What is the time complexity of Fair Candy Swap?

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

What data structures are used to solve Fair Candy Swap?

This problem uses Array, Hash Table, Binary Search, Sorting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Fair Candy Swap?

Insight 1: The core idea is to determine the difference in the total candy amounts and find the values needed to equalize them. We need to find an `x` in Alice's set and `y` in Bob's set such that `sum_alice - x + y = sum_bob - y + x`. Insight 2: The equation can be simplified to `y = (sum_bob - sum_alice + 2 * x) / 2`. This formula is crucial for finding the required candy value from Bob given a specific candy value from Alice.

Which companies ask LeetCode 888. Fair Candy Swap in interviews?

This problem is frequently asked at Odoo, Swiggy.

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Fair Candy Swap - LeetCode 888 Solution

Fair Candy Swap - Complete Solution Guide

Fair Candy Swap is LeetCode problem 888, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Alice and Bob have a different total number of candies. You are given two integer arrays aliceSizes and bobSizes where aliceSizes[i] is the number of candies of the i th box of candy that Alice has and bobSizes[j] is the number of candies of the j th box of candy that Bob has. Since they are friends, they would like to exchange one candy box each so that after the exchange, they both have the same total amount of candy. The total amount of candy a person has is the sum of the number of candies i

Detailed Explanation

The Fair Candy Swap problem requires finding one candy box each for Alice and Bob to exchange, such that after the exchange, both have the same total amount of candy. You are given two arrays, `aliceSizes` and `bobSizes`, representing the sizes of candy boxes Alice and Bob possess, respectively. The goal is to return an array `[x, y]`, where `x` is the size of the candy box Alice gives to Bob, and `y` is the size of the candy box Bob gives to Alice.

Solution Approach

The solution involves calculating the sum of candies for Alice and Bob, then iterating through Alice's candy sizes and using the formula `y = (sum_bob - sum_alice + 2 * x) / 2` to find the corresponding candy size Bob must exchange. A hash set stores Bob's candy sizes, which allows for O(1) average time complexity lookups to determine if `y` exists within Bob's collection. If the value exists in Bob's set, the pair `[x, y]` is the answer, and the algorithm terminates.

Step-by-Step Algorithm

Step 1: Calculate the sum of candy sizes for Alice (`sum_alice`) and Bob (`sum_bob`).
Step 2: Create a hash set (`set_bob`) containing all the candy sizes in Bob's collection. This allows for efficient O(1) lookup.
Step 3: Iterate through each candy size `x` in Alice's collection (`aliceSizes`).
Step 4: For each `x`, calculate the required candy size `y` from Bob using the formula `y = (sum_bob - sum_alice + 2 * x) // 2` (integer division is crucial here!).
Step 5: Check if `y` exists in the `set_bob`. If it does, then `x` and `y` are the solution.
Step 6: If `y` is found in `set_bob`, return the array `[x, y]`. The problem states a solution is guaranteed so no further iteration is needed.
Step 7: The C solution is different. It uses a nested for loop, resulting in a higher Time Complexity of O(n*m).

Key Insights

Insight 1: The core idea is to determine the difference in the total candy amounts and find the values needed to equalize them. We need to find an `x` in Alice's set and `y` in Bob's set such that `sum_alice - x + y = sum_bob - y + x`.
Insight 2: The equation can be simplified to `y = (sum_bob - sum_alice + 2 * x) / 2`. This formula is crucial for finding the required candy value from Bob given a specific candy value from Alice.
Insight 3: Using a set (or hash table) to store Bob's candy sizes enables efficient lookup, which improves the algorithm's performance from O(n*m) to O(n+m).
Insight 4: The problem statement guarantees that at least one solution exists. Therefore, once a valid pair (x, y) is found, we can return it immediately.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(m)

Topics

This problem involves: Array, Hash Table, Binary Search, Sorting.

Companies

Asked at: Odoo, Swiggy.

Fair Candy Swap | LeetCode 888

Easy

Array Hash Table Binary Search Sorting

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

Solution Code

Available in:

class Solution {
    public int[] fairCandySwap(int[] aliceSizes, int[] bobSizes) {
        int sumAlice = 0;
        int sumBob = 0;
        for (int size : aliceSizes) {
            sumAlice += size;
        }
        for (int size : bobSizes) {
            sumBob += size;
        }
        Set<Integer> setBob = new HashSet<>();
        for (int size : bobSizes) {
            setBob.add(size);
        }
        for (int size : aliceSizes) {
            int target = (sumAlice - size + sumBob - size) / 2;
            if (setBob.contains(target)) {
                return new int[]{size, target};
            }
        }
        return new int[]{};
    }
}

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Problem Info

Description

Problem 888: Fair Candy Swap - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Sorting

Companies

OdooSwiggy

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:888

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

7 steps

1
Step 1: Calculate the sum of candy sizes for Alice (`sum_alice`) and Bob (`sum_bob`).
2
Step 2: Create a hash set (`set_bob`) containing all the candy sizes in Bob's collection. This allows for efficient O(1) lookup.
3
Step 3: Iterate through each candy size `x` in Alice's collection (`aliceSizes`).
4
Step 4: For each `x`, calculate the required candy size `y` from Bob using the formula `y = (sum_bob - sum_alice + 2 * x) // 2` (integer division is crucial here!).
5
Step 5: Check if `y` exists in the `set_bob`. If it does, then `x` and `y` are the solution.
6
Step 6: If `y` is found in `set_bob`, return the array `[x, y]`. The problem states a solution is guaranteed so no further iteration is needed.
7
Step 7: The C solution is different. It uses a nested for loop, resulting in a higher Time Complexity of O(n*m).

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(m)

📝

Why?

The time complexity is determined by the dominant operations: calculating the sums of the arrays and searching for the complementary value in the hash set. The space complexity is mainly due to the space used by the hash set to store Bob's candy sizes.

📖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

Sorting and Two Pointers

Sort both `aliceSizes` and `bobSizes`. Then, use two pointers to find the pair (x, y) such that `sum_alice - x + y = sum_bob - y + x`.

Time: O(n log n + m log m)Space: O(1) if sorting in place, O(n) or O(m) otherwise

Trade-offs: This approach avoids using a hash set, but the sorting step makes it slower than the main solution (especially if n and m are large). It can be useful when memory is extremely limited.

✨

Key Insights

1Insight 1: The core idea is to determine the difference in the total candy amounts and find the values needed to equalize them. We need to find an `x` in Alice's set and `y` in Bob's set such that `sum_alice - x + y = sum_bob - y + x`.
2Insight 2: The equation can be simplified to `y = (sum_bob - sum_alice + 2 * x) / 2`. This formula is crucial for finding the required candy value from Bob given a specific candy value from Alice.
3Insight 3: Using a set (or hash table) to store Bob's candy sizes enables efficient lookup, which improves the algorithm's performance from O(n*m) to O(n+m).
4Insight 4: The problem statement guarantees that at least one solution exists. Therefore, once a valid pair (x, y) is found, we can return it immediately.

📚

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!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

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

Edge Cases

Edge case 1: Small input arrays (e.g., `aliceSizes = [1, 1]`, `bobSizes = [2, 2]`). The algorithm handles these cases correctly since it iterates through all possible values in `aliceSizes` and searches for the corresponding value in `bobSizes`.
Edge case 2: Large candy values (e.g., up to 10^5). The integer data type used in the code is sufficient to store these values without overflow issues (assuming 32-bit integers or larger). However, if there were extremely large input values, the code may need to be adapted to use larger data types (e.g., long).
Edge case 3: Duplicate values within `aliceSizes` or `bobSizes`. The algorithm handles duplicates correctly because it iterates through all values and checks if a corresponding value exists in Bob's `set_bob`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using integer division when calculating `y = (sum_bob - sum_alice + 2 * x) / 2`. If floating-point division is used, the result may not be an integer, leading to incorrect results.

✗Common mistake 2: Not using a set or hash table to store Bob's candy sizes. Using nested loops to search for the complementary value in `bobSizes` will result in O(n*m) time complexity, which may cause time limit exceeded errors for larger inputs.

✗Common mistake 3: Incorrectly calculating the target 'y' value. Make sure to derive the correct formula. A common mistake is to forget multiplying x by 2.

✗Common mistake 4: In the C code the line `int diff = (sumBob - sumAlice) / 2;` doesn't account for the possibility of a negative 'diff' value. This works for this particular problem, but is bad practice overall.

✅

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 two departments in a company need to exchange resources (e.g., server time, budget) to balance their workload. This algorithm can help determine the optimal exchange amount for each department.

🏦Real-world scenario 2: Fair distribution of goods: If two groups of people have unequal amounts of a resource (e.g., food, supplies), this algorithm can determine the amount one group needs to give to the other to achieve a more equitable distribution.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: November 16, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

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 are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Fair Candy Swap"?

For this easy-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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