Minimum steps to color the tree with given colors

Given a tree with N nodes which initially have no color and an array color[] of size N which represent the color of each node after the coloring process takes place. The task is to color the tree into the given colors using the smallest possible number of steps. On each step, one can choose a vertex v and a color x, and then color all vertices in the sub-tree of v (including v itself) with color x. Note that root is vertex number 1. 
Examples:

Input: color[] = { 1, 1, 2, 1, 3, 1} 

Output: 4 
Color the sub-tree rooted at node 1 with color 1. 
Then all the vertices have colors 1. 
Now, color the sub-tree rooted at 3 with color 2. 
Finally, color the sub-trees rooted at 5 and 6 with colors 3 and 1 respectively.
Input: color[] = { 1, 2, 3, 2, 2, 3} 

Output: 3 

Approach: Call a DFS function at vertex 1 and initially keep answer as zero. Increment the answer whenever there is a difference in colors of child and parent nodes. 
See the below code for better understanding.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// To store the required answer
int ans = 0;

// To store the graph
vector<int> gr[100005];

// Function to add edges
void Add_Edge(int u, int v)
{
    gr[u].push_back(v);
    gr[v].push_back(u);
}

// Dfs function
void dfs(int child, int par, int color[])
{

    // When there is difference in colors
    if (color[child] != color[par])
        ans++;

    // For all it's child nodes
    for (auto it : gr[child]) {
        if (it == par)
            continue;
        dfs(it, child, color);
    }
}

// Driver code
int main()
{

    // Here zero is for parent of node 1
    int color[] = { 0, 1, 2, 3, 2, 2, 3 };

    // Adding edges in the graph
    Add_Edge(1, 2);
    Add_Edge(1, 3);
    Add_Edge(2, 4);
    Add_Edge(2, 5);
    Add_Edge(3, 6);

    // Dfs call
    dfs(1, 0, color);

    // Required answer
    cout << ans;

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// To store the required answer
static int ans = 0;

// To store the graph
static Vector<Vector<Integer>> gr = new Vector<Vector<Integer>>();

// Function to add edges
static void Add_Edge(int u, int v)
{
    gr.get(u).add(v);
    gr.get(v).add(u);
}

// Dfs function
static void dfs(int child, int par, int color[])
{

    // When there is difference in colors
    if (color[child] != color[par])
        ans++;

    // For all it's child nodes
    for (int i = 0; i < gr.get(child).size(); i++)
    {
        if (gr.get(child).get(i) == par)
            continue;
        dfs(gr.get(child).get(i), child, color);
    }
}

// Driver code
public static void main(String args[])
{
    for(int i = 0; i <= 10; i++)
    gr.add(new Vector<Integer>());

    // Here zero is for parent of node 1
    int color[] = { 0, 1, 2, 3, 2, 2, 3 };

    // Adding edges in the graph
    Add_Edge(1, 2);
    Add_Edge(1, 3);
    Add_Edge(2, 4);
    Add_Edge(2, 5);
    Add_Edge(3, 6);

    // Dfs call
    dfs(1, 0, color);

    // Required answer
    System.out.println( ans);
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach

# To store the required answer
ans = 0

# To store the graph
gr = [[] for i in range(100005)]

# Function to add edges
def Add_Edge(u, v):
    gr[u].append(v)
    gr[v].append(u)

# Dfs function
def dfs(child, par, color):
    global ans

    # When there is difference in colors
    if (color[child] != color[par]):
        ans += 1

    # For all it's child nodes
    for it in gr[child]:
        if (it == par):
            continue
        dfs(it, child, color)

# Driver code

# Here zero is for parent of node 1
color = [0, 1, 2, 3, 2, 2, 3]

# Adding edges in the graph
Add_Edge(1, 2)
Add_Edge(1, 3)
Add_Edge(2, 4)
Add_Edge(2, 5)
Add_Edge(3, 6)

# Dfs call
dfs(1, 0, color)

# Required answer
print(ans)

# This code is contributed
# by mohit kumar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

    // To store the required answer
    static int ans = 0;

    // To store the graph
    static List<List<int>> gr = new List<List<int>>();

    // Function to add edges
    static void Add_Edge(int u, int v)
    {
        gr[u].Add(v);
        gr[v].Add(u);
    }

    // Dfs function
    static void dfs(int child, int par, int []color)
    {

        // When there is difference in colors
        if (color[child] != color[par])
            ans++;

        // For all it's child nodes
        for (int i = 0; i < gr[child].Count; i++)
        {
            if (gr[child][i] == par)
                continue;
            dfs(gr[child][i], child, color);
        }
    }

    // Driver code
    public static void Main(String []args)
    {
        for(int i = 0; i <= 10; i++)
        gr.Add(new List<int>());

        // Here zero is for parent of node 1
        int []color = { 0, 1, 2, 3, 2, 2, 3 };

        // Adding edges in the graph
        Add_Edge(1, 2);
        Add_Edge(1, 3);
        Add_Edge(2, 4);
        Add_Edge(2, 5);
        Add_Edge(3, 6);

        // Dfs call
        dfs(1, 0, color);

        // Required answer
        Console.WriteLine( ans);
    }
}

// This code has been contributed by 29AjayKumar

Javascript

<script>
// Javascript implementation of the approach

// To store the required answer
let ans = 0;

// To store the graph
let gr = [];

// Function to add edges
function Add_Edge(u,v)
{
    gr[u].push(v);
    gr[v].push(u);
}

// Dfs function
function dfs(child,par,color)
{
    // When there is difference in colors
    if (color[child] != color[par])
        ans++;

    // For all it's child nodes
    for (let i = 0; i < gr[child].length; i++)
    {
        if (gr[child][i] == par)
            continue;
        dfs(gr[child][i], child, color);
    }
}

// Driver code
for(let i = 0; i <= 10; i++)
    gr.push([]);

    // Here zero is for parent of node 1
    let color = [ 0, 1, 2, 3, 2, 2, 3 ];

    // Adding edges in the graph
    Add_Edge(1, 2);
    Add_Edge(1, 3);
    Add_Edge(2, 4);
    Add_Edge(2, 5);
    Add_Edge(3, 6);

    // Dfs call
    dfs(1, 0, color);

    // Required answer
    document.write( ans);

// This code is contributed by unknown2108
</script>

Output:

3