mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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168 lines
16 KiB
JSON
168 lines
16 KiB
JSON
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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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"code": "class Solution {\npublic:\n int countPairs(vector<int>& deliciousness) {\n \n }\n};",
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"code": "# @param {Integer[]} deliciousness\n# @return {Integer}\ndef count_pairs(deliciousness)\n \nend",
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"code": "-spec count_pairs(Deliciousness :: [integer()]) -> integer().\ncount_pairs(Deliciousness) ->\n .",
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"code": "defmodule Solution do\n @spec count_pairs(deliciousness :: [integer]) :: integer\n def count_pairs(deliciousness) do\n\n end\nend",
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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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Java 8 features such as lambda expressions and stream API can be used. </p>\\r\\n\\r\\n<p>Most standard library headers are already included automatically for your convenience.</p>\\r\\n<p>Includes <code>Pair</code> class from https://docs.oracle.com/javase/8/javafx/api/javafx/util/Pair.html.</p>\"], \"python\": [\"Python\", \"<p><code>Python 2.7.12</code>.</p>\\r\\n\\r\\n<p>Most libraries are already imported automatically for your convenience, such as <a href=\\\"https://docs.python.org/2/library/array.html\\\" target=\\\"_blank\\\">array</a>, <a href=\\\"https://docs.python.org/2/library/bisect.html\\\" target=\\\"_blank\\\">bisect</a>, <a href=\\\"https://docs.python.org/2/library/collections.html\\\" target=\\\"_blank\\\">collections</a>. 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