6. Reverse alternate nodes in Link List

The problem can be found at the following link: Question Link

My Approach

In this problem, our task is to traverse all even nodes, pick their alternate nodes, reverse those alternate nodes, and then attach this reversed list to the original linked list, thus changing its structure.

Here's the step-by-step approach I used:

• Initialize even and revOdd pointers to track alternative even and reversed nodes.

• Traverse the linked list with an itr pointer.

• For each pair where itr points to the first node and itr->next to the second:

  • Update even->next to skip the second node (itr->next->next) if it exists.

  • Move even to its new place.

  • Reverse the second node and link it to revOdd.

• Lastly, connect the reversed nodes in revOdd to the list's end.

Time and Auxiliary Space Complexity

• Time Complexity: The algorithm traverses the entire linked list once, so the time complexity is O(n), where n is the number of nodes in the linked list.

• Auxiliary Space Complexity: The algorithm uses a constant amount of extra space, so the auxiliary space complexity is O(1).

Code (C++)

class Solution {
public:
    void rearrange(struct Node *odd)
    {
        struct Node* even = odd;
        struct Node* revOdd = NULL;
        struct Node* itr = odd;
        
        while(itr) {
            struct Node* nextItr = itr->next;
            
            if (nextItr) {
                even->next = itr->next->next;
                if (even->next)
                    even = even->next;
                nextItr->next = revOdd;
                revOdd = nextItr;
            } else {
                break;
            }
            itr = even;
        }
        
        even->next = revOdd;
    }
};

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