888. Fair Candy Swap LeetCode Solution

In this guide, you will get 888. Fair Candy Swap LeetCode Solution with the best time and space complexity. The solution to Fair Candy Swap problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Fair Candy Swap solution in C++
Fair Candy Swap solution in Java
Fair Candy Swap solution in Python
Additional Resources

888. Fair Candy Swap LeetCode Solution image

Problem Statement of Fair Candy Swap

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 ith box of candy that Alice has and bobSizes[j] is the number of candies of the jth 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 in each box they have.
Return an integer array answer where answer[0] is the number of candies in the box that Alice must exchange, and answer[1] is the number of candies in the box that Bob must exchange. If there are multiple answers, you may return any one of them. It is guaranteed that at least one answer exists.

Example 1:

Input: aliceSizes = [1,1], bobSizes = [2,2]
Output: [1,2]

Example 2:

Input: aliceSizes = [1,2], bobSizes = [2,3]
Output: [1,2]

Example 3:

Input: aliceSizes = [2], bobSizes = [1,3]
Output: [2,3]

Constraints:

1 <= aliceSizes.length, bobSizes.length <= 104
1 <= aliceSizes[i], bobSizes[j] <= 105
Alice and Bob have a different total number of candies.
There will be at least one valid answer for the given input.

Complexity Analysis

Time Complexity:
Space Complexity:

888. Fair Candy Swap LeetCode Solution in C++

class Solution {
 public:
  vector<int> fairCandySwap(vector<int>& aliceSizes, vector<int>& bobSizes) {
    const int diff = (accumulate(aliceSizes.begin(), aliceSizes.end(), 0) -
                      accumulate(bobSizes.begin(), bobSizes.end(), 0)) /
;
    const unordered_set<int> bobSizesSet{bobSizes.begin(), bobSizes.end()};

    for (const int aliceSize : aliceSizes) {
      const int target = aliceSize - diff;
      if (bobSizesSet.contains(target))
        return {aliceSize, target};
    }

    throw;
  }
};
/* code provided by PROGIEZ */

888. Fair Candy Swap LeetCode Solution in Java

class Solution {
  public int[] fairCandySwap(int[] aliceSizes, int[] bobSizes) {
    final int diff = (Arrays.stream(aliceSizes).sum() - Arrays.stream(bobSizes).sum()) / 2;
    Set<Integer> bobSizesSet = Arrays.stream(bobSizes).boxed().collect(Collectors.toSet());

    for (final int aliceSize : aliceSizes) {
      final int target = aliceSize - diff;
      if (bobSizesSet.contains(target))
        return new int[] {aliceSize, target};
    }

    throw new IllegalArgumentException();
  }
}
// code provided by PROGIEZ

888. Fair Candy Swap LeetCode Solution in Python

class Solution:
  def fairCandySwap(
      self,
      aliceSizes: list[int],
      bobSizes: list[int],
  ) -> list[int]:
    diff = (sum(aliceSizes) - sum(bobSizes)) // 2
    bobSizesSet = set(bobSizes)

    for aliceSize in aliceSizes:
      target = aliceSize - diff
      if target in bobSizesSet:
        return [aliceSize, target]
 # code by PROGIEZ