Problem Description

Alice and Bob each have multiple boxes of candies, where each box contains a certain number of candies. The problem gives us two arrays: aliceSizes and bobSizes, where aliceSizes[i] represents the number of candies in Alice's i-th box, and bobSizes[j] represents the number of candies in Bob's j-th box.

Currently, Alice and Bob have different total amounts of candy. Since they are friends, they want to exchange exactly one box of candy each so that after the exchange, both of them have the same total amount of candy.

The task is to find which boxes they should exchange. You need to return an array answer where:

• answer[0] is the number of candies in the box that Alice gives to Bob
• answer[1] is the number of candies in the box that Bob gives to Alice

The problem guarantees that at least one valid exchange exists, and if there are multiple valid exchanges, you can return any one of them.

For example, if Alice has boxes with [1, 2] candies (total = 3) and Bob has boxes with [1, 3] candies (total = 4), they could exchange Alice's box of 1 candy with Bob's box of 2 candies. After the exchange, Alice would have [2, 2] (total = 4) and Bob would have [1, 1] (total = 2). Wait, that doesn't work! They should exchange Alice's box of 1 candy with Bob's box of 2 candies... Actually, let me recalculate: if Alice gives away 1 and receives 2, her total becomes 3 - 1 + 2 = 4. Bob gives away 2 and receives 1, his total becomes 4 - 2 + 1 = 3. They still don't match. The correct exchange would be Alice's box of 2 candies for Bob's box of 3 candies, making both totals equal to 3.5... which shows this specific example needs different values to work properly.

The key insight is that after the exchange, the sum of candies for both people must be equal, which means each person should have half of the total candies between them.

Intuition

Let's think about what happens when Alice and Bob exchange boxes. If Alice gives a box with a candies and receives a box with b candies from Bob, then:

• Alice's new total = sum(aliceSizes) - a + b
• Bob's new total = sum(bobSizes) - b + a

For them to have equal amounts after the exchange, we need: sum(aliceSizes) - a + b = sum(bobSizes) - b + a

Rearranging this equation: sum(aliceSizes) - sum(bobSizes) = 2a - 2b sum(aliceSizes) - sum(bobSizes) = 2(a - b) a - b = (sum(aliceSizes) - sum(bobSizes)) / 2

This tells us that the difference between the candy count Alice gives and the candy count Bob gives must be exactly half of the difference between their total candies. Let's call this value diff = (sum(aliceSizes) - sum(bobSizes)) / 2.

So for any box Alice chooses to give with a candies, Bob must give back a box with b = a - diff candies. This transforms our problem from checking all possible pairs (which would be O(n*m)) to a simpler lookup problem.

The strategy becomes:

1. Calculate diff once at the beginning
2. Store all of Bob's candy counts in a hash set for O(1) lookup
3. For each of Alice's boxes with a candies, check if Bob has a box with a - diff candies
4. If such a box exists in Bob's collection, we've found our answer: Alice gives a, Bob gives a - diff

This approach is efficient because we only need to iterate through Alice's boxes once, and for each box, we can check in constant time whether Bob has the corresponding box that would make the exchange fair.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The implementation follows the mathematical insight we derived, using a hash table for efficient lookup:

1. Calculate the difference: First, we compute the total candies for both Alice and Bob, find their difference, and divide by 2 to get diff. In the code, this is done with bit shifting: diff = (sum(aliceSizes) - sum(bobSizes)) >> 1, where >> 1 is equivalent to dividing by 2.

2. Create a hash set: We convert Bob's candy sizes into a set s = set(bobSizes). This allows us to check if Bob has a specific candy box size in O(1) time.

3. Find the valid exchange: We iterate through each of Alice's candy boxes. For each box with a candies, we calculate what Bob needs to give back: b = a - diff.

4. Check for validity: Using the walrus operator := for inline assignment, the code checks if b = (a - diff) exists in Bob's set s. If it does, we immediately return [a, b] as our answer.

The algorithm is guaranteed to find an answer because the problem states at least one valid exchange exists. The time complexity is O(n + m) where n and m are the lengths of aliceSizes and bobSizes respectively - we need O(m) to build the hash set and O(n) to iterate through Alice's boxes. The space complexity is O(m) for storing Bob's candy sizes in the set.

The beauty of this solution is that it transforms what could be a brute force O(n*m) nested loop problem into a linear time solution by leveraging the mathematical relationship between the exchanged values and using a hash table for constant-time lookups.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example Input:

• Alice has boxes: [1, 5]
• Bob has boxes: [2, 4]

Step 1: Calculate totals and difference

• Alice's total: 1 + 5 = 6 candies
• Bob's total: 2 + 4 = 6 candies
• Wait, they already have equal amounts! Let's use a different example.

Better Example Input:

• Alice has boxes: [1, 1]
• Bob has boxes: [2, 3]

Step 1: Calculate totals and difference

• Alice's total: 1 + 1 = 2 candies
• Bob's total: 2 + 3 = 5 candies
• Total combined: 2 + 5 = 7 candies
• After exchange, each should have: 7 ÷ 2 = 3.5 candies
• Difference calculation: diff = (2 - 5) ÷ 2 = -3 ÷ 2 = -1.5

Hmm, we get a non-integer difference. Let me try another example.

Working Example Input:

• Alice has boxes: [1, 2]
• Bob has boxes: [1, 3]

Step 1: Calculate totals and difference

• Alice's total: 1 + 2 = 3 candies
• Bob's total: 1 + 3 = 4 candies
• Total combined: 3 + 4 = 7 candies
• After exchange, each should have: 7 ÷ 2 = 3.5 candies
• Still non-integer! Let's use an example that works cleanly.

Final Working Example:

• Alice has boxes: [2, 5]
• Bob has boxes: [1, 4]

Step 1: Calculate totals and difference

• Alice's total: 2 + 5 = 7 candies
• Bob's total: 1 + 4 = 5 candies
• Difference: diff = (7 - 5) ÷ 2 = 2 ÷ 2 = 1

Step 2: Create hash set from Bob's boxes

• Bob's set: {1, 4}

Step 3: Find valid exchange

• Check Alice's first box (2 candies):
  • Bob needs: 2 - 1 = 1 candies
  • Is 1 in Bob's set? Yes! ✓
  • Valid exchange found: Alice gives 2, Bob gives 1

Step 4: Verify the exchange

• After exchange:
  • Alice: 7 - 2 + 1 = 6 candies
  • Bob: 5 - 1 + 2 = 6 candies
  • Both have 6 candies - Success!

The algorithm returns [2, 1] as the answer. Alice exchanges her box of 2 candies for Bob's box of 1 candy, resulting in both having equal total candies.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m + n)

The algorithm consists of:

• Computing the sum of aliceSizes: O(m) where m is the length of aliceSizes
• Computing the sum of bobSizes: O(n) where n is the length of bobSizes
• Creating a set from bobSizes: O(n)
• Iterating through aliceSizes in the worst case: O(m), with each iteration performing:
  • A constant time arithmetic operation a - diff
  • A constant time set lookup operation in s: O(1) on average

Total time complexity: O(m) + O(n) + O(n) + O(m) = O(m + n)

Space Complexity: O(n)

The algorithm uses:

• A set s to store all elements from bobSizes: O(n) space where n is the length of bobSizes
• A few variables (diff, a, b) that use constant space: O(1)

Total space complexity: O(n) + O(1) = O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Division vs Float Division

One common mistake is using float division (/) instead of integer division (//) when calculating the difference. Since we're dealing with candy counts (integers), the difference must be an integer for a valid exchange to exist.

Incorrect:

difference = (sum(aliceSizes) - sum(bobSizes)) / 2  # Returns float

Correct:

difference = (sum(aliceSizes) - sum(bobSizes)) // 2  # Returns integer

2. Confusing the Sign of the Difference

A critical pitfall is misunderstanding what the difference represents and applying it incorrectly. The formula bob_candy = alice_candy - difference might seem counterintuitive at first.

Why this works:

• If Alice has MORE total candies initially, difference is positive
• Alice needs to give away more than she receives
• So Bob's candy must be smaller: bob_candy = alice_candy - positive_difference

Common mistake:

# Wrong: Adding instead of subtracting
bob_candy = alice_candy + difference

3. Not Using a Set for Lookups

Using a list to check if Bob has the required candy size leads to O(m) lookup time per iteration, resulting in O(n*m) overall complexity.

Inefficient:

for alice_candy in aliceSizes:
    bob_candy = alice_candy - difference
    if bob_candy in bobSizes:  # O(m) lookup in list
        return [alice_candy, bob_candy]

Efficient:

bob_sizes_set = set(bobSizes)  # Convert to set first
for alice_candy in aliceSizes:
    bob_candy = alice_candy - difference
    if bob_candy in bob_sizes_set:  # O(1) lookup in set
        return [alice_candy, bob_candy]

4. Forgetting Edge Cases with Negative Values

While the problem typically involves positive candy counts, the calculated bob_candy value could theoretically be zero or negative if the difference is large. Although the problem guarantees a valid solution exists, it's good practice to consider this:

More robust approach:

for alice_candy in aliceSizes:
    bob_candy = alice_candy - difference
    if bob_candy > 0 and bob_candy in bob_sizes_set:
        return [alice_candy, bob_candy]