What is the time complexity of Advantage Shuffle?

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

What data structures are used to solve Advantage Shuffle?

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

What are the key insights for solving Advantage Shuffle?

Insight 1: Sorting both arrays allows us to efficiently compare and strategically assign elements from `nums1` to maximize the advantage. Insight 2: Using a greedy approach, where we try to assign the smallest element in `nums1` that is greater than the current element in `nums2`, or the smallest remaining element if no such element exists, maximizes the advantage.

Which companies ask LeetCode 870. Advantage Shuffle in interviews?

This problem is frequently asked at Point72.

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Advantage Shuffle - LeetCode 870 Solution

Advantage Shuffle - Complete Solution Guide

Advantage Shuffle is LeetCode problem 870, 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 two integer arrays nums1 and nums2 both of the same length. The advantage of nums1 with respect to nums2 is the number of indices i for which nums1[i] > nums2[i] . Return any permutation of nums1 that maximizes its advantage with respect to nums2 . Example 1: Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11] Output: [2,11,7,15] Example 2: Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11] Output: [24,32,8,12] Constraints: 1 <= nums1.length <= 10 5 nums2.length == nums1.length 0 <= nu

Detailed Explanation

The problem asks us to find a permutation of `nums1` such that it maximizes its 'advantage' over `nums2`. The advantage is defined as the number of indices `i` where `nums1[i] > nums2[i]`. In simpler terms, we want to arrange `nums1` so that as many of its elements as possible are greater than the corresponding elements in `nums2`. Both arrays have the same length and contain non-negative integers.

Solution Approach

The provided solution uses a greedy approach combined with sorting to maximize the advantage. First, both `nums1` and `nums2` (or a representation of `nums2` that preserves its original indices) are sorted. Then, the algorithm iterates through the sorted `nums2` in reverse order. For each element in `nums2`, it tries to find the smallest element in `nums1` that is greater than it. If such an element exists, it's assigned to the corresponding index. Otherwise, the smallest element from `nums1` is assigned to that index. The core idea is to use bigger values from `nums1` when they can actually create an advantage, and use smaller values when they can't, so other bigger values can be used for better advantage.

Step-by-Step Algorithm

Step 1: Sort `nums1` in ascending order.
Step 2: Create an array of pairs from `nums2`, where each pair contains the value and its original index in `nums2`. Sort this array of pairs based on the values in ascending order.
Step 3: Initialize an empty result array of the same size as the input arrays.
Step 4: Iterate through the sorted `nums2` pairs in reverse order (from largest to smallest value).
Step 5: For each value in `nums2`, check if there is an element in `nums1` (that hasn't been assigned yet) that is greater than the current value in `nums2`.
Step 6: If such an element exists in `nums1`, assign it to the corresponding index in the result array and remove it from `nums1` (we use a deque for easy removal from both ends). We take the largest available value of nums1 that beats the current value of nums2.
Step 7: If no such element exists in `nums1`, assign the smallest remaining element from `nums1` to the corresponding index in the result array and remove it from `nums1`. The element from `nums1` is useless for this turn, so use the smallest available.
Step 8: Return the result array, which now contains the permutation of `nums1` that maximizes the advantage over `nums2`.

Key Insights

Insight 1: Sorting both arrays allows us to efficiently compare and strategically assign elements from `nums1` to maximize the advantage.
Insight 2: Using a greedy approach, where we try to assign the smallest element in `nums1` that is greater than the current element in `nums2`, or the smallest remaining element if no such element exists, maximizes the advantage.
Insight 3: The indices of the sorted `nums2` need to be tracked, so that after assigning a value to that index, the result can be placed in its original position.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Greedy, Sorting.

Companies

Asked at: Point72.

Advantage Shuffle | LeetCode 870

Medium

Array Two Pointers Greedy Sorting

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.Deque;
import java.util.LinkedList;

class Solution {
    public int[] advantageCount(int[] nums1, int[] nums2) {
        int n = nums1.length;
        
        Deque<Integer> sorted_nums1 = new LinkedList<>();
        Arrays.sort(nums1);
        for (int num : nums1) {
            sorted_nums1.add(num);
        }
        
        Integer[] sorted_nums2_indices = new Integer[n];
        for (int i = 0; i < n; i++) {
            sorted_nums2_indices[i] = i;
        }
        
        Arrays.sort(sorted_nums2_indices, (a, b) -> Integer.compare(nums2[a], nums2[b]));
        
        int[] result = new int[n];
        
        for (int i = n - 1; i >= 0; i--) {
            int original_index = sorted_nums2_indices[i];
            int val = nums2[original_index];
            
            if (sorted_nums1.getLast() > val) {
                result[original_index] = sorted_nums1.removeLast();
            } else {
                result[original_index] = sorted_nums1.removeFirst();
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 870: Advantage Shuffle - Detailed explanation and solution approaches.

Topics

Array Two Pointers Greedy Sorting

Companies

Point72

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:870

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

8 steps

1
Step 1: Sort `nums1` in ascending order.
2
Step 2: Create an array of pairs from `nums2`, where each pair contains the value and its original index in `nums2`. Sort this array of pairs based on the values in ascending order.
3
Step 3: Initialize an empty result array of the same size as the input arrays.
4
Step 4: Iterate through the sorted `nums2` pairs in reverse order (from largest to smallest value).
5
Step 5: For each value in `nums2`, check if there is an element in `nums1` (that hasn't been assigned yet) that is greater than the current value in `nums2`.
6
Step 6: If such an element exists in `nums1`, assign it to the corresponding index in the result array and remove it from `nums1` (we use a deque for easy removal from both ends). We take the largest available value of nums1 that beats the current value of nums2.
7
Step 7: If no such element exists in `nums1`, assign the smallest remaining element from `nums1` to the corresponding index in the result array and remove it from `nums1`. The element from `nums1` is useless for this turn, so use the smallest available.
8
Step 8: Return the result array, which now contains the permutation of `nums1` that maximizes the advantage over `nums2`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the sorting operations (both `nums1` and `nums2`), which take O(n log n) time. The iteration through the sorted arrays takes O(n) time. Thus, the overall time complexity is O(n log n). The space complexity is O(n) because we create a new array `result` of size n, and in the Python and Java solutions, we create an auxiliary array to store sorted indices and values. The C solution also allocates memory for the sorted pairs and return array, contributing to O(n) space complexity. The deque also takes O(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

Binary Search

Instead of using a deque, after sorting `nums1`, we could iterate through sorted `nums2` and perform binary search on `nums1` to find the smallest element greater than the current element in `nums2`. This avoids the need for a deque.

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

Trade-offs: Binary search might have slightly higher constant factors compared to using a deque for removing elements from both ends. The overall asymptotic complexity remains the same.

✨

Key Insights

1Insight 1: Sorting both arrays allows us to efficiently compare and strategically assign elements from `nums1` to maximize the advantage.
2Insight 2: Using a greedy approach, where we try to assign the smallest element in `nums1` that is greater than the current element in `nums2`, or the smallest remaining element if no such element exists, maximizes the advantage.
3Insight 3: The indices of the sorted `nums2` need to be tracked, so that after assigning a value to that index, the result can be placed in its original position.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

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: `nums1` and `nums2` are already sorted in ascending order. The solution will still produce a valid result, as it compares the largest elements and assigns them based on the advantage criteria.
Edge case 2: All elements in `nums1` are smaller than all elements in `nums2`. The solution will assign the smallest elements from `nums1` to each index, resulting in zero advantage, which is the optimal result in this scenario.
Edge case 3: `nums1` and `nums2` contain duplicate elements. The sorting and greedy approach will still work correctly, as it only relies on the relative order of elements and not their uniqueness.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not preserving the original indices of `nums2` after sorting. This will result in incorrect assignment of elements to the final result array.

✗Common mistake 2: Failing to consider the case where no element in `nums1` is greater than the current element in `nums2`. In this case, the smallest element from `nums1` should be used to avoid wasting larger elements unnecessarily.

✗Common mistake 3: Modifying `nums1` directly instead of using a copy or a deque. This can lead to incorrect results if `nums1` is needed for other purposes.

✗Common mistake 4: Not realizing that sorting both arrays enables a greedy strategy that leads to the optimal solution.

✅

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. Imagine `nums1` representing the capabilities of different machines and `nums2` representing the requirements of different tasks. The advantage shuffle can be used to assign machines to tasks in a way that maximizes the number of tasks that can be successfully completed.

🏦Real-world scenario 2: Competitive gaming. Consider a scenario where `nums1` represents the skill levels of your team members and `nums2` represents the skill levels of the opposing team members. The advantage shuffle can be used to strategically assign your team members to matchups against the opposing team members in a way that maximizes the number of wins.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 29, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

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

@AlgoMasterJul 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.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

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 "Advantage Shuffle"?

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

🤖 Need help with this problem?