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Check if a string is present in the given Linked List as a subsequence

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  • Last Updated : 28 Feb, 2022

Given a string S of size N and a linked list, the task is to check if the linked list contains a string as a subsequence. Print Yes if it contains the subsequence otherwise print No.

Example:

Input: S = “bad”, Linked List: b -> r -> a -> d -> NULL
Output: Yes

Input: S = “bad”, Linked List: a -> p -> p -> l -> e -> NULL
Output: No

Approach: This problem can be solved using two pointers one on the string and one on the linked list. Now, follow the below steps to solve this problem:

  1. Create variable i, and initialise it with 0. Also, create a pointer cur that points at the head of the linked list.
  2. Now, run a while loop till i is less than N and cur is not NULL and in each iteration:
    1. Check if S[i] is equal to the data in the node cur or not. If it is increment i to (i+1) and move cur to the next node.
    2. If it isn’t then only move cur to the next node.
  3. If the loop ends, then check if i became N or not. If it was, then print Yes, otherwise print No.

Below is the implementation of the above approach:

C++




// C++ code for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Node of Linked List
class Node {
public:
    char data;
    Node* next;
 
    Node(char d)
    {
        data = d;
        next = NULL;
    }
};
 
// Function to check if the linked list contains
// a string as a subsequence
bool checkSub(Node* head, string S)
{
    Node* cur = head;
    int i = 0, N = S.size();
 
    while (i < N and cur) {
        if (S[i] == cur->data) {
            i += 1;
        }
        cur = cur->next;
    }
 
    if (i == N) {
        return 1;
    }
    return 0;
}
 
// Driver Code
int main()
{
    Node* head = new Node('b');
    head->next = new Node('r');
    head->next->next = new Node('a');
    head->next->next->next = new Node('d');
 
    string S = "bad";
 
    if (checkSub(head, S)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

Java




// Java code for the above approach
import java.util.*;
class GFG
{
 
  // Node of Linked List
  static class Node {
 
    char data;Node next;
 
    Node(char d)
    {
      data = d;
      next = null;
    }
 
  };
 
  // Function to check if the linked list contains
  // a String as a subsequence
  static boolean checkSub(Node head, String S)
  {
    Node cur = head;
    int i = 0, N = S.length();
 
    while (i < N && cur!=null) {
      if (S.charAt(i) == cur.data) {
        i += 1;
      }
      cur = cur.next;
    }
 
    if (i == N) {
      return true;
    }
    return false;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    Node head = new Node('b');
    head.next = new Node('r');
    head.next.next = new Node('a');
    head.next.next.next = new Node('d');
 
    String S = "bad";
 
    if (checkSub(head, S)) {
      System.out.print("Yes");
    }
    else {
      System.out.print("No");
    }
  }
}
 
// This code is contributed by gauravrajput1

Python3




# Python code for the above approach
 
# Node of Linked List
class Node:
    def __init__(self, data):
        self.data = data;
        self.next = None;
 
# Function to check if the linked list contains
# a String as a subsequence
def checkSub(head, S):
    cur = head;
    i = 0;
    N = len(S);
 
    while (i < N and cur != None):
        if (S[i] == cur.data):
            i += 1;
         
        cur = cur.next;
     
    if (i == N):
        return True;
     
    return False;
 
# Driver Code
if __name__ == '__main__':
    head =  Node('b');
    head.next =  Node('r');
    head.next.next =  Node('a');
    head.next.next.next =  Node('d');
 
    S = "bad";
 
    if (checkSub(head, S)):
        print("Yes");
    else:
        print("No");
     
 
# This code is contributed by Rajput-Ji

C#




// C# code for the above approach
using System;
 
public class GFG
{
 
  // Node of Linked List
  class Node {
 
    public char data;
    public Node next;
 
    public Node(char d)
    {
      data = d;
      next = null;
    }
 
  };
 
  // Function to check if the linked list contains
  // a String as a subsequence
  static bool checkSub(Node head, String S)
  {
    Node cur = head;
    int i = 0, N = S.Length;
 
    while (i < N && cur!=null) {
      if (S[i] == cur.data) {
        i += 1;
      }
      cur = cur.next;
    }
 
    if (i == N) {
      return true;
    }
    return false;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    Node head = new Node('b');
    head.next = new Node('r');
    head.next.next = new Node('a');
    head.next.next.next = new Node('d');
 
    String S = "bad";
 
    if (checkSub(head, S)) {
      Console.Write("Yes");
    }    
    else {
      Console.Write("No");
    }
  }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
       // JavaScript code for the above approach
 
       // Node of Linked List
       class Node {
 
           constructor(d) {
               this.data = d;
               this.next = null;
           }
       };
 
       // Function to check if the linked list contains
       // a string as a subsequence
       function checkSub(head, S) {
           let cur = head;
           let i = 0, N = S.length;
 
           while (i < N && cur) {
               if (S[i] == cur.data) {
                   i += 1;
               }
               cur = cur.next;
           }
 
           if (i == N) {
               return 1;
           }
           return 0;
       }
 
       // Driver Code
       let head = new Node('b');
       head.next = new Node('r');
       head.next.next = new Node('a');
       head.next.next.next = new Node('d');
 
       let S = "bad";
 
       if (checkSub(head, S)) {
           document.write("Yes");
       }
       else {
           document.write("No");
       }
 
 // This code is contributed by Potta Lokesh
   </script>

 
 

Output

Yes

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 


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