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GO Classes Test Series 2024 | Mock GATE | Test 13 | Question: 37

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Consider the null-terminated linked list of four integers $\textsf{1->2->3->4->NULL},$ and the variable 'list' points to the head of the linked list. Upon running the provided code, the linked list gets divided into two lists, as illustrated in the diagram below.

The C code that was executed is as follows:

struct ListNode {
    int val;
    struct ListNode *next;
}list, list2;
list2 = list->next->next->next; 
list2->next = list; 
//LINE X 
// LINE Y
list->next->next = NULL; 
list2->next->next = NULL;

To successfully accomplish the task, what should be inserted at $\textsf{LINE X}$ and $\textsf{LINE Y}?$

  1.  LINE X: list->next->next->next = list->next;
 LINE Y: list = list->next->next;
  2.  LINE X: list->next->next = list->next;
 LINE Y: list = list->next->next; 
  3.  LINE X: list = list->next->next;
 LINE Y:  list->next->next->next = list->next; 
  4.  LINE X:list->next->next>next = list->next>next;
 LINE Y: list = list->next->next;
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2 Answers

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Given linked list: list → 1 → 2 → 3 → 4→ NULL where list is a pointer variable pointing to first node of linked list or simply it is head pointer for better understanding.

now in code line a list2 pointer variable is given,

list2 = list → next → next → next

       = 1 → next → next → next

      = 2 → next → next

       = 3 → next

       = 4

list2 is pointing to node having data value 4.

list2 → next = list means 4 is pointing to 1.

so up till this point we got one circular linked list.

now we want two node as    list → 3 → 2 and list2 → 4 → 1

now what we have to change the next pointer of 3 pointing to 2  and for that we have to perform

list → next → next → next = list → next ;

3 → next = 2;

now we have to move the list pointer pointing to 3 and for that we have to perform list = list → next → next;

and you got result.

------------------ Variation --------------------------------------------------------

PS:

We can also do it another way

after making circular list, directly take list pointer and point that to 3

list = list → next → next;

then we can make pointing list node next pointer to 2,

list → next = list2 → next → next;

3 → next = 4 → next → next;

3 → next = 1 → next;

and finally we made 3 → next = 2;

so if it is MSQ then this also be correct way.
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Original linked list : list  > 1 > 2  > 3 > 4 | NULL

In Option A :

list > next  > next > next = list > next ;

This makes the list as

list > 1  > 2  > 3  > 2

list = list > next > next;

This further updates as :

list  > 3  > 2

later this is made NULL terminating linked list by the code.

Hence Option A satisfies the linked list given in the question.

open for any correction :)
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