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Maximize sum of MEX values of each node in an N-ary Tree

  • Difficulty Level : Expert
  • Last Updated : 18 May, 2021

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.

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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 Tree
void 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 1
pair<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 Code
int 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 approach
import 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 Tree
static 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 1
static 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 Code
public 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 Tree
def 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 1
def 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 Code
if __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 approach
using 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 Tree
static 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 1
static 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 Code
public 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 Tree
function 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 1
function 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 nodes
var N = 7;
 
// Given N-1 edges
var edges = [ [ 1, 4 ], [ 1, 5 ], [ 5, 2 ],
              [ 5, 3 ], [ 4, 7 ], [ 7, 6 ] ];
               
// Stores the tree
var tree = Array.from(Array(N+1), ()=> Array());
 
// Generates the tree
makeTree(tree, edges, N);
 
// Returns maximum sum of MEX
// values of each node
document.write( dfs(1, tree)[0]);
 
</script>
Output: 
13

 

Time Complexity: O(N)
Auxiliary Space: O(1)




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