How do you solve Fair Candy Swap on LeetCode?

Fair Candy Swap (LeetCode #888) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the total difference in candies helps to find the required swap efficiently.

What is the time complexity of Fair Candy Swap?

The optimal solution for Fair Candy Swap runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only loop through Alice's sizes once and use a set for fast lookups.

What is the brute force solution for Fair Candy Swap?

The brute force approach for Fair Candy Swap is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of candy boxes from Alice and Bob to see if swapping them results in equal total candies. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Fair Candy Swap?

Fair Candy Swap uses the following concepts: Array, Hash Table, Binary Search, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Fair Candy Swap?

Always start with a brute force solution to understand the problem better. Look for patterns in the problem that can simplify your approach. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Fair Candy Swap?

Not considering the mathematical relationship between the sums, leading to unnecessary iterations. Forgetting to check if the required candy size exists in Bob's set.

#888

Fair Candy Swap

Easy

ArrayHash TableBinary SearchSortingHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses a mathematical approach to find the required swap without checking every pair. We calculate the difference in total candies and use a set for quick lookups.

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Algorithm

4 steps

1Step 1: Calculate the total number of candies for Alice and Bob.
2Step 2: Calculate the difference between the two totals and divide by 2.
3Step 3: Store all of Bob's candy sizes in a set for O(1) lookups.
4Step 4: For each box in Alice's sizes, check if there exists a box in Bob's sizes that satisfies the swapping condition.

solution.py8 lines

1def fairCandySwap(aliceSizes, bobSizes):
2    totalAlice = sum(aliceSizes)
3    totalBob = sum(bobSizes)
4    diff = (totalAlice - totalBob) // 2
5    bobSet = set(bobSizes)
6    for a in aliceSizes:
7        if a - diff in bobSet:
8            return [a, a - diff]

ℹ

Complexity note: The time complexity is O(n) because we only loop through Alice's sizes once and use a set for fast lookups.

1Understanding the total difference in candies helps to find the required swap efficiently.
2Using a set allows for quick lookups, reducing the need for nested loops.

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