mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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189 lines
22 KiB
JSON
189 lines
22 KiB
JSON
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"title": "Minimum Cost to Make Array Equal",
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"content": "<p>You are given two <strong>0-indexed</strong> arrays <code>nums</code> and <code>cost</code> consisting each of <code>n</code> <strong>positive</strong> integers.</p>\n\n<p>You can do the following operation <strong>any</strong> number of times:</p>\n\n<ul>\n\t<li>Increase or decrease <strong>any</strong> element of the array <code>nums</code> by <code>1</code>.</li>\n</ul>\n\n<p>The cost of doing one operation on the <code>i<sup>th</sup></code> element is <code>cost[i]</code>.</p>\n\n<p>Return <em>the <strong>minimum</strong> total cost such that all the elements of the array </em><code>nums</code><em> become <strong>equal</strong></em>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3,5,2], cost = [2,3,1,14]\n<strong>Output:</strong> 8\n<strong>Explanation:</strong> We can make all the elements equal to 2 in the following way:\n- Increase the 0<sup>th</sup> element one time. The cost is 2.\n- Decrease the 1<sup><span style=\"font-size: 10.8333px;\">st</span></sup> element one time. The cost is 3.\n- Decrease the 2<sup>nd</sup> element three times. The cost is 1 + 1 + 1 = 3.\nThe total cost is 2 + 3 + 3 = 8.\nIt can be shown that we cannot make the array equal with a smaller cost.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,2,2,2,2], cost = [4,2,8,1,3]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> All the elements are already equal, so no operations are needed.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length == cost.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i], cost[i] <= 10<sup>6</sup></code></li>\n</ul>\n",
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"translatedTitle": "使数组相等的最小开销",
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"translatedContent": "<p>给你两个下标从 <strong>0</strong> 开始的数组 <code>nums</code> 和 <code>cost</code> ,分别包含 <code>n</code> 个 <strong>正</strong> 整数。</p>\n\n<p>你可以执行下面操作 <strong>任意</strong> 次:</p>\n\n<ul>\n\t<li>将 <code>nums</code> 中 <strong>任意</strong> 元素增加或者减小 <code>1</code> 。</li>\n</ul>\n\n<p>对第 <code>i</code> 个元素执行一次操作的开销是 <code>cost[i]</code> 。</p>\n\n<p>请你返回使 <code>nums</code> 中所有元素 <strong>相等</strong> 的 <strong>最少</strong> 总开销。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre><b>输入:</b>nums = [1,3,5,2], cost = [2,3,1,14]\n<b>输出:</b>8\n<b>解释:</b>我们可以执行以下操作使所有元素变为 2 :\n- 增加第 0 个元素 1 次,开销为 2 。\n- 减小第 1 个元素 1 次,开销为 3 。\n- 减小第 2 个元素 3 次,开销为 1 + 1 + 1 = 3 。\n总开销为 2 + 3 + 3 = 8 。\n这是最小开销。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre><b>输入:</b>nums = [2,2,2,2,2], cost = [4,2,8,1,3]\n<b>输出:</b>0\n<b>解释:</b>数组中所有元素已经全部相等,不需要执行额外的操作。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length == cost.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i], cost[i] <= 10<sup>6</sup></code></li>\n</ul>\n",
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"Changing the elements into one of the numbers already existing in the array nums is optimal.",
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