Given an N-ary tree rooted at 1, the task is to assign values from the range [0, N – 1] to each node in any order such that the sum of MEX values of each node in the tree is maximized and print the maximum possible sum of MEX values of each node in the tree.
The MEX value of node V is defined as the smallest missing positive number in a tree rooted at node V.
Examples:
Input: N = 3, Edges[] = {{1, 2}, {1, 3}}
Output: 4
Explanation:
Assign value 0 to node 2, 1 to node 3 and 2 to node 1.
Therefore, the maximum sum of MEX of all nodes = MEX{1} + MEX{2} + MEX{3} = 3 + 1 + 0 = 4.Input: N = 7, Edges[] = {1, 5}, {1, 4}, {5, 2}, {5, 3}, {4, 7}, {7, 6}}
Output: 13
Explanation:
Assign value 0 to node 6, 1 to node 7, 2 to node 4, 6 to node 1, 5 to node 5, 3 to node 2 and 4 to node 3.
Therefore, the maximum sum of MEX of all nodes = MEX{1} + MEX{2} + MEX{3} + MEX{4} + MEX{5} + MEX{6} + MEX{7} = 7 + 0 + 0 + 3 + 0 + 1 + 0 = 13.
Approach: The idea is to perform DFS Traversal on the given N-ary tree and find the sum of MEX for each subtree in the tree. Follow the steps below to solve the problem:
- Perform Depth First Search(DFS) on tree rooted at node 1.
- Initialize a variable mex with 0 and size with 1.
- Iterate through all children of the current node and perform the following operations:
- Recursively call the children of the current node and store the maximum sum of MEX among all subtree in mex.
- Increase the size of the tree rooted at the current node.
- Increase the value of mex by size.
- After completing the above steps, print the value of mex as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to create an N-ary Treevoid makeTree(vector<int> tree[], pair<int, int> edges[], int N){ // Traverse the edges for (int i = 0; i < N - 1; i++) { int u = edges[i].first; int v = edges[i].second; // Add edges tree[u].push_back(v); }}// Function to get the maximum sum// of MEX values of tree rooted at 1pair<int, int> dfs(int node, vector<int> tree[]){ // Initialize mex int mex = 0; int size = 1; // Iterate through all children // of node for (int u : tree[node]) { // Recursively find maximum sum // of MEX values of each node // in tree rooted at u pair<int, int> temp = dfs(u, tree); // Store the maximum sum of MEX // of among all subtrees mex = max(mex, temp.first); // Increase the size of tree // rooted at current node size += temp.second; } // Resulting MEX for the current // node of the recursive call return { mex + size, size };}// Driver Codeint main(){ // Given N nodes int N = 7; // Given N-1 edges pair<int, int> edges[] = { { 1, 4 }, { 1, 5 }, { 5, 2 }, { 5, 3 }, { 4, 7 }, { 7, 6 } }; // Stores the tree vector<int> tree[N + 1]; // Generates the tree makeTree(tree, edges, N); // Returns maximum sum of MEX // values of each node cout << dfs(1, tree).first; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{static class pair{ int first, second; public pair(int first, int second) { this.first = first; this.second = second; }}// Function to create an N-ary Treestatic void makeTree(Vector<Integer> tree[], pair edges[], int N) { // Traverse the edges for(int i = 0; i < N - 1; i++) { int u = edges[i].first; int v = edges[i].second; // Add edges tree[u].add(v); }}// Function to get the maximum sum// of MEX values of tree rooted at 1static pair dfs(int node, Vector<Integer> tree[]) { // Initialize mex int mex = 0; int size = 1; // Iterate through all children // of node for(int u : tree[node]) { // Recursively find maximum sum // of MEX values of each node // in tree rooted at u pair temp = dfs(u, tree); // Store the maximum sum of MEX // of among all subtrees mex = Math.max(mex, temp.first); // Increase the size of tree // rooted at current node size += temp.second; } // Resulting MEX for the current // node of the recursive call return new pair(mex + size, size);}// Driver Codepublic static void main(String[] args) { // Given N nodes int N = 7; // Given N-1 edges pair edges[] = { new pair(1, 4), new pair(1, 5), new pair(5, 2), new pair(5, 3), new pair(4, 7), new pair(7, 6) }; // Stores the tree @SuppressWarnings("unchecked") Vector<Integer>[] tree = new Vector[N + 1]; for(int i = 0; i < tree.length; i++) tree[i] = new Vector<Integer>(); // Generates the tree makeTree(tree, edges, N); // Returns maximum sum of MEX // values of each node System.out.print((dfs(1, tree).first));}}// This code is contributed by Princi Singh |
Python3
# Python3 program for the above approach# Function to create an N-ary Treedef makeTree(tree, edges, N): # Traverse the edges for i in range(N - 1): u = edges[i][0] v = edges[i][1] # Add edges tree[u].append(v) return tree# Function to get the maximum sum# of MEX values of tree rooted at 1def dfs(node, tree): # Initialize mex mex = 0 size = 1 # Iterate through all children # of node for u in tree[node]: # Recursively find maximum sum # of MEX values of each node # in tree rooted at u temp = dfs(u, tree) # Store the maximum sum of MEX # of among all subtrees mex = max(mex, temp[0]) # Increase the size of tree # rooted at current node size += temp[1] # Resulting MEX for the current # node of the recursive call return [mex + size, size]# Driver Codeif __name__ == '__main__': # Given N nodes N = 7 # Given N-1 edges edges = [ [ 1, 4 ], [ 1, 5 ], [ 5, 2 ], [ 5, 3 ], [ 4, 7 ], [ 7, 6 ] ] # Stores the tree tree = [[] for i in range(N + 1)] # Generates the tree tree = makeTree(tree, edges, N) # Returns maximum sum of MEX # values of each node print(dfs(1, tree)[0])# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{public class pair{ public int first, second; public pair(int first, int second) { this.first = first; this.second = second; }}// Function to create an N-ary Treestatic void makeTree(List<int> []tree, pair []edges, int N) { // Traverse the edges for(int i = 0; i < N - 1; i++) { int u = edges[i].first; int v = edges[i].second; // Add edges tree[u].Add(v); }}// Function to get the maximum sum// of MEX values of tree rooted at 1static pair dfs(int node, List<int> []tree) { // Initialize mex int mex = 0; int size = 1; // Iterate through all children // of node foreach(int u in tree[node]) { // Recursively find maximum sum // of MEX values of each node // in tree rooted at u pair temp = dfs(u, tree); // Store the maximum sum of MEX // of among all subtrees mex = Math.Max(mex, temp.first); // Increase the size of tree // rooted at current node size += temp.second; } // Resulting MEX for the current // node of the recursive call return new pair(mex + size, size);}// Driver Codepublic static void Main(String[] args) { // Given N nodes int N = 7; // Given N-1 edges pair []edges = { new pair(1, 4), new pair(1, 5), new pair(5, 2), new pair(5, 3), new pair(4, 7), new pair(7, 6) }; // Stores the tree List<int>[] tree = new List<int>[N + 1]; for(int i = 0; i < tree.Length; i++) tree[i] = new List<int>(); // Generates the tree makeTree(tree, edges, N); // Returns maximum sum of MEX // values of each node Console.Write((dfs(1, tree).first));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// JavaScript program for the above approach// Function to create an N-ary Treefunction makeTree(tree, edges, N){ // Traverse the edges for (var i = 0; i < N - 1; i++) { var u = edges[i][0]; var v = edges[i][1]; // Add edges tree[u].push(v); }}// Function to get the maximum sum// of MEX values of tree rooted at 1function dfs(node, tree){ // Initialize mex var mex = 0; var size = 1; // Iterate through all children // of node tree[node].forEach(u => { // Recursively find maximum sum // of MEX values of each node // in tree rooted at u var temp = dfs(u, tree); // Store the maximum sum of MEX // of among all subtrees mex = Math.max(mex, temp[0]); // Increase the size of tree // rooted at current node size += temp[1]; }); // Resulting MEX for the current // node of the recursive call return [mex + size, size ];}// Driver Code// Given N nodesvar N = 7;// Given N-1 edgesvar edges = [ [ 1, 4 ], [ 1, 5 ], [ 5, 2 ], [ 5, 3 ], [ 4, 7 ], [ 7, 6 ] ]; // Stores the treevar tree = Array.from(Array(N+1), ()=> Array());// Generates the treemakeTree(tree, edges, N);// Returns maximum sum of MEX// values of each nodedocument.write( dfs(1, tree)[0]);</script> |
Output:
13
Time Complexity: O(N)
Auxiliary Space: O(N)
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