mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
171 lines
20 KiB
JSON
171 lines
20 KiB
JSON
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"questionId": "1830",
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"questionFrontendId": "1711",
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"categoryTitle": "Algorithms",
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"boundTopicId": 543102,
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"title": "Count Good Meals",
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"content": "<p>A <strong>good meal</strong> is a meal that contains <strong>exactly two different food items</strong> with a sum of deliciousness equal to a power of two.</p>\n\n<p>You can pick <strong>any</strong> two different foods to make a good meal.</p>\n\n<p>Given an array of integers <code>deliciousness</code> where <code>deliciousness[i]</code> is the deliciousness of the <code>i<sup>th</sup></code> item of food, return <em>the number of different <strong>good meals</strong> you can make from this list modulo</em> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>Note that items with different indices are considered different even if they have the same deliciousness value.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> deliciousness = [1,3,5,7,9]\n<strong>Output:</strong> 4\n<strong>Explanation: </strong>The good meals are (1,3), (1,7), (3,5) and, (7,9).\nTheir respective sums are 4, 8, 8, and 16, all of which are powers of 2.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> deliciousness = [1,1,1,3,3,3,7]\n<strong>Output:</strong> 15\n<strong>Explanation: </strong>The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= deliciousness.length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= deliciousness[i] <= 2<sup>20</sup></code></li>\n</ul>\n",
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"translatedTitle": "大餐计数",
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"translatedContent": "<p><strong>大餐</strong> 是指 <strong>恰好包含两道不同餐品</strong> 的一餐，其美味程度之和等于 2 的幂。</p>\n\n<p>你可以搭配 <strong>任意</strong> 两道餐品做一顿大餐。</p>\n\n<p>给你一个整数数组 <code>deliciousness</code> ，其中 <code>deliciousness[i]</code> 是第 <code>i<sup></sup></code> 道餐品的美味程度，返回你可以用数组中的餐品做出的不同 <strong>大餐</strong> 的数量。结果需要对 <code>10<sup>9</sup> + 7</code> 取余。</p>\n\n<p>注意，只要餐品下标不同，就可以认为是不同的餐品，即便它们的美味程度相同。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>deliciousness = [1,3,5,7,9]\n<strong>输出：</strong>4\n<strong>解释：</strong>大餐的美味程度组合为 (1,3) 、(1,7) 、(3,5) 和 (7,9) 。\n它们各自的美味程度之和分别为 4 、8 、8 和 16 ，都是 2 的幂。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>deliciousness = [1,1,1,3,3,3,7]\n<strong>输出：</strong>15\n<strong>解释：</strong>大餐的美味程度组合为 3 种 (1,1) ，9 种 (1,3) ，和 3 种 (1,7) 。</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= deliciousness.length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= deliciousness[i] <= 2<sup>20</sup></code></li>\n</ul>\n",
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"Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values",
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"You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers"
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