Compare two strings represented as linked lists
Last Updated :
13 Jul, 2023
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 greater 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 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 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# 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
<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>
Time Complexity: O(N + M)
Auxiliary Space: O(1)
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