mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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175 lines
18 KiB
JSON
175 lines
18 KiB
JSON
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"question": {
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"questionId": "3444",
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"questionFrontendId": "3164",
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"boundTopicId": null,
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"title": "Find the Number of Good Pairs II",
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"titleSlug": "find-the-number-of-good-pairs-ii",
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"content": "<p>You are given 2 integer arrays <code>nums1</code> and <code>nums2</code> of lengths <code>n</code> and <code>m</code> respectively. You are also given a <strong>positive</strong> integer <code>k</code>.</p>\n\n<p>A pair <code>(i, j)</code> is called <strong>good</strong> if <code>nums1[i]</code> is divisible by <code>nums2[j] * k</code> (<code>0 <= i <= n - 1</code>, <code>0 <= j <= m - 1</code>).</p>\n\n<p>Return the total number of <strong>good</strong> pairs.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums1 = [1,3,4], nums2 = [1,3,4], k = 1</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">5</span></p>\n\n<p><strong>Explanation:</strong></p>\nThe 5 good pairs are <code>(0, 0)</code>, <code>(1, 0)</code>, <code>(1, 1)</code>, <code>(2, 0)</code>, and <code>(2, 2)</code>.</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums1 = [1,2,4,12], nums2 = [2,4], k = 3</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The 2 good pairs are <code>(3, 0)</code> and <code>(3, 1)</code>.</p>\n</div>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= n, m <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums1[i], nums2[j] <= 10<sup>6</sup></code></li>\n\t<li><code>1 <= k <= 10<sup>3</sup></code></li>\n</ul>\n",
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"difficulty": "Medium",
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"exampleTestcases": "[1,3,4]\n[1,3,4]\n1\n[1,2,4,12]\n[2,4]\n3",
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"categoryTitle": "Algorithms",
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"code": "class Solution {\npublic:\n long long numberOfPairs(vector<int>& nums1, vector<int>& nums2, int k) {\n \n }\n};",
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"code": "class Solution {\n public long numberOfPairs(int[] nums1, int[] nums2, int k) {\n \n }\n}",
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"code": "class Solution(object):\n def numberOfPairs(self, nums1, nums2, k):\n \"\"\"\n :type nums1: List[int]\n :type nums2: List[int]\n :type k: int\n :rtype: int\n \"\"\"\n ",
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"code": "public class Solution {\n public long NumberOfPairs(int[] nums1, int[] nums2, int k) {\n \n }\n}",
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"code": "/**\n * @param {number[]} nums1\n * @param {number[]} nums2\n * @param {number} k\n * @return {number}\n */\nvar numberOfPairs = function(nums1, nums2, k) {\n \n};",
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"code": "class Solution {\n\n /**\n * @param Integer[] $nums1\n * @param Integer[] $nums2\n * @param Integer $k\n * @return Integer\n */\n function numberOfPairs($nums1, $nums2, $k) {\n \n }\n}",
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"code": "# @param {Integer[]} nums1\n# @param {Integer[]} nums2\n# @param {Integer} k\n# @return {Integer}\ndef number_of_pairs(nums1, nums2, k)\n \nend",
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"lang": "Scala",
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"code": "object Solution {\n def numberOfPairs(nums1: Array[Int], nums2: Array[Int], k: Int): Long = {\n \n }\n}",
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"langSlug": "rust",
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"code": "impl Solution {\n pub fn number_of_pairs(nums1: Vec<i32>, nums2: Vec<i32>, k: i32) -> i64 {\n \n }\n}",
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"code": "(define/contract (number-of-pairs nums1 nums2 k)\n (-> (listof exact-integer?) (listof exact-integer?) exact-integer? exact-integer?)\n )",
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"lang": "Erlang",
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"langSlug": "erlang",
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"code": "-spec number_of_pairs(Nums1 :: [integer()], Nums2 :: [integer()], K :: integer()) -> integer().\nnumber_of_pairs(Nums1, Nums2, K) ->\n .",
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"lang": "Elixir",
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"code": "defmodule Solution do\n @spec number_of_pairs(nums1 :: [integer], nums2 :: [integer], k :: integer) :: integer\n def number_of_pairs(nums1, nums2, k) do\n \n end\nend",
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"hints": [
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"Let <code>f[v]</code> be the number of occurrences of <code>v/k</code> in nums2.",
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"For each value <code>v</code> in nums1, enumerating all its factors <code>d</code> (in <code>sqrt(v)</code> time) and sum up all the <code>f[d]</code> to get the final answer.",
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"It is also possible to improve the complexity from <code>v * sqrt(v)</code> to <code>v * log(v)</code> - How?"
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