g1701_1800.s1782_count_pairs_of_nodes.Solution Maven / Gradle / Ivy

Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation
Java-based LeetCode algorithm problem solutions, regularly updated
There is a newer version: 1.38
package g1701_1800.s1782_count_pairs_of_nodes;

// #Hard #Binary_Search #Two_Pointers #Graph
// #2022_04_30_Time_128_ms_(86.96%)_Space_175.4_MB_(39.13%)

import java.util.HashMap;
import java.util.Map;

/**
 * 1782 - Count Pairs Of Nodes\.
 *
 * Hard
 *
 * You are given an undirected graph defined by an integer `n`, the number of nodes, and a 2D integer array `edges`, the edges in the graph, where edges[i] = [u_i, v_i] indicates that there is an **undirected** edge between u_i and v_i. You are also given an integer array `queries`.
 *
 * Let `incident(a, b)` be defined as the **number of edges** that are connected to **either** node `a` or `b`.
 *
 * The answer to the j^th query is the **number of pairs** of nodes `(a, b)` that satisfy **both** of the following conditions:
 *
 * *   `a < b`
 * *   `incident(a, b) > queries[j]`
 *
 * Return _an array_ `answers` _such that_ `answers.length == queries.length` _and_ `answers[j]` _is the answer of the_ j^th _query_.
 *
 * Note that there can be **multiple edges** between the same two nodes.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/06/08/winword_2021-06-08_00-58-39.png)
 *
 * **Input:** n = 4, edges = \[\[1,2],[2,4],[1,3],[2,3],[2,1]], queries = [2,3]
 *
 * **Output:** [6,5]
 *
 * **Explanation:** The calculations for incident(a, b) are shown in the table above. The answers for each of the queries are as follows: 
 *
 * - answers[0] = 6. All the pairs have an incident(a, b) value greater than 2. 
 *
 * - answers[1] = 5. All the pairs except (3, 4) have an incident(a, b) value greater than 3.
 *
 * **Example 2:**
 *
 * **Input:** 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]
 *
 * **Output:** [10,10,9,8,6]
 *
 * **Constraints:**
 *
 * *   2 <= n <= 2 * 10⁴
 * *   1 <= edges.length <= 10⁵
 * *   1 <= u_i, v_i <= n
 * *   u_i != v_i
 * *   `1 <= queries.length <= 20`
 * *   `0 <= queries[j] < edges.length`
**/
public class Solution {
    public int[] countPairs(int n, int[][] edges, int[] queries) {
        Map edgeCount = new HashMap<>();
        int[] degree = new int[n];
        for (int[] e : edges) {
            int u = e[0] - 1;
            int v = e[1] - 1;
            degree[u]++;
            degree[v]++;
            int eId = Math.min(u, v) * n + Math.max(u, v);
            edgeCount.put(eId, edgeCount.getOrDefault(eId, 0) + 1);
        }
        Map degreeCount = new HashMap<>();
        int maxDegree = 0;
        for (int d : degree) {
            degreeCount.put(d, degreeCount.getOrDefault(d, 0) + 1);
            maxDegree = Math.max(maxDegree, d);
        }
        int[] count = new int[2 * maxDegree + 1];
        for (Map.Entry d1 : degreeCount.entrySet()) {
            for (Map.Entry d2 : degreeCount.entrySet()) {
                count[d1.getKey() + d2.getKey()] +=
                        (d1 == d2)
                                ? d1.getValue() * (d1.getValue() - 1)
                                : d1.getValue() * d2.getValue();
            }
        }
        for (int i = 0; i < count.length; i++) {
            count[i] /= 2;
        }
        for (Map.Entry e : edgeCount.entrySet()) {
            int u = e.getKey() / n;
            int v = e.getKey() % n;
            count[degree[u] + degree[v]]--;
            count[degree[u] + degree[v] - e.getValue()]++;
        }
        for (int i = count.length - 2; i >= 0; i--) {
            count[i] += count[i + 1];
        }
        int[] res = new int[queries.length];
        for (int q = 0; q < queries.length; q++) {
            res[q] = ((queries[q] + 1) >= count.length) ? 0 : count[queries[q] + 1];
        }
        return res;
    }
}