How do you solve Rearranging Fruits on LeetCode?

Rearranging Fruits (LeetCode #2561) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Both baskets must have the same total count of each fruit type for them to be rearranged into identical baskets.

What is the brute force solution for Rearranging Fruits?

The brute force approach for Rearranging Fruits is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible swaps between the two baskets to see if we can make them equal. This is straightforward but inefficient since it examines every possible combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Rearranging Fruits?

Rearranging Fruits uses the following concepts: Array, Hash Table, Greedy, Sort. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Rearranging Fruits?

Always consider edge cases, such as when baskets already match or when they cannot be made equal. Explain your thought process clearly; this shows your understanding and can earn you points even if your solution isn't optimal. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Rearranging Fruits?

Failing to check if the total count of any fruit type is odd, which makes it impossible to balance. Not using frequency maps, leading to inefficient solutions.

#2561

Rearranging Fruits

Hard

ArrayHash TableGreedySortHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses frequency maps to count occurrences of each fruit cost in both baskets. This allows us to determine if it's possible to make the baskets equal and calculate the minimum cost efficiently.

⚙️

Algorithm

4 steps

1Step 1: Create frequency maps for both baskets to count occurrences of each fruit cost.
2Step 2: Check if the sum of frequencies for each unique fruit cost is odd; if so, return -1 (impossible to balance).
3Step 3: Calculate the minimum fruit cost from both baskets to use in swap calculations.
4Step 4: For each unique fruit cost, calculate the total number of swaps needed and the associated costs.

solution.py15 lines

1from collections import Counter
2
3def minCost(basket1, basket2):
4    count1 = Counter(basket1)
5    count2 = Counter(basket2)
6    total_cost = 0
7    min_fruit = min(min(basket1), min(basket2))
8
9    for fruit in set(count1.keys()).union(set(count2.keys())):
10        total_count = count1[fruit] + count2[fruit]
11        if total_count % 2 != 0:
12            return -1
13        total_cost += abs(count1[fruit] - count2[fruit]) // 2 * min_fruit
14
15    return total_cost

ℹ

Complexity note: This complexity is efficient because we only traverse each basket a limited number of times to create frequency maps and calculate costs.

1Both baskets must have the same total count of each fruit type for them to be rearranged into identical baskets.
2The minimum fruit cost is crucial for calculating the cost of swaps effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.