm. The space complexity is O(1) because we are only using a constant amount of extra space to store the `weak_characters_count` and `max_defense` variables. Note that the C implementation with `qsort` has O(1) space assuming the implementation is in place. However, in some languages the sorting could require O(log n) space

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Compare each character with every other character to see if it's weak.

Time: O(n^2)Space: O(1)

Trade-offs: This approach is less efficient than the sorting-based approach for larger inputs because of the nested loops. It avoids sorting, but the increased time complexity makes it impractical for large datasets. It is simpler to implement, though.

✨

Key Insights

1Insight 1: Sorting the characters based on attack allows us to efficiently compare defenses, as we only need to consider characters that have a greater attack value.
2Insight 2: When attack values are equal, sorting defenses in ascending order ensures that characters with the same attack will not incorrectly be counted as weak against each other.
3Insight 3: Maintaining the maximum defense seen so far enables us to quickly determine if a character is weak by comparing its defense against this maximum value.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Input with only two characters where one is weak. The algorithm correctly identifies and counts the weak character because of the sorting and max_defense comparison.
Edge case 2: Input with multiple characters having the same attack value. The sorting by defense in ascending order ensures that only truly weaker characters are counted.
Edge case 3: Input with no weak characters. The algorithm correctly returns 0 because no character's defense will be less than max_defense after processing the array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the characters correctly or at all. Sorting is crucial for the algorithm's efficiency and correctness. Without proper sorting, the algorithm may incorrectly identify weak characters.

✗Common mistake 2: Incorrectly comparing the attack and defense values. The problem statement explicitly states that both attack and defense values must be strictly greater for a character to be considered weak. Forgetting the 'strictly greater' constraint can lead to wrong results.

✗Common mistake 3: Forgetting that equal attack values need special handling. If the attacks are equal, just sorting based on the attack value will not be sufficient and will lead to an incorrect result.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Game development - Identifying weaker game units/characters based on their stats compared to others.

🏦Real-world scenario 2: Employee performance analysis - Determining employees who are 'weak' based on multiple performance metrics compared to their peers, such as sales figures and customer satisfaction ratings.

🎮Real-world scenario 3: Identifying underperforming products based on features and customer reviews compared to competitors

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "The Number of Weak Characters in the Game"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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