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leetcode-cn/problem (Chinese)/将数组变相同的最小代价 [minimum-cost-to-make-arrays-identical].html Normal file
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<p>给你两个长度都为 <code>n</code>&nbsp;的整数数组&nbsp;<code>arr</code> 和&nbsp;<code>brr</code>&nbsp;以及一个整数&nbsp;<code>k</code>&nbsp;。你可以对 <code>arr</code>&nbsp;执行以下操作任意次：</p>
<ul>
<li>将&nbsp;<code>arr</code>&nbsp;分割成若干个&nbsp;<strong>连续的</strong>&nbsp;子数组，并将这些子数组按任意顺序重新排列。这个操作的代价为&nbsp;<code>k</code>&nbsp;。</li>
<li>
<p>选择 <code>arr</code>&nbsp;中的任意一个元素，将它增加或者减少一个正整数&nbsp;<code>x</code>&nbsp;。这个操作的代价为 <code>x</code>&nbsp;。</p>
</li>
</ul>
<p>请你返回将 <code>arr</code>&nbsp;变为 <code>brr</code>&nbsp;的 <strong>最小</strong>&nbsp;总代价。</p>
<p><strong>子数组</strong>&nbsp;是一个数组中一段连续 <strong>非空</strong>&nbsp;的元素序列。</p>
<p>&nbsp;</p>
<p><strong class="example">示例 1：</strong></p>
<div class="example-block">
<p><span class="example-io"><b>输入：</b>arr = [-7,9,5], brr = [7,-2,-5], k = 2</span></p>
<p><span class="example-io"><b>输出：</b>13</span></p>
<p><b>解释：</b></p>
<ul>
<li>将&nbsp;<code>arr</code>&nbsp;分割成两个连续子数组：<code>[-7]</code> 和&nbsp;<code>[9, 5]</code>&nbsp;然后将它们重新排列成&nbsp;<code>[9, 5, -7]</code>&nbsp;，代价为 2 。</li>
<li>将&nbsp;<code>arr[0]</code>&nbsp;减小 2 ，数组变为&nbsp;<code>[7, 5, -7]</code>&nbsp;，操作代价为 2 。</li>
<li>将&nbsp;<code>arr[1]</code>&nbsp;减小 7 ，数组变为&nbsp;<code>[7, -2, -7]</code>&nbsp;，操作代价为 7 。</li>
<li>将&nbsp;<code>arr[2]</code>&nbsp;增加 2 ，数组变为&nbsp;<code>[7, -2, -5]</code>&nbsp;，操作代价为 2 。</li>
</ul>
<p>将两个数组变相等的总代价为&nbsp;<code>2 + 2 + 7 + 2 = 13</code>&nbsp;。</p>
</div>
<p><strong class="example">示例 2：</strong></p>
<div class="example-block">
<p><span class="example-io"><b>输入：</b>arr = [2,1], brr = [2,1], k = 0</span></p>
<p><span class="example-io"><b>输出：</b>0</span></p>
<p><b>解释：</b></p>
<p>由于数组已经相等，不需要进行任何操作，所以总代价为 0 。</p>
</div>
<p>&nbsp;</p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 &lt;= arr.length == brr.length &lt;= 10<sup>5</sup></code></li>
<li><code>0 &lt;= k &lt;= 2 * 10<sup>10</sup></code></li>
<li><code>-10<sup>5</sup> &lt;= arr[i] &lt;= 10<sup>5</sup></code></li>
<li><code>-10<sup>5</sup> &lt;= brr[i] &lt;= 10<sup>5</sup></code></li>
</ul>