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In this approach, we first calculate the total number of candies Alice and Bob have, respectively. Let this be sumA for Alice and sumB for Bob. If Alice swaps a candy box with x candies and Bob swaps a box with y candies, the condition to balance their total candies after swap can be derived as: sumA - x + y = sumB - y + x. Simplifying this, we get y = x + (sumB - sumA) / 2. Using this equation, we can loop through one person's candy sizes array and for each candy size, compute the necessary size from the other person's array and check if it exists using a set for quick lookup.
Time Complexity: O(n + m), where n is the length of aliceSizes and m is the length of bobSizes. This is because we iterate over each list once and use constant time checks for the existence of an element in Bob's collection.
Space Complexity: O(m), due to the additional space used for storing Bob's candy sizes in a set.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public int[] FairCandySwap(int[] aliceSizes, int[] bobSizes) {
6 int sumA = 0, sumB = 0;
7 foreach (int x in aliceSizes) sumA += x;
8 foreach (int y in bobSizes) sumB += y;
9 int delta = (sumB - sumA) / 2;
10 HashSet<int> setB = new HashSet<int>(bobSizes);
11 foreach (int x in aliceSizes) {
12 if (setB.Contains(x + delta)) {
13 return new int[]{x, x + delta};
}
}
return null; // This line should not be reached
}
}The C# solution uses the HashSet collection to efficiently manage Bob's candy sizes. After calculating the balance required (delta), Alice's candy sizes are checked to see if a valid exchange can be made using these calculations.
This approach involves first sorting both the candy size arrays of Alice and Bob. The idea is to use a two-pointer technique to find a suitable pair of candy boxes such that swapping them would balance Alice's and Bob's total candies. Once both lists are sorted, use two pointers, starting at the beginnings of Alice's and Bob's lists, and adjust them based on the comparison of their sums to find the correct swap.
Time Complexity: O(n log n + m log m), due to the sorting step. The subsequent linear scan is O(n + m).
Space Complexity: O(1), with only a few additional variables used.
After sorting the candy sizes, this Python solution makes use of a single pass with two pointers to find the valid swap. The sort guarantees that once a valid swap is found, it is also optimal due to sorted order ensuring smaller and larger elements follow a linear order.