Minimum steps to color the tree with given colors

Last Updated : 11 Jul, 2025

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# 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>