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Reverse Nodes In Even Length Groups  Leetcode Solution
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LeetCode: Reverse Nodes In Even Length Groups Leetcode Solution
Difficulty: Medium
Topics: linkedlist
Problem Description:
Given a linked list, reverse the nodes of each evenindexed group and leave the oddindexed groups as is.
Example:
Input: 1>2>3>4>5>6>7>8 Output: 1>4>3>2>5>8>7>6
Explanation: The evenindexed groups are (1, 2), (3, 4), (5, 6), and (7, 8). We swap the nodes within each group to get the result.
Solution:
First, we create a helper function that can reverse the nodes of a linked list given the head and tail nodes. We use this helper function to reverse the nodes in each evenindexed group.
Next, we traverse the linked list, keeping track of the evenindexed nodes (indexed starting at 0). When we reach the end of an evenindexed group, we call the helper function to reverse the nodes within that group. We repeat this process until we reach the end of the list.
Algorithm:

Define a function to reverse the nodes of a linked list given the head and tail nodes. The function should return the new head and tail nodes.

Define a function to reverse nodes in the linked list at even indices. This functions walks to the 2n+1th Node starting from the beginning of the list.

Define a new function that walks to the end of each evenindexed group and reverses the nodes within that group using the reverse function defined in step 1.

Traverse the linked list, keeping track of the evenindexed nodes (indexed starting at 0). When we reach the end of an evenindexed group (i.e. the (2n+1)th node, assuming 0indexing), we reverse the nodes within that group using the helper function.

Return the new head of the modified linked list.
Pseudocode:
// Helper function to reverse the nodes of a linked list given the head and tail nodes function reverse(head, tail): prev = null current = head while (prev != tail): next = current.next current.next = prev prev = current current = next return (tail, head)
// Function to reverse nodes in evenindexed groups function reverseEvenGroups(head): if (head == null): return null even = head.next if (even == null): return head odd = head while (even != null): (even, odd).next = (odd, even).next odd.next = even.next even.next = odd even = odd.next if (even == null): break odd = odd.next even = even.next return head
// Main function to reverse nodes in evenindexed groups function reverseNodesInEvenLengthGroups(head): if (head == null): return null even = head.next if (even == null): return head odd = head while (even != null): odd.next = reverseEvenGroups(odd.next) odd = even even = even.next if (even == null): break odd = odd.next even = even.next return head
Time complexity:
The time complexity of this algorithm is O(n), where n is the number of nodes in the linked list. We traverse the linked list once, and for each evenindexed group, we reverse the nodes in that group using the reverse function, which takes O(k) time, where k is the number of nodes in the group (which is at most n/2, since we only consider evenindexed groups). Therefore, the total time complexity of the algorithm is O(n).
Reverse Nodes In Even Length Groups Solution Code
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