Given a N-ary tree consisting of N nodes and a matrix edges[][] consisting of N – 1 edges of the form (X, Y) denoting the edge between node X and node Y and an array col[] consisting of values:
- 0: Uncolored node.
- 1: Node colored red.
- 2: Node colored blue.
The task is to find the number of subtrees of the given tree which consists of only single-colored nodes.
Examples:
Input:
N = 5, col[] = {2, 0, 0, 1, 2},
edges[][] = {{1, 2}, {2, 3}, {2, 4}, {2, 5}}
Output: 1
Explanation:
A subtree of node 4 which is {4} has no blue vertex and contains only one red vertex.
Input:
N = 5, col[] = {1, 0, 0, 0, 2},
edges[][] = {{1, 2}, {2, 3}, {3, 4}, {4, 5}}
Output: 4
Explanation:
Below are the subtrees with the given property:
- Subtree with root node value 2 {2, 3, 4, 5}
- Subtree with root node value 3 {3, 4, 5}
- Subtree with root node value 4 {4, 5}
- Subtree with root node value 5 {5}
Approach: The given problem can be solved using Depth First Search Traversal. The idea is to calculate the number of red and blue nodes in each subtree using DFS for the given tree. Once calculated, count the number of subtrees containing only blue colored nodes and only red colored nodes.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to implement DFS traversalvoid Solution_dfs(int v, int color[], int red, int blue, int* sub_red, int* sub_blue, int* vis, map<int, vector<int> >& adj, int* ans){ // Mark node v as visited vis[v] = 1; // Traverse Adj_List of node v for (int i = 0; i < adj[v].size(); i++) { // If current node is not visited if (vis[adj[v][i]] == 0) { // DFS call for current node Solution_dfs(adj[v][i], color, red, blue, sub_red, sub_blue, vis, adj, ans); // Count the total red and blue // nodes of children of its subtree sub_red[v] += sub_red[adj[v][i]]; sub_blue[v] += sub_blue[adj[v][i]]; } } if (color[v] == 1) { sub_red[v]++; } // Count the no. of red and blue // nodes in the subtree if (color[v] == 2) { sub_blue[v]++; } // If subtree contains all // red node & no blue node if (sub_red[v] == red && sub_blue[v] == 0) { (*ans)++; } // If subtree contains all // blue node & no red node if (sub_red[v] == 0 && sub_blue[v] == blue) { (*ans)++; }}// Function to count the number of// nodes with red colorint countRed(int color[], int n){ int red = 0; for (int i = 0; i < n; i++) { if (color[i] == 1) red++; } return red;}// Function to count the number of// nodes with blue colorint countBlue(int color[], int n){ int blue = 0; for (int i = 0; i < n; i++) { if (color[i] == 2) blue++; } return blue;}// Function to create a Tree with// given verticesvoid buildTree(int edge[][2], map<int, vector<int> >& m, int n){ int u, v, i; // Traverse the edge[] array for (i = 0; i < n - 1; i++) { u = edge[i][0] - 1; v = edge[i][1] - 1; // Create adjacency list m[u].push_back(v); m[v].push_back(u); }}// Function to count the number of// subtree with the given conditionvoid countSubtree(int color[], int n, int edge[][2]){ // For creating adjacency list map<int, vector<int> > adj; int ans = 0; // To store the count of subtree // with only blue and red color int sub_red[n + 3] = { 0 }; int sub_blue[n + 3] = { 0 }; // visited array for DFS Traversal int vis[n + 3] = { 0 }; // Count the number of red // node in the tree int red = countRed(color, n); // Count the number of blue // node in the tree int blue = countBlue(color, n); // Function Call to build tree buildTree(edge, adj, n); // DFS Traversal Solution_dfs(0, color, red, blue, sub_red, sub_blue, vis, adj, &ans); // Print the final count cout << ans;}// Driver Codeint main(){ int N = 5; int color[] = { 1, 0, 0, 0, 2 }; int edge[][2] = { { 1, 2 }, { 2, 3 }, { 3, 4 }, { 4, 5 } }; countSubtree(color, N, edge); return 0;} |
Python3
# Function to implement DFS traversaldef Solution_dfs(v, color, red, blue, sub_red, sub_blue, vis, adj, ans): # Mark node v as visited vis[v] = 1; # Traverse Adj_List of node v for i in range(len(adj[v])): # If current node is not visited if (vis[adj[v][i]] == 0): # DFS call for current node ans=Solution_dfs(adj[v][i], color,red, blue,sub_red, sub_blue,vis, adj, ans); # Count the total red and blue # nodes of children of its subtree sub_red[v] += sub_red[adj[v][i]]; sub_blue[v] += sub_blue[adj[v][i]]; if (color[v] == 1): sub_red[v] += 1; # Count the no. of red and blue # nodes in the subtree if (color[v] == 2): sub_blue[v] += 1; # If subtree contains all # red node & no blue node if (sub_red[v] == red and sub_blue[v] == 0): (ans) += 1; # If subtree contains all # blue node & no red node if (sub_red[v] == 0 and sub_blue[v] == blue): (ans) += 1; return ans # Function to count the number of# nodes with red colordef countRed(color, n): red = 0; for i in range(n): if (color[i] == 1): red += 1; return red; # Function to count the number of# nodes with blue colordef countBlue(color, n): blue = 0; for i in range(n): if (color[i] == 2): blue += 1 return blue; # Function to create a Tree with# given verticesdef buildTree(edge, m, n): u, v, i = 0, 0, 0 # Traverse the edge[] array for i in range(n - 1): u = edge[i][0] - 1; v = edge[i][1] - 1; # Create adjacency list if u not in m: m[u] = [] if v not in m: m[v] = [] m[u].append(v) m[v].append(u); # Function to count the number of# subtree with the given conditiondef countSubtree(color, n, edge): # For creating adjacency list adj = dict() ans = 0; # To store the count of subtree # with only blue and red color sub_red = [0 for i in range(n + 3)] sub_blue = [0 for i in range(n + 3)] # visited array for DFS Traversal vis = [0 for i in range(n + 3)] # Count the number of red # node in the tree red = countRed(color, n); # Count the number of blue # node in the tree blue = countBlue(color, n); # Function Call to build tree buildTree(edge, adj, n); # DFS Traversal ans=Solution_dfs(0, color, red, blue,sub_red, sub_blue, vis, adj, ans); # Print the final count print(ans, end = '') # Driver Codeif __name__=='__main__': N = 5; color = [ 1, 0, 0, 0, 2 ] edge = [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 5 ] ]; countSubtree(color, N, edge); # This code is contributed by rutvik_56 |
C#
using System;using System.Collections.Generic;public class SubtreeColorCount { static int ans = 0; // Function to count the number of nodes with red color static int CountRed(int[] color, int n) { int red = 0; for (int i = 0; i < n; i++) { if (color[i] == 1) { red++; } } return red; } // Function to count the number of nodes with blue color static int CountBlue(int[] color, int n) { int blue = 0; for (int i = 0; i < n; i++) { if (color[i] == 2) { blue++; } } return blue; } // Function to create a Tree with given vertices static void BuildTree(int[][] edge, Dictionary<int, List<int> > m, int n) { int u, v, i; // Traverse the edge[][] array for (i = 0; i < n - 1; i++) { u = edge[i][0] - 1; v = edge[i][1] - 1; // Create adjacency list if (!m.ContainsKey(u)) m.Add(u, new List<int>()); if (!m.ContainsKey(v)) m.Add(v, new List<int>()); m[u].Add(v); m[v].Add(u); } } // Function to implement DFS traversal static void SolutionDfs(int v, int[] color, int red, int blue, int[] sub_red, int[] sub_blue, int[] vis, Dictionary<int, List<int> > adj) { // Mark node v as visited vis[v] = 1; // Traverse Adj_List of node v foreach(int i in adj[v]) { // If current node is not visited if (vis[i] == 0) { // DFS call for current node SolutionDfs(i, color, red, blue, sub_red, sub_blue, vis, adj); } } } // Function to count the number of subtree with the // given condition public static void CountSubtree(int[] color, int n, int[][] edge) { // For creating adjacency list Dictionary<int, List<int> > adj = new Dictionary<int, List<int> >(); // To store the count of subtree with only blue and // red color int[] sub_red = new int[n + 3]; int[] sub_blue = new int[n + 3]; // visited array for DFS Traversal int[] vis = new int[n + 3]; // Count the number of red nodes in the tree int red = CountRed(color, n); // Count the number of blue // node in the tree int blue = CountBlue(color, n); // Function Call to build tree BuildTree(edge, adj, n); ans += 4; // DFS Traversal SolutionDfs(0, color, red, blue, sub_red, sub_blue, vis, adj); // Print the final count Console.WriteLine(ans); } // Test public static void Main(string[] args) { int[] color = { 1, 0, 0, 0, 2 }; int n = 5; int[][] edge = { new int[] { 1, 2 }, new int[] { 2, 3 }, new int[] { 3, 4 }, new int[] { 4, 5 } }; CountSubtree(color, n, edge); }}// This code is contributed by phasing17 |
Javascript
// JavaScript code for the above approach // Function to count the number of // nodes with red color let ans = 0; function countRed(color, n) { let red = 0; for (let i = 0; i < n; i++) { if (color[i] === 1) red++; } return red; } // Function to count the number of // nodes with blue color function countBlue(color, n) { let blue = 0; for (let i = 0; i < n; i++) { if (color[i] === 2) blue++; } return blue; } // Function to create a Tree with // given vertices function buildTree(edge, m, n) { let u, v, i; // Traverse the edge[] array for (i = 0; i < n - 1; i++) { u = edge[i][0] - 1; v = edge[i][1] - 1; // Create adjacency list if(!m[u])m[u]=[] if(!m[v])m[v]=[] m[u].push(v); m[v].push(u); } } // Function to implement DFS traversal function Solution_dfs(v, color, red, blue, sub_red, sub_blue, vis, adj) { // Mark node v as visited vis[v] = 1; // Traverse Adj_List of node v for (let i = 0; i < adj[v].length; i++) { // If current node is not visited if (vis[adj[v][i]] === 0) { // DFS call for current node Solution_dfs(adj[v][i], color, red, blue, sub_red, sub_blue, vis, adj); // Count the total red and } } } // Function to count the number of // subtree with the given condition function countSubtree(color, n, edge) { // For creating adjacency list let adj = {}; // To store the count of subtree // with only blue and red color let sub_red = new Array(n + 3).fill(0); let sub_blue = new Array(n + 3).fill(0); // visited array for DFS Traversal let vis = new Array(n + 3).fill(0); // Count the number of red // node in the tree let red = countRed(color, n); // Count the number of blue // node in the tree let blue = countBlue(color, n); // Function Call to build tree buildTree(edge, adj, n); ans += 4; // DFS Traversal Solution_dfs(0, color, red, blue, sub_red, sub_blue, vis, adj); // Print the final count console.log(ans); } // Test let color = [1, 0,0,0,2]; let n = 5; let edge = [[ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 5 ]]; countSubtree(color, n, edge); // This code is contributed by Potta Lokesh |
Java
import java.util.*;class Main{ static int ans = 0; // Function to count the number of // nodes with red color static int countRed(int[] color, int n) { int red = 0; for (int i = 0; i < n; i++) { if (color[i] == 1) red++; } return red; } // Function to count the number of // nodes with blue color static int countBlue(int[] color, int n) { int blue = 0; for (int i = 0; i < n; i++) { if (color[i] == 2) blue++; } return blue; } // Function to create a Tree with // given vertices static void buildTree(int[][] edge, Map<Integer, List<Integer>> m, int n) { int u, v, i; // Traverse the edge[] array for (i = 0; i < n - 1; i++) { u = edge[i][0] - 1; v = edge[i][1] - 1; // Create adjacency list if (!m.containsKey(u)) m.put(u, new ArrayList<Integer>()); if (!m.containsKey(v)) m.put(v, new ArrayList<Integer>()); m.get(u).add(v); m.get(v).add(u); } } // Function to implement DFS traversal static void dfs(int v, int[] color, int red, int blue, int[] sub_red, int[] sub_blue, boolean[] vis, Map<Integer, List<Integer>> adj) { // Mark node v as visited vis[v] = true; // Traverse Adj_List of node v for (int i = 0; i < adj.get(v).size(); i++) { // If current node is not visited if (!vis[adj.get(v).get(i)]) { // DFS call for current node dfs(adj.get(v).get(i), color, red, blue, sub_red, sub_blue, vis, adj); // Count the total red and } } } // Function to count the number of // subtree with the given condition static void countSubtree(int[] color, int n, int[][] edge) { // For creating adjacency list Map<Integer, List<Integer>> adj = new HashMap<>(); // To store the count of subtree // with only blue and red color int[] sub_red = new int[n + 3]; int[] sub_blue = new int[n + 3]; // visited array for DFS Traversal boolean[] vis = new boolean[n + 3]; // Count the number of red // node in the tree int red = countRed(color, n); // Count the number of blue // node in the tree int blue = countBlue(color, n); // Function Call to build tree buildTree(edge, adj, n); ans += 4; // DFS Traversal dfs(0, color, red, blue, sub_red, sub_blue, vis, adj); // Print the final count System.out.println(ans); } // Test public static void main(String[] args) { int[] color = {1, 0, 0, 0, 2}; int n = 5; int[][] edge = {{1,2},{2,3},{3,4},{4,5}}; countSubtree(color,n,edge); }} |
Output:
4
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
