Compare two Linked List of Strings
Last Updated :
15 Mar, 2023
Given two linked lists L1 and L2 in which in every node a string is stored. The task is to check whether the strings combining all the nodes are similar or not.
Examples:
Input: L1 = ["He", "llo", "wor", "ld"],
L2 = ["H", "e", "ll", "owo", "r", "ld"]
Output: true
Explanation: both lists makes the string of "Helloworld".
Input: L1 = ["w", "o", "l", "d"],
L2 = ["wo", "d", "rl"]
Output: false
Explanation: L1 makes "world" but L2 makes "wodrl" both are different.
Input: L1 = ["w", "", "orl", "d"],
L2 = ["worl", "", "", "", "d"]
Output: true
Explanation: both lists makes the string of "world".
Naive Approach: This is the simple approach. Traverse both the lists and store their values in another string, then compare both strings if equal return true else return false.
Below is the implementation of the above approach.
C++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Structure of Node of linked list
class Node {
public:
string s;
Node* next;
Node(string s)
{
this->s = s;
next = NULL;
}
};
// Function to compare if two linkedlists
// are similar or not
bool compare(Node* list1, Node* list2)
{
// Declare string variables to store
// the strings formed from the linked lists
string s1, s2;
while (list1 != NULL) {
s1 += list1->s;
list1 = list1->next;
}
while (list2 != NULL) {
s2 += list2->s;
list2 = list2->next;
}
return s1 == s2;
}
// Driver Code
int main()
{
Node* n1 = new Node("w");
Node* head1 = n1;
Node* n2 = new Node("");
Node* n3 = new Node("orl");
Node* n4 = new Node("d");
Node* n5 = new Node("worl");
Node* head2 = n5;
Node* n6 = new Node("");
Node* n7 = new Node("");
Node* n8 = new Node("nd");
n1->next = n2;
n2->next = n3;
n3->next = n4;
n5->next = n6;
n6->next = n7;
n7->next = n8;
if (compare(head1, head2) == true)
cout << "true";
else
cout << "false";
return 0;
}
// Java program for the above approach
import java.util.*;
public class GFG {
// Structure of Node of linked list
public static class Node {
public String s;
public Node next;
public Node(String s) {
this.s = s;
next = null;
}
};
// Function to compare if two linkedlists
// are similar or not
public static boolean compare(Node list1, Node list2) {
// Declare String variables to store
// the Strings formed from the linked lists
String s1 = "", s2 = "";
while (list1 != null) {
s1 += list1.s;
list1 = list1.next;
}
while (list2 != null) {
s2 += list2.s;
list2 = list2.next;
}
return s1.equals(s2);
}
// Driver Code
public static void main(String[] args) {
Node n1 = new Node("w");
Node head1 = n1;
Node n2 = new Node("");
Node n3 = new Node("orl");
Node n4 = new Node("d");
Node n5 = new Node("worl");
Node head2 = n5;
Node n6 = new Node("");
Node n7 = new Node("");
Node n8 = new Node("nd");
n1.next = n2;
n2.next = n3;
n3.next = n4;
n5.next = n6;
n6.next = n7;
n7.next = n8;
if (compare(head1, head2) == true)
System.out.println("true");
else
System.out.println("false");
}
}
// this code is contributed by bhardwajji
# Python code for the above approach
# Structure of Node of linked list
class Node:
def __init__(self,str):
self.s = str
self.next = None
# Function to compare if two linkedlists
# are similar or not
def compare(list1, list2):
# Declare string variables to store
# the strings formed from the linked lists
s1,s2 = "",""
while (list1 != None):
s1 += list1.s
list1 = list1.next
while (list2 != None):
s2 += list2.s
list2 = list2.next
return s1 == s2
# Driver Code
n1 = Node("w")
head1 = n1
n2 = Node("")
n3 = Node("orl")
n4 = Node("d")
n5 = Node("worl")
head2 = n5
n6 = Node("")
n7 = Node("")
n8 = Node("nd")
n1.next = n2
n2.next = n3
n3.next = n4
n5.next = n6
n6.next = n7
n7.next = n8
if (compare(head1, head2) == True):
print("true")
else:
print("false")
# This code is contributed by shinjanpatra
// C# program for above approach
using System;
using System.Collections.Generic;
class GFG{
// Structure of Node of linked list
public class Node {
public String s;
public Node next;
public Node(String s)
{
this.s = s;
next = null;
}
};
// Function to compare if two linkedlists
// are similar or not
static bool compare(Node list1, Node list2)
{
// Declare String variables to store
// the Strings formed from the linked lists
String s1 ="", s2="";
while (list1 != null) {
s1 += list1.s;
list1 = list1.next;
}
while (list2 != null) {
s2 += list2.s;
list2 = list2.next;
}
return s1 == s2;
}
// Driver Code
public static void Main()
{
Node n1 = new Node("w");
Node head1 = n1;
Node n2 = new Node("");
Node n3 = new Node("orl");
Node n4 = new Node("d");
Node n5 = new Node("worl");
Node head2 = n5;
Node n6 = new Node("");
Node n7 = new Node("");
Node n8 = new Node("nd");
n1.next = n2;
n2.next = n3;
n3.next = n4;
n5.next = n6;
n6.next = n7;
n7.next = n8;
if (compare(head1, head2) == true)
Console.Write("true");
else
Console.Write("false");
}
}
// This code is contributed by jana_sayantan.
<script>
// JavaScript code for the above approach
// Structure of Node of linked list
class Node {
constructor(str) {
this.s = str;
this.next = null;
}
};
// Function to compare if two linkedlists
// are similar or not
function compare(list1, list2)
{
// Declare string variables to store
// the strings formed from the linked lists
let s1 = "", s2 = "";
while (list1 != null) {
s1 += list1.s;
list1 = list1.next;
}
while (list2 != null) {
s2 += list2.s;
list2 = list2.next;
}
return s1 == s2;
}
// Driver Code
let n1 = new Node("w");
let head1 = n1;
let n2 = new Node("");
let n3 = new Node("orl");
let n4 = new Node("d");
let n5 = new Node("worl");
let head2 = n5;
let n6 = new Node("");
let n7 = new Node("");
let n8 = new Node("nd");
n1.next = n2;
n2.next = n3;
n3.next = n4;
n5.next = n6;
n6.next = n7;
n7.next = n8;
if (compare(head1, head2) == true)
document.write("true");
else
document.write("false");
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N+M) where N and M are lengths of the strings.
Auxiliary Space: O(N+M)
Efficient Approach: This problem can be solved by using Two pointers approach. Follow the steps below to solve the given problem.
- Traverse both the lists while maintaining two pointers for the characters.
- Compare each of the characters separately. If different, return false otherwise, continue to compare.
- If the pointer's value becomes the size of a string in a node, then move to the next node.
- Repeat the steps until the nodes do not become NULL.
Below is the implementation of the above approach.
C++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Structure of Node of a linked list
class Node {
public:
string s;
Node* next;
Node(string s)
{
this->s = s;
next = NULL;
}
};
// Compare function
bool compare(Node* list1, Node* list2)
{
int i = 0, j = 0;
while (list1 != NULL && list2 != NULL) {
while (i < list1->s.size() &&
j < list2->s.size()) {
if (list1->s[i] != list2->s[j])
return false;
i++;
j++;
}
if (i == list1->s.size()) {
i = 0;
list1 = list1->next;
}
if (j == list2->s.size()) {
j = 0;
list2 = list2->next;
}
}
return list1 == NULL && list2 == NULL;
}
// Driver Code
int main()
{
Node* n1 = new Node("w");
Node* head1 = n1;
Node* n2 = new Node("");
Node* n3 = new Node("orl");
Node* n4 = new Node("d");
Node* n5 = new Node("worl");
Node* head2 = n5;
Node* n6 = new Node("");
Node* n7 = new Node("");
Node* n8 = new Node("nd");
n1->next = n2;
n2->next = n3;
n3->next = n4;
n5->next = n6;
n6->next = n7;
n7->next = n8;
if (compare(head1, head2) == true)
cout << "true";
else
cout << "false";
return 0;
}
// Java program for above approach
public class Compare {
// Structure of Node of a linked list
static class Node {
String s;
Node next;
Node(String s)
{
this.s = s;
next = null;
}
}
// Compare function
static boolean compare(Node list1, Node list2)
{
int i = 0, j = 0;
while (list1 != null && list2 != null) {
while (i < list1.s.length()
&& j < list2.s.length()) {
if (list1.s.charAt(i) != list2.s.charAt(j))
return false;
i++;
j++;
}
if (i == list1.s.length()) {
i = 0;
list1 = list1.next;
}
if (j == list2.s.length()) {
j = 0;
list2 = list2.next;
}
}
return list1 == null && list2 == null;
}
// Driver Code
public static void main(String[] args)
{
Node n1 = new Node("w");
Node head1 = n1;
Node n2 = new Node("");
Node n3 = new Node("orl");
Node n4 = new Node("d");
Node n5 = new Node("worl");
Node head2 = n5;
Node n6 = new Node("");
Node n7 = new Node("");
Node n8 = new Node("nd");
n1.next = n2;
n2.next = n3;
n3.next = n4;
n5.next = n6;
n6.next = n7;
n7.next = n8;
if (compare(head1, head2) == true)
System.out.println("true");
else
System.out.println("false");
}
}
// This code is contributed by Lovely Jain
# Python program recursively print all sentences that can be
# Structure of Node of a linked list
class Node:
def __init__(self,s):
self.s = s
self.next = None
# Compare function
def compare(list1, list2):
i,j = 0,0
while (list1 != None and list2 != None):
while (i < len(list1.s) and
j < len(list2.s)):
if (list1.s[i] != list2.s[j]):
return False
i += 1
j += 1
if (i == len(list1.s)):
i = 0
list1 = list1.next
if (j == len(list2.s)):
j = 0
list2 = list2.next
return list1 == None and list2 == None
# Driver Code
n1 = Node("w")
head1 = n1
n2 = Node("")
n3 = Node("orl")
n4 = Node("d")
n5 = Node("worl")
head2 = n5
n6 = Node("")
n7 = Node("")
n8 = Node("nd")
n1.next = n2
n2.next = n3
n3.next = n4
n5.next = n6
n6.next = n7
n7.next = n8
if (compare(head1, head2) == True):
print("true")
else:
print("false")
# This code is contributed by shinjanpatra
// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// Compare function
static bool compare(Node list1, Node list2)
{
int i = 0, j = 0;
while (list1 != null && list2 != null) {
while (i < list1.s.Length && j < list2.s.Length) {
if (list1.s[i] != list2.s[j])
return false;
i++;
j++;
}
if (i == list1.s.Length) {
i = 0;
list1 = list1.next;
}
if (j == list2.s.Length) {
j = 0;
list2 = list2.next;
}
}
return list1 == null && list2 == null;
}
// Driver code
public static void Main(string[] args){
Node n1 = new Node("w");
Node head1 = n1;
Node n2 = new Node("");
Node n3 = new Node("orl");
Node n4 = new Node("d");
Node n5 = new Node("worl");
Node head2 = n5;
Node n6 = new Node("");
Node n7 = new Node("");
Node n8 = new Node("nd");
n1.next = n2;
n2.next = n3;
n3.next = n4;
n5.next = n6;
n6.next = n7;
n7.next = n8;
if (compare(head1, head2) == true)
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
// Structure of Node of a linked list
public class Node{
public String s;
public Node next;
public Node(String s)
{
this.s = s;
next = null;
}
}
// This code is contributed by subhamgoyal2014.
<script>
// JavaScript program recursively print all sentences that can be
// Structure of Node of a linked list
class Node {
constructor(s)
{
this.s = s;
this.next = null;
}
}
// Compare function
function compare(list1, list2)
{
let i = 0, j = 0;
while (list1 != null && list2 != null) {
while (i < list1.s.length &&
j < list2.s.length) {
if (list1.s[i] != list2.s[j])
return false;
i++;
j++;
}
if (i == list1.s.length) {
i = 0;
list1 = list1.next;
}
if (j == list2.s.length) {
j = 0;
list2 = list2.next;
}
}
return list1 == null && list2 == null;
}
// Driver Code
let n1 = new Node("w");
let head1 = n1;
let n2 = new Node("");
let n3 = new Node("orl");
let n4 = new Node("d");
let n5 = new Node("worl");
let head2 = n5;
let n6 = new Node("");
let n7 = new Node("");
let n8 = new Node("nd");
n1.next = n2;
n2.next = n3;
n3.next = n4;
n5.next = n6;
n6.next = n7;
n7.next = n8;
if (compare(head1, head2) == true)
document.write("true");
else
document.write("false");
// This code is contributed by shinjanpatra
</script>
Time Complexity: O(N+M) where N and M are length of the strings.
Auxiliary Space: O(1).
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