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Alice and Bob have a different total number of candies. To be fair, each of them should exchange one of their own candies so that each person ends up with the same total amount of candies. The problem is to determine a pair of candies that they can swap to achieve this fair distribution.

Given two integer arrays A and B where A[i] and B[i] represent the size of the i-th candy from Alice’s and Bob’s collection respectively, return an integer array [x, y] such that:

If there are multiple answers, you may return any one of them.

The solution should fit in O(N) time complexity, where N is the size of the input arrays.

Clarifying Questions

  1. Input Size: Is there a limit on the size of the arrays A and B?
  2. Elements in Arrays: Can A and B have negative values or zeroes?
  3. Multiple Solutions: If there are multiple valid pairs, is any pair permissible to return?
  4. Empty Arrays: Can there be cases where A or B are empty?

Strategy

The key to solving this problem is to use the fact that the sum of candies in Alice’s collection minus the sum of candies in Bob’s collection is a key factor in finding the fair swap.

Steps to Approach:

  1. Compute the total sum of candies for Alice (sumA) and Bob (sumB).

  2. Determine the difference delta such that if Alice gives away x and receives y, it will balance out the total candies for both:

    [ \text{sumA} - x + y = \text{sumB} - y + x ]

    From this, we derive:

    [ 2(y - x) = \text{sumB} - \text{sumA} ]

    Therefore,

    [ y = x + \left(\frac{\text{sumB} - \text{sumA}}{2}\right) ]

  3. Use a set to store Bob’s candy sizes for quick lookup.
  4. Iterate through Alice’s candy sizes and for each candy x, compute y and check if y exists in Bob’s set.

Code Implementation:

def fairCandySwap(A, B):
    sumA = sum(A)
    sumB = sum(B)
    delta = (sumB - sumA) // 2
    setB = set(B)
    
    for x in A:
        y = x + delta
        if y in setB:
            return [x, y]

Time Complexity:

  1. Calculating the sums of A and B takes O(N + M), where N and M are the lengths of A and B respectively.
  2. Creating the set from B takes O(M).
  3. Iterating through A and checking membership in the set takes O(N).

Thus, the total time complexity is O(N + M).

This approach satisfies the time complexity requirement and ensures a valid solution for the problem.

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