Reverse linked list 2- interviewbit

#1

I understood this solution using Gfg but it is giving TLE. How can i improve the solution.

ide->https://ide.codingblocks.com/s/243527

ques->https://www.interviewbit.com/problems/reverse-link-list-ii/

#2

@mikkyimran hey check this pseudo code:

ListNode* Solution::reverseBetween(ListNode* A, int B, int C) {
    if(B != 1){ // if B is not the first element 
        A->next = reverseBetween(A->next, B-1, C-1);// pass the next element decreasing B and C
        return A;
    }
    else{ // if B is the first element
        ListNode *prev=NULL, *curr=A, *next=NULL;
        int count=0;
        while(count<C){ // reverse the list till C 
            next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
            count++;
        }
        A->next = curr;// Append the first list element to the C+1 element
        return prev;// return the new start
    }
}
#3

How can i optimise my current solution?

#4

@mikkyimran The solution which I have given is optimised one.

#5

@rishabhmahajan546 can u pls explain with an example . i did a dry run but was not able to get the answer.

#6

@mikkyimran hey I am using recursion in this code we will call recursion for head->next and it will reverse from m to n and then we have to handle the case when first element is itself m ,so for that hmne simple reverse wala logic lgaya hai iterative one from m to n.Hope you get it

#7

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