rearranging-fruits

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Algorithms

You have two fruit baskets containing <code>n</code> fruits each. You are given two 0-indexed integer arrays <code>basket1</code> and <code>basket2</code> 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:

<ul>
	<li>Chose two indices <code>i</code> and <code>j</code>, and swap the <code>ith&nbsp;</code>fruit of <code>basket1</code> with the <code>jth</code>&nbsp;fruit of <code>basket2</code>.</li>
	<li>The cost of the swap is <code>min(basket1[i],basket2[j])</code>.</li>
</ul>

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 <code>-1</code> if impossible.

&nbsp;
Example 1:

<pre>
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.
</pre>

Example 2:

<pre>
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.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>basket1.length == basket2.length</code></li>
	<li><code>1 &lt;= basket1.length &lt;= 105</code></li>
	<li><code>1 &lt;= basket1[i],basket2[i]&nbsp;&lt;= 109</code></li>
</ul>

minCost

Rearranging Fruits

The Latest Time to Catch a Bus

the-latest-time-to-catch-a-bus

坐上公交的最晚时间

Minimum Number of Operations to Make Arrays Similar

minimum-number-of-operations-to-make-arrays-similar

使数组相似的最少操作次数

Greedy

Array

Hash Table

你有两个果篮，每个果篮中有 <code>n</code> 个水果。给你两个下标从 0 开始的整数数组 <code>basket1</code> 和 <code>basket2</code> ，用以表示两个果篮中每个水果的交换成本。你想要让两个果篮相等。为此，可以根据需要多次执行下述操作：

<ul>
	<li>选中两个下标 <code>i</code> 和 <code>j</code> ，并交换 <code>basket1</code> 中的第 <code>i</code> 个水果和 <code>basket2</code> 中的第 <code>j</code> 个水果。</li>
	<li>交换的成本是 <code>min(basket1i,basket2j)</code> 。</li>
</ul>

根据果篮中水果的成本进行排序，如果排序后结果完全相同，则认为两个果篮相等。

返回使两个果篮相等的最小交换成本，如果无法使两个果篮相等，则返回 <code>-1</code> 。

&nbsp;

示例 1：

<pre>
输入：basket1 = [4,2,2,2], basket2 = [1,4,1,2]
输出：1
解释：交换 basket1 中下标为 1 的水果和 basket2 中下标为 0 的水果，交换的成本为 1 。此时，basket1 = [4,1,2,2] 且 basket2 = [2,4,1,2] 。重排两个数组，发现二者相等。
</pre>

示例 2：

<pre>
输入：basket1 = [2,3,4,1], basket2 = [3,2,5,1]
输出：-1
解释：可以证明无法使两个果篮相等。
</pre>

&nbsp;

提示：

<ul>
	<li><code>basket1.length == bakste2.length</code></li>
	<li><code>1 &lt;= basket1.length &lt;= 105</code></li>
	<li><code>1 &lt;= basket1i,basket2i &lt;= 109</code></li>
</ul>