Given a N-ary Tree consisting of nodes valued in the range [0, N – 1] and an array arr[] where each node i is associated to value arr[i], the task is to print the maximum value associated with any node at each level of the given N-ary Tree.
Examples:
Input: N = 8, Edges[][] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}},
arr[] = {4, 2, 3, -5, -1, 3, -2, 6}
Output: 4 3 6
Explanation:
Below is the given N-ary Tree:
The Max of all nodes of the 0th level is 4.
The Max of all nodes of the 1st level is 3.
The Max of all nodes of the 2nd level is 6.Input: N = 10, Edges[][] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}, {6, 8}, {6, 9}},
arr[] = {1, 2, -1, 3, 4, 5, 8, 6, 12, 7}
Output: 1 3 8 12
Explanation:
Below is the given N-ary Tree:
The Max of all nodes of the 0th level is 1.
The Max of all nodes of the 1st level is 3.
The Max of all nodes of the 2nd level is 8.
The Max of all nodes of the 3rd level is 12.
Approach: This problem can be solved by performing the Level Order Traversal of the given tree. While traversing the tree, process nodes of each level separately. For every level being processed, compute the maximum value of all nodes in the level. Follow the steps below:
- Store all the child nodes of the current level in a Queue and pop the nodes of the current level one by one.
- Find the maximum value of all the popped nodes of the current level.
- Print the maximum value obtained in the above step.
- Follow the above steps for each level of the given Tree and print the maximum value for each level respectively.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum value// at each level of N-ary treeint maxAtLevel(int N, int M, vector<int> Value, int Edges[][2]){ // Stores the adjacency list vector<int> adj[N]; // Create the adjacency list for (int i = 0; i < M; i++) { int u = Edges[i][0]; int v = Edges[i][1]; adj[u].push_back(v); } // Perform level order traversal // of nodes at each level queue<int> q; // Push the root node q.push(0); // Iterate until queue is empty while (!q.empty()) { // Get the size of queue int count = q.size(); int maxVal = 0; // Iterate for all the nodes // in the queue currently while (count--) { // Dequeue an node from queue int temp = q.front(); q.pop(); maxVal = max(maxVal, Value[temp]); // Enqueue the children of // dequeued node for (int i = 0; i < adj[temp].size(); i++) { q.push(adj[temp][i]); } } // Print the result cout << maxVal << " "; }}// Driver Codeint main(){ // Number of nodes int N = 10; // Edges of the N-ary tree int Edges[][2] = { { 0, 1 }, { 0, 2 }, { 0, 3 }, { 1, 4 }, { 1, 5 }, { 3, 6 }, { 6, 7 }, { 6, 8 }, { 6, 9 } }; // Given cost vector<int> Value = { 1, 2, -1, 3, 4, 5, 8, 6, 12, 7 }; // Function Call maxAtLevel(N, N - 1, Value, Edges); return 0;} |
Java
// Java program for // the above approachimport java.util.*;class GFG{// Function to find the maximum value// at each level of N-ary treestatic void maxAtLevel(int N, int M, int []Value, int Edges[][]){ // Stores the adjacency list Vector<Integer> []adj = new Vector[N]; for (int i = 0; i < adj.length; i++) adj[i] = new Vector<Integer>(); // Create the adjacency list for (int i = 0; i < M; i++) { int u = Edges[i][0]; int v = Edges[i][1]; adj[u].add(v); } // Perform level order traversal // of nodes at each level Queue<Integer> q = new LinkedList<>(); // Push the root node q.add(0); // Iterate until queue is empty while (!q.isEmpty()) { // Get the size of queue int count = q.size(); int maxVal = 0; // Iterate for all the nodes // in the queue currently while (count-- > 0) { // Dequeue an node from queue int temp = q.peek(); q.remove(); maxVal = Math.max(maxVal, Value[temp]); // Enqueue the children of // dequeued node for (int i = 0; i < adj[temp].size(); i++) { q.add(adj[temp].get(i)); } } // Print the result System.out.print(maxVal + " "); }}// Driver Codepublic static void main(String[] args){ // Number of nodes int N = 10; // Edges of the N-ary tree int Edges[][] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}, {6, 8}, {6, 9}}; // Given cost int []Value = {1, 2, -1, 3, 4, 5, 8, 6, 12, 7}; // Function Call maxAtLevel(N, N - 1, Value, Edges);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function to find the maximum value# at each level of N-ary treedef maxAtLevel(N, M, Value, Edges): # Stores the adjacency list adj = [[] for i in range(N)] # Create the adjacency list for i in range(M): u = Edges[i][0] v = Edges[i][1] adj[u].append(v) # Perform level order traversal # of nodes at each level q = [] # Push the root node q.append(0) # Iterate until queue is empty while (len(q)): # Get the size of queue count = len(q) maxVal = 0 # Iterate for: all the nodes # in the queue currently while (count): # Dequeue an node from queue temp = q[0] q.remove(q[0]) maxVal = max(maxVal, Value[temp]) # Enqueue the children of # dequeued node for i in range(len(adj[temp])): q.append(adj[temp][i]) count -= 1 # Print the result print(maxVal, end = " ")# Driver Codeif __name__ == '__main__': # Number of nodes N = 10 # Edges of the N-ary tree Edges = [ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 1, 4 ], [ 1, 5 ], [ 3, 6 ], [ 6, 7 ], [ 6, 8 ], [ 6, 9 ] ] # Given cost Value = [ 1, 2, -1, 3, 4, 5, 8, 6, 12, 7 ] # Function Call maxAtLevel(N, N - 1, Value, Edges)# This code is contributed by ipg2016107 |
C#
// C# program for // the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the // maximum value at each // level of N-ary treestatic void maxAtLevel(int N, int M, int []Value, int [,]Edges){ // Stores the adjacency list List<int> []adj = new List<int>[N]; for (int i = 0; i < adj.Length; i++) adj[i] = new List<int>(); // Create the adjacency list for (int i = 0; i < M; i++) { int u = Edges[i, 0]; int v = Edges[i, 1]; adj[u].Add(v); } // Perform level order traversal // of nodes at each level Queue<int> q = new Queue<int>(); // Push the root node q.Enqueue(0); // Iterate until queue is empty while (q.Count != 0) { // Get the size of queue int count = q.Count; int maxVal = 0; // Iterate for all the nodes // in the queue currently while (count-- > 0) { // Dequeue an node from queue int temp = q.Peek(); q.Dequeue(); maxVal = Math.Max(maxVal, Value[temp]); // Enqueue the children of // dequeued node for (int i = 0; i < adj[temp].Count; i++) { q.Enqueue(adj[temp][i]); } } // Print the result Console.Write(maxVal + " "); }}// Driver Codepublic static void Main(String[] args){ // Number of nodes int N = 10; // Edges of the N-ary tree int [,]Edges = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}, {6, 8}, {6, 9}}; // Given cost int []Value = {1, 2, -1, 3, 4, 5, 8, 6, 12, 7}; // Function Call maxAtLevel(N, N - 1, Value, Edges);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach// Function to find the maximum value// at each level of N-ary treefunction maxAtLevel(N, M, Value, Edges){ // Stores the adjacency list let adj = new Array(N); for(let i = 0; i < adj.length; i++) adj[i] = []; // Create the adjacency list for(let i = 0; i < M; i++) { let u = Edges[i][0]; let v = Edges[i][1]; adj[u].push(v); } // Perform level order traversal // of nodes at each level let q = []; // Push the root node q.push(0); // Iterate until queue is empty while (q.length > 0) { // Get the size of queue let count = q.length; let maxVal = 0; // Iterate for all the nodes // in the queue currently while (count-- > 0) { // Dequeue an node from queue let temp = q[0]; q.shift(); maxVal = Math.max(maxVal, Value[temp]); // Enqueue the children of // dequeued node for(let i = 0; i < adj[temp].length; i++) { q.push(adj[temp][i]); } } // Print the result document.write(maxVal + " "); }}// Driver code// Number of nodeslet N = 10;// Edges of the N-ary treelet Edges = [ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 1, 4 ], [ 1, 5 ], [ 3, 6 ], [ 6, 7 ], [ 6, 8 ], [ 6, 9 ] ];// Given costlet Value = [ 1, 2, -1, 3, 4, 5, 8, 6, 12, 7 ];// Function CallmaxAtLevel(N, N - 1, Value, Edges);// This code is contributed by suresh07</script> |
Output:
1 3 8 12
Time Complexity: O(N)
Auxiliary Space: O(N)
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