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Compare two strings represented as linked lists

  • Difficulty Level : Basic
  • Last Updated : 24 Dec, 2021

Given two strings, represented as linked lists (every character is a node in a linked list). Write a function compare() that works similar to strcmp(), i.e., it returns 0 if both strings are the same, 1 if the first linked list is lexicographically greater, and -1 if the second string is lexicographically greater.

Examples: 

Input: list1 = g->e->e->k->s->a
          list2 = g->e->e->k->s->b
Output: -1
Explanation: “geeksb” is lexicographically greater than “geeksa”.

Input: list1 = g->e->e->k->s->a
          list2 = g->e->e->k->s
Output: 1
Explanation: “geeksa” is gerater than “geeks”.

Input: list1 = g->e->e->k->s
          list2 = g->e->e->k->s
Output: 0
Explanation: Both the strings are “geeks”.

 

Naive Approach: Traverse the linked list and form the strings. Now compare the strings to check if both the strings are same or not.

Time Complexity: O(N + M) where N and M are the lengths of the linked lists.
Auxiliary Space: O(N + M)

Efficient Approach: An efficient approach is to use two pointer algorithm. Follow the steps mentioned below:

  • Create two pointers and point to the starting of the linked lists.
  • Traverse both the lists. If the characters are same keep traversing and increment both the pointers.
  • If there is a mismatch:
    • If character at that node of list1 is lexicographically greater print 1 and stop iteration.
    • If character at that node of list2 is lexicographically greater print -1 and stop iteration.
  • If traversal reaches the end of one of the list:
    • If the end of the other list is also reached both strings are same.
    • Else the other one is the lexicographically bigger one.

Below is the implementation of the above approach.

C++




// C++ program to compare two strings
// represented as linked lists
#include<bits/stdc++.h>
using namespace std;
  
// Linked list Node structure
struct Node
{
    char c;
    struct Node *next;
};
  
// Function to create newNode in a linkedlist
Node* newNode(char c)
{
    Node *temp = new Node;
    temp->c = c;
    temp->next = NULL;
    return temp;
};
 
int compare(Node *list1, Node *list2)
{   
    // Traverse both lists. Stop when
    // either end of a linked list is reached
    // or current characters don't match
    while (list1 && list2 &&
           list1->c == list2->c)
    {        
        list1 = list1->next;
        list2 = list2->next;
    }
     
    // If both lists are not empty,
    // compare mismatching characters
    if (list1 && list2)
        return (list1->c > list2->c)? 1: -1;
 
    // If either of the two lists
    // has reached end
    if (list1 && !list2) return 1;
    if (list2 && !list1) return -1;
     
    // If none of the above conditions is true,
    // both lists have reached end
    return 0;
}
 
// Driver code
int main()
{
    Node *list1 = newNode('g');
    list1->next = newNode('e');
    list1->next->next = newNode('e');
    list1->next->next->next = newNode('k');
    list1->next->next->next->next =
        newNode('s');
    list1->next->next->next->next->next =
        newNode('b');
 
    Node *list2 = newNode('g');
    list2->next = newNode('e');
    list2->next->next = newNode('e');
    list2->next->next->next = newNode('k');
    list2->next->next->next->next =
        newNode('s');
    list2->next->next->next->next->next =
        newNode('a');
 
    cout << compare(list1, list2);
    return 0;
}

Java




// Java program to compare two strings represented as a linked list
 
// Linked List Class
class LinkedList {
 
    Node head;  // head of list
    static Node a, b;
 
    /* Node Class */
    static class Node {
 
        char data;
        Node next;
 
        // Constructor to create a new node
        Node(char d) {
            data = d;
            next = null;
        }
    }
 
    int compare(Node node1, Node node2) {
 
        if (node1 == null && node2 == null) {
            return 1;
        }
        while (node1 != null && node2 != null && node1.data == node2.data) {
            node1 = node1.next;
            node2 = node2.next;
        }
 
        // if the list are different in size
        if (node1 != null && node2 != null) {
            return (node1.data > node2.data ? 1 : -1);
        }
 
        // if either of the list has reached end
        if (node1 != null && node2 == null) {
            return 1;
        }
        if (node1 == null && node2 != null) {
            return -1;
        }
        return 0;
    }
 
    public static void main(String[] args) {
 
        LinkedList list = new LinkedList();
        Node result = null;
 
        list.a = new Node('g');
        list.a.next = new Node('e');
        list.a.next.next = new Node('e');
        list.a.next.next.next = new Node('k');
        list.a.next.next.next.next = new Node('s');
        list.a.next.next.next.next.next = new Node('b');
 
        list.b = new Node('g');
        list.b.next = new Node('e');
        list.b.next.next = new Node('e');
        list.b.next.next.next = new Node('k');
        list.b.next.next.next.next = new Node('s');
        list.b.next.next.next.next.next = new Node('a');
 
        int value;
        value = list.compare(a, b);
        System.out.println(value);
 
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python




# Python program to compare two strings represented as
# linked lists
 
# A linked list node structure
class Node:
 
    # Constructor to create a new node
    def __init__(self, key):
        self.c = key ;
        self.next = None
 
def compare(list1, list2):
     
    # Traverse both lists. Stop when either end of linked
    # list is reached or current characters don't match
    while(list1 and list2 and list1.c == list2.c):
        list1 = list1.next
        list2 = list2.next
 
    # If both lists are not empty, compare mismatching
    # characters
    if(list1 and list2):
        return 1 if list1.c > list2.c else -1
 
    # If either of the two lists has reached end
    if (list1 and not list2):
        return 1
 
    if (list2 and not list1):
        return -1
    return 0
 
# Driver program
 
list1 = Node('g')
list1.next = Node('e')
list1.next.next = Node('e')
list1.next.next.next = Node('k')
list1.next.next.next.next = Node('s')
list1.next.next.next.next.next = Node('b')
 
list2 = Node('g')
list2.next = Node('e')
list2.next.next = Node('e')
list2.next.next.next = Node('k')
list2.next.next.next.next = Node('s')
list2.next.next.next.next.next = Node('a')
 
print compare(list1, list2)
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#




// C# program to compare two
// strings represented as a linked list
using System;
 
// Linked List Class
public class LinkedList
{
 
    public Node head; // head of list
    public Node a, b;
 
    /* Node Class */
    public class Node
    {
 
        public char data;
        public Node next;
 
        // Constructor to create a new node
        public Node(char d)
        {
            data = d;
            next = null;
        }
    }
 
    int compare(Node node1, Node node2)
    {
 
        if (node1 == null && node2 == null)
        {
            return 1;
        }
        while (node1 != null &&
                node2 != null &&
                node1.data == node2.data)
        {
            node1 = node1.next;
            node2 = node2.next;
        }
 
        // if the list are different in size
        if (node1 != null && node2 != null)
        {
            return (node1.data > node2.data ? 1 : -1);
        }
 
        // if either of the list has reached end
        if (node1 != null && node2 == null)
        {
            return 1;
        }
        if (node1 == null && node2 != null)
        {
            return -1;
        }
        return 0;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        LinkedList list = new LinkedList();
 
        list.a = new Node('g');
        list.a.next = new Node('e');
        list.a.next.next = new Node('e');
        list.a.next.next.next = new Node('k');
        list.a.next.next.next.next = new Node('s');
        list.a.next.next.next.next.next = new Node('b');
 
        list.b = new Node('g');
        list.b.next = new Node('e');
        list.b.next.next = new Node('e');
        list.b.next.next.next = new Node('k');
        list.b.next.next.next.next = new Node('s');
        list.b.next.next.next.next.next = new Node('a');
 
        int value;
        value = list.compare(list.a, list.b);
        Console.WriteLine(value);
    }
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
 
// Javascript program to compare two
// strings represented as a linked list
 
// Linked List Class
var head; // head of list
    var a, b;
 
    /* Node Class */
     class Node {
 
 
        // Constructor to create a new node
         constructor(d) {
                this.data = d;
                this.next = null;
            }
    }
 
    function compare(node1,  node2) {
 
        if (node1 == null && node2 == null) {
            return 1;
        }
        while (node1 != null && node2 != null &&
        node1.data == node2.data) {
            node1 = node1.next;
            node2 = node2.next;
        }
 
        // if the list are different in size
        if (node1 != null && node2 != null) {
            return (node1.data > node2.data ? 1 : -1);
        }
 
        // if either of the list has reached end
        if (node1 != null && node2 == null) {
            return 1;
        }
        if (node1 == null && node2 != null) {
            return -1;
        }
        return 0;
    }
 
     
 
var result = null;
 
        a = new Node('g');
        a.next = new Node('e');
        a.next.next = new Node('e');
        a.next.next.next = new Node('k');
        a.next.next.next.next = new Node('s');
        a.next.next.next.next.next = new Node('b');
 
        b = new Node('g');
        b.next = new Node('e');
        b.next.next = new Node('e');
        b.next.next.next = new Node('k');
        b.next.next.next.next = new Node('s');
        b.next.next.next.next.next = new Node('a');
 
        var value;
        value = compare(a, b);
        document.write(value);
 
 
// This code contributed by gauravrajput1
 
</script>
Output
1

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

 


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