Given two arrays u and v, representing a graph such that there is an undirected edge from u[i] to v[i] (0 ? v[i], u[i] < N) and each node has some value val[i] (0 ? i < N). For each node, if the nodes connected directly to it are sorted in descending order according to their values (in case of equal values, sort it according to their indices in ascending order), print the number of the node at kth position. If total nodes are < k then print -1.
Examples:
Input: u[] = {0, 0, 1}, v[] = {2, 1, 2}, val[] = {2, 4, 3}, k = 2
Output:
2
0
0
For 0 node, the nodes directly connected to it are 1 and 2
having values 4 and 3 respectively, thus node with 2nd largest value is 2.
For 1 node, the nodes directly connected to it are 0 and 2
having values 2 and 3 respectively, thus node with 2nd largest value is 0.
For 2 node, the nodes directly connected to it are 0 and 1
having values 2 and 4 respectively, thus node with 2nd largest value is 0.Input: u[] = {0, 2}, v[] = {2, 1}, val[] = {2, 4, 3}, k = 2
Output:
-1
-1
0
Approach: The idea is to store the nodes directly connected to a node along with their values in a vector and sort them in increasing order, and the kth largest value for a node, having n number of nodes directly connected to it, will be (n – k)th node from last.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print Kth node for each nodevoid findKthNode(int u[], int v[], int n, int val[], int V, int k){ // Vector to store nodes directly // connected to ith node along with // their values vector<pair<int, int> > g[V]; // Add edges to the vector along with // the values of the node for (int i = 0; i < n; i++) { g[u[i]].push_back(make_pair(val[v[i]], v[i])); g[v[i]].push_back(make_pair(val[u[i]], u[i])); } // Sort neighbors of every node // and find the Kth node for (int i = 0; i < V; i++) { if (g[i].size() > 0) sort(g[i].begin(), g[i].end()); // Get the kth node if (k <= g[i].size()) printf("%d\n", g[i][g[i].size() - k].second); // If total nodes are < k else printf("-1\n"); } return;}// Driver codeint main(){ int V = 3; int val[] = { 2, 4, 3 }; int u[] = { 0, 0, 1 }; int v[] = { 2, 1, 2 }; int n = sizeof(u) / sizeof(int); int k = 2; findKthNode(u, v, n, val, V, k); return 0;} |
Java
// Java implementation of the approach import java.util.*;class GFG{// pair classstatic class pair{ int first,second; pair(int a,int b) { first = a; second = b; }}// Function to print Kth node for each node static void findKthNode(int u[], int v[], int n, int val[], int V, int k) { // Vector to store nodes directly // connected to ith node along with // their values Vector<Vector<pair >> g = new Vector<Vector<pair>>(); for(int i = 0; i < V; i++) g.add(new Vector<pair>()); // Add edges to the Vector along with // the values of the node for (int i = 0; i < n; i++) { g.get(u[i]).add(new pair(val[v[i]], v[i])); g.get(v[i]).add(new pair(val[u[i]], u[i])); } // Sort neighbors of every node // and find the Kth node for (int i = 0; i < V; i++) { if (g.get(i).size() > 0) Collections.sort(g.get(i),new Comparator<pair>() { public int compare(pair p1, pair p2) { return p1.first - p2.first; } }); // Get the kth node if (k <= g.get(i).size()) System.out.printf("%d\n", g.get(i).get(g.get(i).size() - k).second); // If total nodes are < k else System.out.printf("-1\n"); } return; } // Driver code public static void main(String args[]){ int V = 3; int val[] = { 2, 4, 3 }; int u[] = { 0, 0, 1 }; int v[] = { 2, 1, 2 }; int n = u.length; int k = 2; findKthNode(u, v, n, val, V, k); }} // This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the approach # Function to print Kth node for each node def findKthNode(u, v, n, val, V, k): # Vector to store nodes directly connected # to ith node along with their values g = [[] for i in range(V)] # Add edges to the vector along # with the values of the node for i in range(0, n): g[u[i]].append((val[v[i]], v[i])) g[v[i]].append((val[u[i]], u[i])) # Sort neighbors of every node # and find the Kth node for i in range(0, V): if len(g[i]) > 0: g[i].sort() # Get the kth node if k <= len(g[i]): print(g[i][-k][1]) # If total nodes are < k else: print("-1") return# Driver code if __name__ == "__main__": V = 3 val = [2, 4, 3] u = [0, 0, 1] v = [2, 1, 2] n, k = len(u), 2 findKthNode(u, v, n, val, V, k) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approach using System;using System.Collections;using System.Collections.Generic; class GFG{ // pair classclass pair{ public int first,second; public pair(int a,int b) { first = a; second = b; }} class sortHelper : IComparer{ int IComparer.Compare(object a, object b) { pair first = (pair)a; pair second = (pair)b; return first.first - second.first; }} // Function to print Kth node for each node static void findKthNode(int []u, int []v, int n, int []val, int V, int k) { // Vector to store nodes directly // connected to ith node along with // their values ArrayList g = new ArrayList(); for(int i = 0; i < V; i++) g.Add(new ArrayList()); // Add edges to the Vector along with // the values of the node for (int i = 0; i < n; i++) { ((ArrayList)g[u[i]]).Add(new pair(val[v[i]], v[i])); ((ArrayList)g[v[i]]).Add(new pair(val[u[i]], u[i])); } // Sort neighbors of every node // and find the Kth node for (int i = 0; i < V; i++) { if (((ArrayList)g[i]).Count > 0) { ArrayList tmp = (ArrayList)g[i]; tmp.Sort(new sortHelper()); g[i] = tmp; } // Get the kth node if (k <= ((ArrayList)g[i]).Count) Console.Write(((pair)((ArrayList)g[i])[((ArrayList)g[i]).Count - k]).second+"\n"); // If total nodes are < k else Console.Write("-1\n"); } return; } // Driver code public static void Main(string []args){ int V = 3; int []val = { 2, 4, 3 }; int []u = { 0, 0, 1 }; int []v = { 2, 1, 2 }; int n = u.Length; int k = 2; findKthNode(u, v, n, val, V, k); }} // This code is contributed by Pratham76 |
Javascript
<script>// Javascript implementation of the approach// Function to print Kth node for each nodefunction findKthNode(u, v, n, val, V, k){ // Vector to store nodes directly // connected to ith node along with // their values var g = Array.from(Array(V), () => Array()); // Add edges to the vector along with // the values of the node for(var i = 0; i < n; i++) { g[u[i]].push([val[v[i]], v[i]]); g[v[i]].push([val[u[i]], u[i]]); } // Sort neighbors of every node // and find the Kth node for(var i = 0; i < V; i++) { if (g[i].length > 0) g[i].sort(); // Get the kth node if (k <= g[i].length) document.write( g[i][g[i].length - k][1] + "<br>"); // If total nodes are < k else document.write("-1<br>"); } return;}// Driver codevar V = 3;var val = [ 2, 4, 3 ];var u = [ 0, 0, 1 ];var v = [ 2, 1, 2 ];var n = u.length;var k = 2;findKthNode(u, v, n, val, V, k);// This code is contributed by rutvik_56</script> |
Output:
2 0 0
Time Complexity: O(N)
Auxiliary Space: O(N)
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