Solution: Rearranging Fruits

Explore how to apply greedy techniques to rearrange fruits in two baskets at minimal cost. This lesson guides you through counting frequencies, identifying excess elements, and strategically swapping fruits to balance the baskets efficiently. You will understand the algorithm's complexity and implement a solution that ensures minimal swap expenses.

We'll cover the following...

Statement
Solution
- Time complexity
- Space complexity

Statement

Given two 0-indexed integer arrays, basket1 and basket2, representing the cost of each fruit in the basket. Each basket contains $n$ fruits. Make the two baskets identical, i.e., both arrays should have the same costs.

To achieve this, perform the following operation as many times as necessary:

Select two indexes, $i$ and $j$ , and swap the fruit at index $i$ in basket1 with the fruit at index $j$ in basket2.
The cost of this swap is min(basket1[i], basket2[j]).

The two baskets are considered identical if, after sorting the fruits by cost, both baskets contain exactly the same costs.

Return the minimum cost required to make the baskets identical, or $-1$ ...