2561. Rearranging Fruits

Problem

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:

• Chose 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.**

  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 == bakste2.length

• 1 <= basket1.length <= 105

• 1 <= basket1[i],basket2[i]&nbsp;<= 109

Solution (Java)

class Solution {
    public long minCost(int[] basket1, int[] basket2) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int c: basket1) 
            cnt.put(c, cnt.getOrDefault(c, 0) + 1);
        for (int c: basket2) 
            cnt.put(c, cnt.getOrDefault(c, 0) - 1);
        List<Integer> last = new ArrayList<>();
        for (Map.Entry<Integer, Integer> e: cnt.entrySet()) {
            int v = e.getValue();
            // if v is odd, an even distribution is never possible
            if (v % 2 != 0) return -1;
            // the count of transferred k is |v|/2
            for (int i = 0; i < Math.abs(v) / 2; ++i)
                last.add(e.getKey());
        }
        // find the min of two input arrays as the intermediate
        int minx = Math.min(Arrays.stream(basket1).min().getAsInt(),
                            Arrays.stream(basket2).min().getAsInt());
        // Use quickselect instead of sort can get a better complexity
        Collections.sort(last);
        long res = 0;
        // The first half may be the cost
        for (int i = 0; i < last.size() / 2; ++i) 
            res += Math.min(last.get(i), minx * 2);
        return res;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).