What is the time complexity of Divide Players Into Teams of Equal Skill?

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

What data structures are used to solve Divide Players Into Teams of Equal Skill?

This problem uses Array, Hash Table, Two Pointers, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Divide Players Into Teams of Equal Skill?

Insight 1: Sorting the array allows us to efficiently find pairs that sum up to the same target value using a two-pointer approach. Insight 2: The sum of the skills of the first and last elements after sorting provides the target skill sum for each team. If any pair doesn't match this sum, a valid team arrangement is impossible.

Which companies ask LeetCode 2491. Divide Players Into Teams of Equal Skill in interviews?

This problem is frequently asked at Expedia, IBM, PayPal.

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Divide Players Into Teams of Equal Skill - LeetCode 2491 Solution

Divide Players Into Teams of Equal Skill - Complete Solution Guide

Divide Players Into Teams of Equal Skill is LeetCode problem 2491, 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 given a positive integer array skill of even length n where skill[i] denotes the skill of the i th player. Divide the players into n / 2 teams of size 2 such that the total skill of each team is equal . The chemistry of a team is equal to the product of the skills of the players on that team. Return the sum of the chemistry of all the teams, or return -1 if there is no way to divide the players into teams such that the total skill of each team is equal. Example 1: Input: skill = [3,2,5,1

Detailed Explanation

The problem requires dividing a given array of player skills into teams of two, such that each team has the same total skill. The goal is to find the sum of the 'chemistry' of all teams, where the chemistry of a team is the product of the skills of the two players in that team. If it's impossible to form teams with equal total skill, the function should return -1.

Solution Approach

The provided solution uses a two-pointer approach after sorting the skill array. It initializes two pointers, one at the beginning and one at the end of the sorted array. It calculates the sum of the elements pointed to by these pointers, which becomes the target sum. It then iterates, moving the pointers inwards, and verifies that the sum of the skills at these pointers is equal to the target sum. If they are not equal, it means the input cannot be divided into valid teams, so it returns -1. If they are equal, it calculates the chemistry of the team and adds it to the total chemistry.

Step-by-Step Algorithm

Step 1: Sort the skill array in ascending order.
Step 2: Initialize two pointers, 'left' at the beginning (index 0) and 'right' at the end (index n-1) of the array.
Step 3: Calculate the target sum by adding the skills at the 'left' and 'right' pointers: target_sum = skill[left] + skill[right].
Step 4: Initialize a variable 'total_chemistry' to 0. This will accumulate the sum of chemistry values for each team.
Step 5: Iterate while 'left' is less than 'right'.
Step 6: Inside the loop, calculate the current sum: current_sum = skill[left] + skill[right].
Step 7: If current_sum is not equal to target_sum, return -1, indicating that it's not possible to divide the players into valid teams.
Step 8: If current_sum is equal to target_sum, calculate the chemistry of the team by multiplying the skills at the 'left' and 'right' pointers and add it to 'total_chemistry': total_chemistry += (long long)skill[left] * skill[right]. The explicit cast to long long handles potential overflow.
Step 9: Increment 'left' and decrement 'right' to move the pointers towards the center of the array.
Step 10: After the loop completes, return 'total_chemistry'.

Key Insights

Insight 1: Sorting the array allows us to efficiently find pairs that sum up to the same target value using a two-pointer approach.
Insight 2: The sum of the skills of the first and last elements after sorting provides the target skill sum for each team. If any pair doesn't match this sum, a valid team arrangement is impossible.
Insight 3: It's crucial to use long long (or similar data type) to store the 'total_chemistry' to avoid potential integer overflow, as the problem constraints allow skill values up to 1000 and up to n/2 teams, resulting in products that could exceed the maximum value of an int.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Two Pointers, Sorting.

Companies

Asked at: Expedia, IBM, PayPal.

Divide Players Into Teams of Equal Skill | LeetCode 2491

Medium

Array Hash Table Two Pointers Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long dividePlayers(int[] skill) {
        Arrays.sort(skill);
        int n = skill.length;
        int left = 0;
        int right = n - 1;
        
        if (n == 2) {
            return (long)skill[0] * skill[1];
        }

        int target_sum = skill[left] + skill[right];
        long total_chemistry = 0;

        while (left < right) {
            int current_sum = skill[left] + skill[right];
            if (current_sum != target_sum) {
                return -1;
            }
            
            total_chemistry += (long)skill[left] * skill[right];
            left += 1;
            right -= 1;
            
        }

        return total_chemistry;
    }
}

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Problem Info

Description

Problem 2491: Divide Players Into Teams of Equal Skill - Detailed explanation and solution approaches.

Topics

Array Hash Table Two Pointers Sorting

Companies

ExpediaIBMPayPal

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2491

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

10 steps

1
Step 1: Sort the skill array in ascending order.
2
Step 2: Initialize two pointers, 'left' at the beginning (index 0) and 'right' at the end (index n-1) of the array.
3
Step 3: Calculate the target sum by adding the skills at the 'left' and 'right' pointers: target_sum = skill[left] + skill[right].
4
Step 4: Initialize a variable 'total_chemistry' to 0. This will accumulate the sum of chemistry values for each team.
5
Step 5: Iterate while 'left' is less than 'right'.
6
Step 6: Inside the loop, calculate the current sum: current_sum = skill[left] + skill[right].
7
Step 7: If current_sum is not equal to target_sum, return -1, indicating that it's not possible to divide the players into valid teams.
8
Step 8: If current_sum is equal to target_sum, calculate the chemistry of the team by multiplying the skills at the 'left' and 'right' pointers and add it to 'total_chemistry': total_chemistry += (long long)skill[left] * skill[right]. The explicit cast to long long handles potential overflow.
9
Step 9: Increment 'left' and decrement 'right' to move the pointers towards the center of the array.
10
Step 10: After the loop completes, return 'total_chemistry'.

💡

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 dominant operation is sorting the array, which takes O(n log n) time using efficient sorting algorithms like merge sort or quicksort. The two-pointer iteration takes O(n) time, but this is less significant than the sorting time. The space complexity is O(1) because the sorting is done in-place or uses a very small amount of extra space, and the algorithm only uses a few constant-size variables.

📖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

Hash Table Approach

Use a hash table (or dictionary) to count the frequency of each skill value. Then, iterate through the skill values. For each skill value, check if the 'target_sum - skill' value exists in the hash table. This approach allows for finding the pairs with target sum efficiently.

Time: O(n)Space: O(n)

Trade-offs: The hash table approach avoids sorting, resulting in O(n) time complexity. However, it requires O(n) extra space to store the frequency map. This could be a good choice if space is not a major concern and you prioritize minimizing the execution time, especially for very large arrays. The sorting approach is preferable if space is heavily constrained.

✨

Key Insights

1Insight 1: Sorting the array allows us to efficiently find pairs that sum up to the same target value using a two-pointer approach.
2Insight 2: The sum of the skills of the first and last elements after sorting provides the target skill sum for each team. If any pair doesn't match this sum, a valid team arrangement is impossible.
3Insight 3: It's crucial to use long long (or similar data type) to store the 'total_chemistry' to avoid potential integer overflow, as the problem constraints allow skill values up to 1000 and up to n/2 teams, resulting in products that could exceed the maximum value of an int.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Sorting

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

Edge Cases

Edge case 1: Empty or null skill array. While not explicitly handled in the code provided, you would likely want to include a check at the start to ensure the input is valid and return -1 if the array is empty or null.
Edge case 2: Array with odd length. The problem states that the length of the array will always be even so this case is not considered. If the length was odd the code should handle this (likely returning -1) or throw an error.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer Overflow: Not using a large enough data type (like `long long` in C++ or Java) for `total_chemistry` can lead to incorrect results due to overflow when multiplying the skill values, especially with the given constraints.

✗Common mistake 2: Forgetting to sort the array: The two-pointer approach relies on the array being sorted to guarantee finding matching pairs. Omitting the sorting step will likely lead to incorrect results or infinite loops.

✗Common mistake 3: Not checking for the invalid case (returning -1): Not correctly identifying and handling cases where it's impossible to form teams with equal skill sums will lead to incorrect outputs.

✅

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: Resource allocation: You have a set of resources with varying costs or benefits. You need to pair them up to achieve a target value for each pair, and maximize the overall efficiency (chemistry) of these pairings.

🏦Real-world scenario 2: Matching algorithms: In scenarios like online dating or job matching, you might want to pair individuals based on a target compatibility score and maximize the overall happiness (chemistry) of successful matches.

💡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: October 30, 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 13

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

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Divide Players Into Teams of Equal Skill"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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