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39 changes: 39 additions & 0 deletions greedy_methods/rearranging_fruits.py
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# it's a leetcode question no. 2561 where You have two fruit baskets containing n fruits each. You are given two 0-indexed integer arrays basket1 and basket2 representing the cost of fruit in each basket.

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# You want to make both baskets equal. To do so, you can use the following operation as many times as you want:

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# Chose two indices i and j, and swap the ith fruit of basket1 with the jth fruit of basket2.

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# The cost of the swap is min(basket1[i],basket2[j]).
# Two baskets are considered equal if sorting them according to the fruit cost makes them exactly the same baskets.

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# Return the minimum cost to make both the baskets equal or -1 if impossible.

from typing import List

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from collections import defaultdict

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class Solution:
def minCost(self, basket1: List[int], basket2: List[int]) -> int:

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As there is no test file in this pull request nor any test function or class in the file greedy_methods/rearranging_fruits.py, please provide doctest for the function minCost

Variable and function names should follow the snake_case naming convention. Please update the following name accordingly: minCost

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As there is no test file in this pull request nor any test function or class in the file greedy_methods/rearranging_fruits.py, please provide doctest for the function minCost

Variable and function names should follow the snake_case naming convention. Please update the following name accordingly: minCost

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Variable and function names should follow the snake_case naming convention. Please update the following name accordingly: minCost

As there is no test file in this pull request nor any test function or class in the file greedy_methods/rearranging_fruits.py, please provide doctest for the function minCost

n = len(basket1)
freq = defaultdict(int)
mn = float("inf")
for i in range(n):
freq[basket1[i]] += 1
freq[basket2[i]] -= 1
mn = min(mn, basket1[i], basket2[i])

to_swap = []
for j, k in freq.items():
if k % 2 != 0:
return -1
to_swap += [j] * (abs(k) // 2)

to_swap.sort()
res = 0
for i in range(len(to_swap) // 2):
res += min(to_swap[i], 2 * mn)

return res


s = Solution()
print(s.minCost([4, 2, 2, 2], [1, 4, 1, 2])) # Output: 1
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