In this article, we are going to see how to check a single linked list whether it's palindrome or not? This problem can be featured in interview coding rounds. Submitted by Radib Kar, on January 11, 2019

## Problem statement

Given a singly linked list, determine if it is a palindrome. Write a Boolean function to output true or false based on result.

Input Example:

```    Example 1:
Input: 1->2->3->NULL
Output: false

Example 2:
Input: 1->2->2->1->NULL
Output: true
```

## Solution

For a single linked we don't have a prev pointer to traverse back from the end of the list and to check with corresponding start positions, i.e., what we normally do to check a string to be palindrome. But for a single linked list we only can traverse forward.

## Algorithm

```1.  Declare a global temp variable
ListNode* temp;
//function which pass head pointer to check whether it's palindrome
End Function

3.  Recursive function :
boolcheckPalindrome(ListNode* p)
a.  Check for base case
IF p==NULL
return true;
b.  bool result=checkPalindrome(p->next) &(temp->data==p->data);
c.  Traverse tempby one step.
temp=temp->next;
return result;
END FUNCTION
```

Attention here!!!

For any statement like, bool result=checkPalindrome(p->next) &(temp->data==p->data);

Where there is two expression linked with ‘&' or ‘&&' operator the first expression is evaluated first and in case if it's true, only the second expression is evaluated. If the first expression is false, second expression is not at all evaluated.

So for this particular statement,

checkPalindrome(p->next) is kept being evaluated until recursion reaches the base case & (temp->data==p->data) is not at all evaluated until the first expression results true .

Let's evaluate how the function works to check whether it's a palindrome.

For example 2:

```The list is 1->2->2->1->NULL

For the rest of the evaluation a node is represented by
its value for better visualization & understanding
checkPalindrome(1)   //record no 0
base case not matched
bool result = checkPalindrome(1->next) &[(temp->data==p->data)]
[]=not evaluated part ,bold and underlined too
Call to checkPalindrome(1->next)->checkPalindrome(2)
----------------------------------------------------------

checkPalindrome(2) //record no 1
base case not matched
bool result = checkPalindrome(2->next) &[(temp->data==p->data)]
[]=not evaluated part ,bold and underlined too
Call to checkPalindrome(2->next)->checkPalindrome(2)
----------------------------------------------------------

checkPalindrome(2) //record no 2
base case not matched
bool result = checkPalindrome(2->next) &[(temp->data==p->data)]
[]=not evaluated part ,bold and underlined too
Call to checkPalindrome(2->next)->checkPalindrome(1)
----------------------------------------------------------

checkPalindrome(1) //record no 3
base case not matched
bool result = checkPalindrome(1->next) &[(temp->data==p->data)]
[]=not evaluated part ,bold and underlined too
Call to checkPalindrome(1->next)->checkPalindrome(NULL)
----------------------------------------------------------

checkPalindrome(NULL) //record no 4
base case matched
it returns true
----------------------------------------------------------

At record no 3:
bool result = checkPalindrome(1->next) &[(temp->data==p->data)]
checkPalindrome(1->next) is true , hence next expression evaluated
Now, temp is the global variable which was initialized head
in the isPalindrome function, temp hasn't been changed yet
Temp->data=1
P at record 3 is 1 (data part 1)
Thus temp->data==p->data
Hence result=true & true= true
Now temp=temp->next=2
And this record returns true to the
caller function(checkPalindrome(2->next)) (At record 2)
----------------------------------------------------------

At record no 2:
bool result = checkPalindrome(2->next) &[(temp->data==p->data)]
checkPalindrome(2->next) is true (returned from record no 3),
hence next expression is evaluated
Temp->data=2
P at record 2 is 2 (data part 2)
Thus temp->data==p->data
Hence result=true & true= true
Now temp=temp->next=2
And this record returns true to the caller
functioncheckPalindrome(2->next) (At record 1)
----------------------------------------------------------

At record no 1:
bool result = checkPalindrome(2->next) &[(temp->data==p->data)]
checkPalindrome(2->next) is true (returned from record no 2) ,
hence next expression is evaluated
Temp->data=2
P at record 1 is 2 (data part 2)
Thus temp->data==p->data
Hence result=true & true= true
Now temp=temp->next=1
And this record returns true to the caller
functioncheckPalindrome(1->next) (At record 0)
----------------------------------------------------------

At record no 0:
bool result = checkPalindrome(1->next) &[(temp->data==p->data)]
checkPalindrome(1->next) is true , hence next expression evaluated
Temp->data=1
P at record 0 is 1 (data part 1)
Thus temp->data==p->data
Hence result=true & true= true
Now temp=temp->next=NULL
And this record returns true to the

Thus it returns true.
Of course 1->2->2->1->NULL is a valid palindrome.
Similarly you can check for 1->2->3->NULL. Do it yourself!!

Hints:
(It's not a Palindrome)
It will return false at some record & so on.
```

C++ implementation

```#include <bits/stdc++.h>
using namespace std;

class ListNode{ //list node structure
public:
int data;
ListNode* next;
};

ListNode* creatnode(int d){ //create node for single linked list
ListNode* temp=(ListNode*)malloc(sizeof(ListNode));
temp->data=d;
temp->next=NULL;
return temp;
}

printf("displayig the converted list...\n");
while(current!=NULL){ //traverse until current node isn't NULL
if(current->next)
printf("%d->",current->data);
else
printf("%d->NULL\n",current->data);
current=current->next; // go to next node
}
}

ListNode* temp; //global temp variable
bool checkPalindrome(ListNode* p){ //recursive function to check for palindrome
if(p==NULL)
return true;
bool result=checkPalindrome(p->next) & (temp->data==p->data);
temp=temp->next;
return result;
}

bool isPalindrome(ListNode* head) { //functio to check for palindrome
}

int main(){
printf("creating the linked list by inserting new nodes at the end\n");
printf("enter 0 to stop building the list, else enter any integer\n");
int k;
ListNode* curr,*temp;
cout<<"enter list to check whether palindrom...\n";
scanf("%d",&k);
ListNode* head=creatnode(k); //buliding list, first node
scanf("%d",&k);

///////////////////inserting at the end//////////////////////
while(k){
curr=creatnode(k);
temp->next=curr;//appending each node
temp=temp->next;
scanf("%d",&k);
}
cout<<"displaying list1...\n";
cout<<"Palindrome list\n";
else
cout<<"Not palindrome\n";

return 0;
}
```

Output

```First run:
enter 0 to stop building the list, else enter any integer
enter list to check whether palindrom...
1 2 2 1 0
displaying list1...
displayig the converted list...
1->2->2->1->NULL
Palindrome list

Second run:
enter 0 to stop building the list, else enter any integer
enter list to check whether palindrom...
1 2 3 2 0
displaying list1...
displayig the converted list...
1->2->3->2->NULL
Not palindrome
```
The text passage describes a problem statement and solution for checking if a singly linked list is a palindrome. It includes examples, an algorithm, and a C++ implementation with output.
no_question

the passage does not contain a mathematical question or a question that requires a calculation. it is a programming problem that involves logic and data structures, not a mathematical question.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
In this article, we are going to see how to check a single linked list whether it's palindrome or not? This problem can be featured in interview coding rounds. Submitted by Radib Kar, on January 11, 2019

## Problem statement

Given a singly linked list, determine if it is a palindrome. Write a Boolean function to output true or false based on result.

Input Example:

```    Example 1:
Input: 1->2->3->NULL
Output: false

Example 2:
Input: 1->2->2->1->NULL
Output: true
```

## Solution

For a single linked we don't have a prev pointer to traverse back from the end of the list and to check with corresponding start positions, i.e., what we normally do to check a string to be palindrome. But for a single linked list we only can traverse forward.

## Algorithm

```1.  Declare a global temp variable
ListNode* temp;
//function which pass head pointer to check whether it's palindrome
End Function

3.  Recursive function :
boolcheckPalindrome(ListNode* p)
a.  Check for base case
IF p==NULL
return true;
b.  bool result=checkPalindrome(p->next) &(temp->data==p->data);
c.  Traverse tempby one step.
temp=temp->next;
return result;
END FUNCTION
```

Attention here!!!

For any statement like, bool result=checkPalindrome(p->next) &(temp->data==p->data);

Where there is two expression linked with ‘&' or ‘&&' operator the first expression is evaluated first and in case if it's true, only the second expression is evaluated. If the first expression is false, second expression is not at all evaluated.

So for this particular statement,

checkPalindrome(p->next) is kept being evaluated until recursion reaches the base case & (temp->data==p->data) is not at all evaluated until the first expression results true .

Let's evaluate how the function works to check whether it's a palindrome.

For example 2:

```The list is 1->2->2->1->NULL

For the rest of the evaluation a node is represented by
its value for better visualization & understanding
checkPalindrome(1)   //record no 0
base case not matched
bool result = checkPalindrome(1->next) &[(temp->data==p->data)]
[]=not evaluated part ,bold and underlined too
Call to checkPalindrome(1->next)->checkPalindrome(2)
----------------------------------------------------------

checkPalindrome(2) //record no 1
base case not matched
bool result = checkPalindrome(2->next) &[(temp->data==p->data)]
[]=not evaluated part ,bold and underlined too
Call to checkPalindrome(2->next)->checkPalindrome(2)
----------------------------------------------------------

checkPalindrome(2) //record no 2
base case not matched
bool result = checkPalindrome(2->next) &[(temp->data==p->data)]
[]=not evaluated part ,bold and underlined too
Call to checkPalindrome(2->next)->checkPalindrome(1)
----------------------------------------------------------

checkPalindrome(1) //record no 3
base case not matched
bool result = checkPalindrome(1->next) &[(temp->data==p->data)]
[]=not evaluated part ,bold and underlined too
Call to checkPalindrome(1->next)->checkPalindrome(NULL)
----------------------------------------------------------

checkPalindrome(NULL) //record no 4
base case matched
it returns true
----------------------------------------------------------

At record no 3:
bool result = checkPalindrome(1->next) &[(temp->data==p->data)]
checkPalindrome(1->next) is true , hence next expression evaluated
Now, temp is the global variable which was initialized head
in the isPalindrome function, temp hasn't been changed yet
Temp->data=1
P at record 3 is 1 (data part 1)
Thus temp->data==p->data
Hence result=true & true= true
Now temp=temp->next=2
And this record returns true to the
caller function(checkPalindrome(2->next)) (At record 2)
----------------------------------------------------------

At record no 2:
bool result = checkPalindrome(2->next) &[(temp->data==p->data)]
checkPalindrome(2->next) is true (returned from record no 3),
hence next expression is evaluated
Temp->data=2
P at record 2 is 2 (data part 2)
Thus temp->data==p->data
Hence result=true & true= true
Now temp=temp->next=2
And this record returns true to the caller
functioncheckPalindrome(2->next) (At record 1)
----------------------------------------------------------

At record no 1:
bool result = checkPalindrome(2->next) &[(temp->data==p->data)]
checkPalindrome(2->next) is true (returned from record no 2) ,
hence next expression is evaluated
Temp->data=2
P at record 1 is 2 (data part 2)
Thus temp->data==p->data
Hence result=true & true= true
Now temp=temp->next=1
And this record returns true to the caller
functioncheckPalindrome(1->next) (At record 0)
----------------------------------------------------------

At record no 0:
bool result = checkPalindrome(1->next) &[(temp->data==p->data)]
checkPalindrome(1->next) is true , hence next expression evaluated
Temp->data=1
P at record 0 is 1 (data part 1)
Thus temp->data==p->data
Hence result=true & true= true
Now temp=temp->next=NULL
And this record returns true to the

Thus it returns true.
Of course 1->2->2->1->NULL is a valid palindrome.
Similarly you can check for 1->2->3->NULL. Do it yourself!!

Hints:
(It's not a Palindrome)
It will return false at some record & so on.
```

C++ implementation

```#include <bits/stdc++.h>
using namespace std;

class ListNode{ //list node structure
public:
int data;
ListNode* next;
};

ListNode* creatnode(int d){ //create node for single linked list
ListNode* temp=(ListNode*)malloc(sizeof(ListNode));
temp->data=d;
temp->next=NULL;
return temp;
}

printf("displayig the converted list...\n");
while(current!=NULL){ //traverse until current node isn't NULL
if(current->next)
printf("%d->",current->data);
else
printf("%d->NULL\n",current->data);
current=current->next; // go to next node
}
}

ListNode* temp; //global temp variable
bool checkPalindrome(ListNode* p){ //recursive function to check for palindrome
if(p==NULL)
return true;
bool result=checkPalindrome(p->next) & (temp->data==p->data);
temp=temp->next;
return result;
}

bool isPalindrome(ListNode* head) { //functio to check for palindrome
}

int main(){
printf("creating the linked list by inserting new nodes at the end\n");
printf("enter 0 to stop building the list, else enter any integer\n");
int k;
ListNode* curr,*temp;
cout<<"enter list to check whether palindrom...\n";
scanf("%d",&k);
ListNode* head=creatnode(k); //buliding list, first node
scanf("%d",&k);

///////////////////inserting at the end//////////////////////
while(k){
curr=creatnode(k);
temp->next=curr;//appending each node
temp=temp->next;
scanf("%d",&k);
}
cout<<"displaying list1...\n";
cout<<"Palindrome list\n";
else
cout<<"Not palindrome\n";

return 0;
}
```

Output

```First run:
enter 0 to stop building the list, else enter any integer
enter list to check whether palindrom...
1 2 2 1 0
displaying list1...
displayig the converted list...
1->2->2->1->NULL
Palindrome list

Second run:
enter 0 to stop building the list, else enter any integer
enter list to check whether palindrom...
1 2 3 2 0
displaying list1...
displayig the converted list...
1->2->3->2->NULL
Not palindrome
```
user
The text passage describes a problem statement and solution for checking if a singly linked list is a palindrome. It includes examples, an algorithm, and a C++ implementation with output.

TYPE: NO_QUESTION

The passage does not contain a mathematical question or a question that requires a calculation. It is a programming problem that involves logic and data structures, not a mathematical question.