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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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int* fairCandySwap(int* aliceSizes, int alen, int* bobSizes, int blen,
The C code initializes the required sums for Alice and Bob. Then it computes the difference in candies needed (delta) and uses an array to act as a set for Bob's candy sizes. Finally, it attempts to find a suitable swap by iterating over Alice's candy sizes and checking if Bob has the necessary candy size using the array.
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.
#include <algorithm>
std::vector<int> fairCandySwap(std::vector<int>& aliceSizes, std::vector<int>& bobSizes) {
std::sort(aliceSizes.begin(), aliceSizes.end());
std::sort(bobSizes.begin(), bobSizes.end());
int sumA = 0, sumB = 0;
for (int x : aliceSizes) sumA += x;
for (int y : bobSizes) sumB += y;
int delta = (sumB - sumA) / 2;
int i = 0, j = 0;
while (i < aliceSizes.size() && j < bobSizes.size()) {
int diff = bobSizes[j] - aliceSizes[i];
if (diff == delta) {
return {aliceSizes[i], bobSizes[j]};
} else if (diff < delta) {
j++;
} else {
i++;
}
}
return {}; // This line should not be reached
}In this C++ solution, after sorting both lists, two pointers traverse each list. They aim to find when the difference between elements pointed at is equal to delta. The current pointer advances in Bob's list if the difference is too small, or in Alice's list if too large.