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++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to implement DFS traversal void 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 color 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 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 void 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 condition void 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 Code int 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;
} |
# Function to implement DFS traversal def 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 color def 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 color def 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 vertices def 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 condition def 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 Code if __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 |
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 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 |
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);
}
} |
4
Time Complexity: O(N)
Auxiliary Space: O(N)