Minimum number of groups of nodes such that no ancestor is present in the same group - GeeksforGeeks

07 Jun, 2021
Given a tree of N nodes. The task is to form the minimum number of groups of nodes such that every node belong to exactly one group, and none of its ancestors are in the same group. The parent of each node is given (-1 if a node does not have a parent).Examples:  
Input: par[] = {-1, 1, 2, 1, 4} Output: 3 The three groups can be: Group 1: {1} Group 2: {2, 4} Group 3: {3, 5}Input: par[] = {-1, 1, 1, 2, 2, 5, 6} Output: 5  
 
Approach: The groups can be made by grouping nodes on the same level together (A node and any of it’s ancestors cannot be on the same level). So the minimum number of groups will be the maximum depth of the tree.Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the depth of the treeint findDepth(int x, vector<int> child[]){    int mx = 0;     // Find the maximum depth of all its children    for (auto ch : child[x])        mx = max(mx, findDepth(ch, child));     // Add 1 for the depth of the current node    return mx + 1;} // Function to return// the minimum number of groups requiredint minimumGroups(int n, int par[]){    vector<int> child[n + 1];     // For every node create a list of its children    for (int i = 1; i <= n; i++)        if (par[i] != -1)            child[par[i]].push_back(i);    int res = 0;     for (int i = 1; i <= n; i++)         // If the node is root        // perform dfs starting with this node        if (par[i] == -1)            res = max(res, findDepth(i, child));     return res;} // Driver codemain(){    int par[] = { 0, -1, 1, 1, 2, 2, 5, 6 };    int n = sizeof(par) / sizeof(par[0]) - 1;    cout << minimumGroups(n, par);}
// Java implementation of the approachimport java.util.*; class GFG{     // Function to return the depth of the tree    static int findDepth(int x, Vector<Integer> child[])    {        int mx = 0;         // Find the maximum depth of all its children        for (Integer ch : child[x])            mx = Math.max(mx, findDepth(ch, child));         // Add 1 for the depth of the current node        return mx + 1;    }     // Function to return    // the minimum number of groups required    static int minimumGroups(int n, int par[])    {        Vector<Integer>[] child = new Vector[n + 1];        for (int i = 0; i <= n; i++)        {            child[i] = new Vector<Integer>();        }                 // For every node create a list of its children        for (int i = 1; i <= n; i++)            if (par[i] != -1)                child[par[i]].add(i);        int res = 0;         for (int i = 1; i <= n; i++)             // If the node is root            // perform dfs starting with this node            if (par[i] == -1)                res = Math.max(res, findDepth(i, child));         return res;    }     // Driver code    public static void main(String[] args)    {        int par[] = { 0, -1, 1, 1, 2, 2, 5, 6 };        int n = par.length - 1;        System.out.print(minimumGroups(n, par));    }} // This code is contributed by PrinciRaj1992
# Python3 implementation of the approach # Function to return the depth of the treedef findDepth(x, child):    mx = 0         # Find the maximum depth    # of all its children    for ch in child[x]:        mx = max(mx, findDepth(ch, child))             # Add 1 for the depth    # of the current node    return mx + 1 # Function to return the minimum # number of groups requireddef minimumGroups(n, par):    child = [[] for i in range(n + 1)]         # For every node create a list    # of its children    for i in range(0, n):        if (par[i] != -1):            child[par[i]].append(i)    res = 0    for i in range(0, n):                 # If the node is root        # perform dfs starting with this node        if(par[i] == -1):            res = max(res, findDepth(i, child))    return res # Driver Codearray = [0, -1, 1, 1, 2, 2, 5, 6]print(minimumGroups(len(array), array)) # This code is contributed# by SidharthPanicker
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG{     // Function to return the depth of the tree    static int findDepth(int x, List<int> []child)    {        int mx = 0;         // Find the maximum depth of all its children        foreach (int ch in child[x])            mx = Math.Max(mx, findDepth(ch, child));         // Add 1 for the depth of the current node        return mx + 1;    }     // Function to return    // the minimum number of groups required    static int minimumGroups(int n, int []par)    {        List<int>[] child = new List<int>[n + 1];        for (int i = 0; i <= n; i++)        {            child[i] = new List<int>();        }                 // For every node create a list of its children        for (int i = 1; i <= n; i++)            if (par[i] != -1)                child[par[i]].Add(i);        int res = 0;         for (int i = 1; i <= n; i++)             // If the node is root            // perform dfs starting with this node            if (par[i] == -1)                res = Math.Max(res, findDepth(i, child));         return res;    }     // Driver code    public static void Main(String[] args)    {        int []par = { 0, -1, 1, 1, 2, 2, 5, 6 };        int n = par.Length - 1;        Console.Write(minimumGroups(n, par));    }} // This code is contributed by Rajput-Ji
<script> // Javascript implementation of the approach // Function to return the depth of the treefunction findDepth(x, child){    var mx = 0;     // Find the maximum depth of all its children    child[x].forEach(ch => {                 mx = Math.max(mx, findDepth(ch, child));    });     // Add 1 for the depth of the current node    return mx + 1;} // Function to return// the minimum number of groups requiredfunction minimumGroups(n, par){    var child = Array.from(Array(n+1), ()=>Array());     // For every node create a list of its children    for (var i = 1; i <= n; i++)        if (par[i] != -1)            child[par[i]].push(i);    var res = 0;     for (var i = 1; i <= n; i++)         // If the node is root        // perform dfs starting with this node        if (par[i] == -1)            res = Math.max(res, findDepth(i, child));     return res;} // Driver codevar par = [0, -1, 1, 1, 2, 2, 5, 6 ];var n = par.length - 1;document.write( minimumGroups(n, par)); // This code is contributed by famously.</script>
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SidharthPanicker
princiraj1992
Rajput-Ji
famously
DFS
Tree
DFS
Tree
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