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

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:

Create variable i, and initialise it with 0. Also, create a pointer cur that points at the head of the linked list.
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.
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# 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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Check if a string is present in the given Linked List as a subsequence

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