Count of Root to Leaf Paths consisting of at most M consecutive Nodes having value K

Given an Acyclic Undirected Graph in the form of a Binary Tree with the root at vertex 1 and values at each vertex [1, N] denoted by the array arr[], the task is to find the number of root to leaf paths that contain at most m consecutive nodes with value K.

Example:

Input: arr[] = {1, 0, 1, 0, 0, 1, 0}, K = 1, M = 2 
 

Output:
Explanation: 
Path 1 : 1 -> 2 -> 4 contains maximum 1 consecutive K 
Path 2 : 1 -> 2 -> 5 contains maximum 1 consecutive K 
Path 3 : 1 -> 3 -> 6 contains maximum 3 consecutive K 
Path 4 : 1 -> 3 -> 7 contains maximum 2 consecutive K 
Since the given value of M is 2, therefore there are 3 paths that contains atmost 2 consecutive K.
Input: arr[] = {2, 1, 3, 2, 1, 2, 1, 4, 3, 5, 2}, K = 2, M = 2 
 



            2
         /     \
        1       3
      /   \    /  \
     2     1  2    1
   /  \   / \
  4    3 5   2


Output:
 

Approach: 
The problem can be solved using Depth First Search approach: 

  • Depth First Search can be used to traverse all the paths from the root vertex.
  • Every time if the value at the present node is K, increment the count.
  • Otherwise, set count to 0.
  • If the count exceeds M, then return.
  • Otherwise, traverse its neighboring nodes and repeat the above steps.
  • Finally, print the value of the count obtained.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Initialize the adjacency
// list and visited array
vector<int> adj[100005];
int visited[100005] = { 0 };
int ans = 0;
 
// Function to find the number of root to
// leaf paths that contain atmost m
// consecutive nodes with value k
void dfs(int node, int count, int m,
         int arr[], int k)
{
    // Mark the current node
    // as visited
    visited[node] = 1;
 
    // If value at current node is k
    if (arr[node - 1] == k) {
 
        // Increment counter
        count++;
    }
    else {
        count = 0;
    }
 
    // If count is greater than m
    // return from that path
    if (count > m) {
        return;
    }
 
    // Path is allowed if size of present node
    // becomes 0 i.e it has no child root and
    // no more than m consecutive 1's
    if (adj[node].size() == 1 && node != 1) {
        ans++;
    }
 
    for (auto x : adj[node]) {
        if (!visited[x]) {
            dfs(x, count, m, arr, k);
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 1, 3, 2, 1, 2, 1 };
    int N = 7, K = 2, M = 2;
 
    // Desigining the tree
    adj[1].push_back(2);
    adj[2].push_back(1);
 
    adj[1].push_back(3);
    adj[3].push_back(1);
 
    adj[2].push_back(4);
    adj[4].push_back(2);
 
    adj[2].push_back(5);
    adj[5].push_back(2);
 
    adj[3].push_back(6);
    adj[6].push_back(3);
 
    adj[3].push_back(7);
    adj[7].push_back(3);
 
    // Counter counts no.
    // of consecutive nodes
    int counter = 0;
 
    dfs(1, counter, M, arr, K);
 
    cout << ans << "\n";
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Initialize the adjacency
// list and visited array
@SuppressWarnings("unchecked")
static Vector<Integer> []adj = new Vector[100005];
static int []visited = new int[100005];
static int ans = 0;
 
// Function to find the number of root to
// leaf paths that contain atmost m
// consecutive nodes with value k
static void dfs(int node, int count, int m,
                int arr[], int k)
{
     
    // Mark the current node
    // as visited
    visited[node] = 1;
 
    // If value at current node is k
    if (arr[node - 1] == k)
    {
         
        // Increment counter
        count++;
    }
    else
    {
        count = 0;
    }
 
    // If count is greater than m
    // return from that path
    if (count > m)
    {
        return;
    }
 
    // Path is allowed if size of present node
    // becomes 0 i.e it has no child root and
    // no more than m consecutive 1's
    if (adj[node].size() == 1 && node != 1)
    {
        ans++;
    }
 
    for(int x : adj[node])
    {
        if (visited[x] == 0)
        {
            dfs(x, count, m, arr, k);
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 1, 3, 2, 1, 2, 1 };
    int N = 7, K = 2, M = 2;
     
    for(int i = 0; i < adj.length; i++)
        adj[i] = new Vector<Integer>();
         
    // Desigining the tree
    adj[1].add(2);
    adj[2].add(1);
 
    adj[1].add(3);
    adj[3].add(1);
 
    adj[2].add(4);
    adj[4].add(2);
 
    adj[2].add(5);
    adj[5].add(2);
 
    adj[3].add(6);
    adj[6].add(3);
 
    adj[3].add(7);
    adj[7].add(3);
 
    // Counter counts no.
    // of consecutive nodes
    int counter = 0;
 
    dfs(1, counter, M, arr, K);
 
    System.out.print(ans + "\n");
}
}
 
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 Program to implement
# the above approach
 
# Initialize the adjacency
# list and visited array
adj = [[] for i in range(100005)]
visited = [ 0 for i in range(100005)]
ans = 0;
  
# Function to find the number of root to
# leaf paths that contain atmost m
# consecutive nodes with value k
def dfs(node, count, m, arr, k):
     
    global ans
     
    # Mark the current node
    # as visited
    visited[node] = 1;
  
    # If value at current
    # node is k
    if (arr[node - 1] == k):
  
        # Increment counter
        count += 1;
     
    else:
        count = 0;   
  
    # If count is greater than m
    # return from that path
    if (count > m):
        return;   
  
    # Path is allowed if size
    # of present node becomes 0
    # i.e it has no child root and
    # no more than m consecutive 1's
    if (len(adj[node]) == 1 and node != 1):
        ans += 1
     
    for x in adj[node]:       
        if (not visited[x]):
            dfs(x, count, m, arr, k);
 
# Driver code
if __name__ == "__main__":
         
    arr = [2, 1, 3, 2, 1, 2, 1]
    N = 7
    K = 2
    M = 2
  
    # Desigining the tree
    adj[1].append(2);
    adj[2].append(1);
  
    adj[1].append(3);
    adj[3].append(1);
  
    adj[2].append(4);
    adj[4].append(2);
  
    adj[2].append(5);
    adj[5].append(2);
  
    adj[3].append(6);
    adj[6].append(3);
  
    adj[3].append(7);
    adj[7].append(3);
  
    # Counter counts no.
    # of consecutive nodes
    counter = 0;
  
    dfs(1, counter, M, arr, K);   
    print(ans)       
 
# This code is contributed by rutvik_56

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Initialize the adjacency
// list and visited array
static List<int> []adj = new List<int>[100005];
static int []visited = new int[100005];
static int ans = 0;
 
// Function to find the number of root to
// leaf paths that contain atmost m
// consecutive nodes with value k
static void dfs(int node, int count, int m,
                int []arr, int k)
{
     
    // Mark the current node
    // as visited
    visited[node] = 1;
 
    // If value at current node is k
    if (arr[node - 1] == k)
    {
         
        // Increment counter
        count++;
    }
    else
    {
        count = 0;
    }
 
    // If count is greater than m
    // return from that path
    if (count > m)
    {
        return;
    }
 
    // Path is allowed if size of present node
    // becomes 0 i.e it has no child root and
    // no more than m consecutive 1's
    if (adj[node].Count == 1 && node != 1)
    {
        ans++;
    }
 
    foreach(int x in adj[node])
    {
        if (visited[x] == 0)
        {
            dfs(x, count, m, arr, k);
        }
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 2, 1, 3, 2, 1, 2, 1 };
    int K = 2, M = 2;
     
    for(int i = 0; i < adj.Length; i++)
        adj[i] = new List<int>();
         
    // Desigining the tree
    adj[1].Add(2);
    adj[2].Add(1);
 
    adj[1].Add(3);
    adj[3].Add(1);
 
    adj[2].Add(4);
    adj[4].Add(2);
 
    adj[2].Add(5);
    adj[5].Add(2);
 
    adj[3].Add(6);
    adj[6].Add(3);
 
    adj[3].Add(7);
    adj[7].Add(3);
 
    // Counter counts no.
    // of consecutive nodes
    int counter = 0;
 
    dfs(1, counter, M, arr, K);
 
    Console.Write(ans + "\n");
}
}
 
// This code is contributed by Amit Katiyar

chevron_right


Output: 

4


 

Time Complexity: O(V + E) 
Auxiliary Space: O(V)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.