leetcode-problemset/leetcode-cn/originData/closest-dessert-cost.json

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        "question": {
            "questionId": "1900",
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            "title": "Closest Dessert Cost",
            "titleSlug": "closest-dessert-cost",
            "content": "<p>You would like to make dessert and are preparing to buy the ingredients. You have <code>n</code> ice cream base flavors and <code>m</code> types of toppings to choose from. You must follow these rules when making your dessert:</p>\n\n<ul>\n\t<li>There must be <strong>exactly one</strong> ice cream base.</li>\n\t<li>You can add <strong>one or more</strong> types of topping or have no toppings at all.</li>\n\t<li>There are <strong>at most two</strong> of <strong>each type</strong> of topping.</li>\n</ul>\n\n<p>You are given three inputs:</p>\n\n<ul>\n\t<li><code>baseCosts</code>, an integer array of length <code>n</code>, where each <code>baseCosts[i]</code> represents the price of the <code>i<sup>th</sup></code> ice cream base flavor.</li>\n\t<li><code>toppingCosts</code>, an integer array of length <code>m</code>, where each <code>toppingCosts[i]</code> is the price of <strong>one</strong> of the <code>i<sup>th</sup></code> topping.</li>\n\t<li><code>target</code>, an integer representing your target price for dessert.</li>\n</ul>\n\n<p>You want to make a dessert with a total cost as close to <code>target</code> as possible.</p>\n\n<p>Return <em>the closest possible cost of the dessert to </em><code>target</code>. If there are multiple, return <em>the <strong>lower</strong> one.</em></p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> baseCosts = [1,7], toppingCosts = [3,4], target = 10\n<strong>Output:</strong> 10\n<strong>Explanation:</strong> Consider the following combination (all 0-indexed):\n- Choose base 1: cost 7\n- Take 1 of topping 0: cost 1 x 3 = 3\n- Take 0 of topping 1: cost 0 x 4 = 0\nTotal: 7 + 3 + 0 = 10.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> baseCosts = [2,3], toppingCosts = [4,5,100], target = 18\n<strong>Output:</strong> 17\n<strong>Explanation:</strong> Consider the following combination (all 0-indexed):\n- Choose base 1: cost 3\n- Take 1 of topping 0: cost 1 x 4 = 4\n- Take 2 of topping 1: cost 2 x 5 = 10\n- Take 0 of topping 2: cost 0 x 100 = 0\nTotal: 3 + 4 + 10 + 0 = 17. You cannot make a dessert with a total cost of 18.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> baseCosts = [3,10], toppingCosts = [2,5], target = 9\n<strong>Output:</strong> 8\n<strong>Explanation:</strong> It is possible to make desserts with cost 8 and 10. Return 8 as it is the lower cost.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == baseCosts.length</code></li>\n\t<li><code>m == toppingCosts.length</code></li>\n\t<li><code>1 &lt;= n, m &lt;= 10</code></li>\n\t<li><code>1 &lt;= baseCosts[i], toppingCosts[i] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= target &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "最接近目标价格的甜点成本",
            "translatedContent": "<p>你打算做甜点，现在需要购买配料。目前共有 <code>n</code> 种冰激凌基料和 <code>m</code> 种配料可供选购。而制作甜点需要遵循以下几条规则：</p>\n\n<ul>\n\t<li>必须选择 <strong>一种</strong> 冰激凌基料。</li>\n\t<li>可以添加 <strong>一种或多种</strong> 配料，也可以不添加任何配料。</li>\n\t<li>每种类型的配料 <strong>最多两份</strong> 。</li>\n</ul>\n\n<p>给你以下三个输入：</p>\n\n<ul>\n\t<li><code>baseCosts</code> ，一个长度为 <code>n</code> 的整数数组，其中每个 <code>baseCosts[i]</code> 表示第 <code>i</code> 种冰激凌基料的价格。</li>\n\t<li><code>toppingCosts</code>，一个长度为 <code>m</code> 的整数数组，其中每个 <code>toppingCosts[i]</code> 表示 <strong>一份</strong> 第 <code>i</code> 种冰激凌配料的价格。</li>\n\t<li><code>target</code> ，一个整数，表示你制作甜点的目标价格。</li>\n</ul>\n\n<p>你希望自己做的甜点总成本尽可能接近目标价格 <code>target</code> 。</p>\n\n<p>返回最接近<em> </em><code>target</code> 的甜点成本。如果有多种方案，返回 <strong>成本相对较低</strong> 的一种。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>baseCosts = [1,7], toppingCosts = [3,4], target = 10\n<strong>输出：</strong>10\n<strong>解释：</strong>考虑下面的方案组合（所有下标均从 0 开始）：\n- 选择 1 号基料：成本 7\n- 选择 1 份 0 号配料：成本 1 x 3 = 3\n- 选择 0 份 1 号配料：成本 0 x 4 = 0\n总成本：7 + 3 + 0 = 10 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>baseCosts = [2,3], toppingCosts = [4,5,100], target = 18\n<strong>输出：</strong>17\n<strong>解释：</strong>考虑下面的方案组合（所有下标均从 0 开始）：\n- 选择 1 号基料：成本 3\n- 选择 1 份 0 号配料：成本 1 x 4 = 4\n- 选择 2 份 1 号配料：成本 2 x 5 = 10\n- 选择 0 份 2 号配料：成本 0 x 100 = 0\n总成本：3 + 4 + 10 + 0 = 17 。不存在总成本为 18 的甜点制作方案。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>baseCosts = [3,10], toppingCosts = [2,5], target = 9\n<strong>输出：</strong>8\n<strong>解释：</strong>可以制作总成本为 8 和 10 的甜点。返回 8 ，因为这是成本更低的方案。\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>baseCosts = [10], toppingCosts = [1], target = 1\n<strong>输出：</strong>10\n<strong>解释：</strong>注意，你可以选择不添加任何配料，但你必须选择一种基料。</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == baseCosts.length</code></li>\n\t<li><code>m == toppingCosts.length</code></li>\n\t<li><code>1 <= n, m <= 10</code></li>\n\t<li><code>1 <= baseCosts[i], toppingCosts[i] <= 10<sup>4</sup></code></li>\n\t<li><code>1 <= target <= 10<sup>4</sup></code></li>\n</ul>\n",
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