mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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195 lines
19 KiB
JSON
195 lines
19 KiB
JSON
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"questionId": "1891",
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"title": "Count Pairs Of Nodes",
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"content": "<p>You are given an undirected graph defined by an integer <code>n</code>, the number of nodes, and a 2D integer array <code>edges</code>, the edges in the graph, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is an <strong>undirected</strong> edge between <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code>. You are also given an integer array <code>queries</code>.</p>\n\n<p>Let <code>incident(a, b)</code> be defined as the <strong>number of edges</strong> that are connected to <strong>either</strong> node <code>a</code> or <code>b</code>.</p>\n\n<p>The answer to the <code>j<sup>th</sup></code> query is the <strong>number of pairs</strong> of nodes <code>(a, b)</code> that satisfy <strong>both</strong> of the following conditions:</p>\n\n<ul>\n\t<li><code>a < b</code></li>\n\t<li><code>incident(a, b) > queries[j]</code></li>\n</ul>\n\n<p>Return <em>an array </em><code>answers</code><em> such that </em><code>answers.length == queries.length</code><em> and </em><code>answers[j]</code><em> is the answer of the </em><code>j<sup>th</sup></code><em> query</em>.</p>\n\n<p>Note that there can be <strong>multiple edges</strong> between the same two nodes.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/08/winword_2021-06-08_00-58-39.png\" style=\"width: 529px; height: 305px;\" />\n<pre>\n<strong>Input:</strong> n = 4, edges = [[1,2],[2,4],[1,3],[2,3],[2,1]], queries = [2,3]\n<strong>Output:</strong> [6,5]\n<strong>Explanation:</strong> The calculations for incident(a, b) are shown in the table above.\nThe answers for each of the queries are as follows:\n- answers[0] = 6. All the pairs have an incident(a, b) value greater than 2.\n- answers[1] = 5. All the pairs except (3, 4) have an incident(a, b) value greater than 3.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 5, edges = [[1,5],[1,5],[3,4],[2,5],[1,3],[5,1],[2,3],[2,5]], queries = [1,2,3,4,5]\n<strong>Output:</strong> [10,10,9,8,6]\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= n <= 2 * 10<sup>4</sup></code></li>\n\t<li><code>1 <= edges.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= u<sub>i</sub>, v<sub>i</sub> <= n</code></li>\n\t<li><code>u<sub>i </sub>!= v<sub>i</sub></code></li>\n\t<li><code>1 <= queries.length <= 20</code></li>\n\t<li><code>0 <= queries[j] < edges.length</code></li>\n</ul>\n",
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"code": "class Solution(object):\n def countPairs(self, n, edges, queries):\n \"\"\"\n :type n: int\n :type edges: List[List[int]]\n :type queries: List[int]\n :rtype: List[int]\n \"\"\"\n ",
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"code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* countPairs(int n, int** edges, int edgesSize, int* edgesColSize, int* queries, int queriesSize, int* returnSize) {\n \n}",
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"code": "class Solution {\n\n /**\n * @param Integer $n\n * @param Integer[][] $edges\n * @param Integer[] $queries\n * @return Integer[]\n */\n function countPairs($n, $edges, $queries) {\n \n }\n}",
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"code": "# @param {Integer} n\n# @param {Integer[][]} edges\n# @param {Integer[]} queries\n# @return {Integer[]}\ndef count_pairs(n, edges, queries)\n \nend",
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"code": "object Solution {\n def countPairs(n: Int, edges: Array[Array[Int]], queries: Array[Int]): Array[Int] = {\n \n }\n}",
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"code": "impl Solution {\n pub fn count_pairs(n: i32, edges: Vec<Vec<i32>>, queries: Vec<i32>) -> Vec<i32> {\n \n }\n}",
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"code": "(define/contract (count-pairs n edges queries)\n (-> exact-integer? (listof (listof exact-integer?)) (listof exact-integer?) (listof exact-integer?))\n )",
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"code": "-spec count_pairs(N :: integer(), Edges :: [[integer()]], Queries :: [integer()]) -> [integer()].\ncount_pairs(N, Edges, Queries) ->\n .",
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"code": "defmodule Solution do\n @spec count_pairs(n :: integer, edges :: [[integer]], queries :: [integer]) :: [integer]\n def count_pairs(n, edges, queries) do\n \n end\nend",
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"hints": [
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"We want to count pairs (x,y) such that degree[x] + degree[y] - occurrences(x,y) > k",
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"Think about iterating on x, and counting the number of valid y to pair with x.",
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"You can consider at first that the (- occurrences(x,y)) isn't there, or it is 0 at first for all y. Count the valid y this way.",
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"Then you can iterate on the neighbors of x, let that neighbor be y, and update occurrences(x,y).",
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"When you update occurrences(x,y), the left-hand side decreases. Once it reaches k, then y is not valid for x anymore, so you should decrease the answer by 1."
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