LeetCode #2561 — HARD

Rearranging Fruits

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Choose two indices i and j, and swap the i^th fruit of basket1 with the j^th fruit of basket2.
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.

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

Example 1:

Input: basket1 = [4,2,2,2], basket2 = [1,4,1,2]
Output: 1
Explanation: Swap index 1 of basket1 with index 0 of basket2, which has cost 1. Now basket1 = [4,1,2,2] and basket2 = [2,4,1,2]. Rearranging both the arrays makes them equal.

Example 2:

Input: basket1 = [2,3,4,1], basket2 = [3,2,5,1]
Output: -1
Explanation: It can be shown that it is impossible to make both the baskets equal.

Constraints:

basket1.length == basket2.length
1 <= basket1.length <= 10⁵
1 <= basket1[i], basket2[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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. You want to make both baskets equal. To do so, you can use the following operation as many times as you want: Choose two indices i and j, and swap the ith fruit of basket1 with the jth fruit of basket2. 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. Return the minimum cost to make both the baskets equal or -1 if impossible.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Greedy

Example 1

[4,2,2,2]
[1,4,1,2]

Example 2

[2,3,4,1]
[3,2,5,1]

Core Insight

What unlocks the optimal approach

Create two frequency maps for both arrays, and find the minimum element among all elements of both arrays.
Check if the sum of frequencies of an element in both arrays is odd, if so return -1
Store the elements that need to be swapped in a vector, and sort it.
Can we reduce swapping cost with the help of minimum element?
Calculate the minimum cost of swapping.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2561: Rearranging Fruits
class Solution {
    public long minCost(int[] basket1, int[] basket2) {
        int n = basket1.length;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            cnt.merge(basket1[i], 1, Integer::sum);
            cnt.merge(basket2[i], -1, Integer::sum);
        }
        int mi = 1 << 30;
        List<Integer> nums = new ArrayList<>();
        for (var e : cnt.entrySet()) {
            int x = e.getKey(), v = e.getValue();
            if (v % 2 != 0) {
                return -1;
            }
            for (int i = Math.abs(v) / 2; i > 0; --i) {
                nums.add(x);
            }
            mi = Math.min(mi, x);
        }
        Collections.sort(nums);
        int m = nums.size();
        long ans = 0;
        for (int i = 0; i < m / 2; ++i) {
            ans += Math.min(nums.get(i), mi * 2);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2561: Rearranging Fruits
func minCost(basket1 []int, basket2 []int) (ans int64) {
	cnt := map[int]int{}
	for i, a := range basket1 {
		cnt[a]++
		cnt[basket2[i]]--
	}
	mi := 1 << 30
	nums := []int{}
	for x, v := range cnt {
		if v%2 != 0 {
			return -1
		}
		for i := abs(v) / 2; i > 0; i-- {
			nums = append(nums, x)
		}
		mi = min(mi, x)
	}
	sort.Ints(nums)
	m := len(nums)
	for i := 0; i < m/2; i++ {
		ans += int64(min(nums[i], mi*2))
	}
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2561: Rearranging Fruits
class Solution:
    def minCost(self, basket1: List[int], basket2: List[int]) -> int:
        cnt = Counter()
        for a, b in zip(basket1, basket2):
            cnt[a] += 1
            cnt[b] -= 1
        mi = min(cnt)
        nums = []
        for x, v in cnt.items():
            if v % 2:
                return -1
            nums.extend([x] * (abs(v) // 2))
        nums.sort()
        m = len(nums) // 2
        return sum(min(x, mi * 2) for x in nums[:m])

// Accepted solution for LeetCode #2561: Rearranging Fruits
use std::collections::HashMap;

impl Solution {
    pub fn min_cost(basket1: Vec<i32>, basket2: Vec<i32>) -> i64 {
        let n = basket1.len();
        let mut cnt: HashMap<i32, i32> = HashMap::new();

        for i in 0..n {
            *cnt.entry(basket1[i]).or_insert(0) += 1;
            *cnt.entry(basket2[i]).or_insert(0) -= 1;
        }

        let mut mi = i32::MAX;
        let mut nums = Vec::new();

        for (x, v) in cnt {
            if v % 2 != 0 {
                return -1;
            }
            for _ in 0..(v.abs() / 2) {
                nums.push(x);
            }
            mi = mi.min(x);
        }

        nums.sort();

        let m = nums.len();
        let mut ans = 0;

        for i in 0..(m / 2) {
            ans += nums[i].min(mi * 2) as i64;
        }

        ans
    }
}

// Accepted solution for LeetCode #2561: Rearranging Fruits
function minCost(basket1: number[], basket2: number[]): number {
    const n = basket1.length;
    const cnt: Map<number, number> = new Map();
    for (let i = 0; i < n; i++) {
        cnt.set(basket1[i], (cnt.get(basket1[i]) || 0) + 1);
        cnt.set(basket2[i], (cnt.get(basket2[i]) || 0) - 1);
    }
    let mi = Number.MAX_SAFE_INTEGER;
    const nums: number[] = [];
    for (const [x, v] of cnt.entries()) {
        if (v % 2 !== 0) {
            return -1;
        }
        for (let i = 0; i < Math.abs(v) / 2; i++) {
            nums.push(x);
        }
        mi = Math.min(mi, x);
    }

    nums.sort((a, b) => a - b);
    const m = nums.length;
    let ans = 0;
    for (let i = 0; i < m / 2; i++) {
        ans += Math.min(nums[i], mi * 2);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.