The O(n) Club: Reverse Nodes in K-Group: Now You’re Juggling Pointers and Sanity

The O(n) Club: Reverse Nodes in K-Group — Now You’re Juggling Pointers and Sanity

⚡ TL;DR

Reverse your singly-linked list in groups of k nodes. If there’s a leftover group that’s smaller than k, pretend you didn’t even see it. Brute force = stack or array (which works, but makes you look like you’re prepping for LinkedIn’s “how NOT to do tech interviews” seminar). Real adults use pointer magic:

// Brute force (everyone's first bad idea)
public ListNode reverseKGroup(ListNode head, int k) {
    Stack
<listnode> stack = new Stack<>();
    ListNode dummy = new ListNode(0);
    ListNode curr = head, tail = dummy;
    while (curr != null) {
        int count = 0;
        ListNode temp = curr;
        while (temp != null && count < k) {
            stack.push(temp);
            temp = temp.next;
            count++;
        }
        if (count == k) {
            while (!stack.isEmpty()) {
                tail.next = stack.pop();
                tail = tail.next;
            }
            tail.next = temp;
            curr = temp;
        } else {
            tail.next = curr; // Just give up
            break;
        }
    }
    return dummy.next;
}
</listnode>

🧠 How Devs Usually Mess This Up

Oh, the carnage. Common fails include:

• Reversing every group — including the runt at the end. The last mini-group must stay in witness protection, unreversed.
• Forgetting about links. Skip a pointer update and your list is now two lists (or zero). Hope you weren’t attached.
• Setting prev = null. Now you’ve cut your beautiful reversed group off from civilization. Next time, use group boundaries for a happier ending.

🔍 What’s Actually Going On

Imagine a delivery line where every batch of k packages flips places for dramatic effect—except the last handful. You let them slide, because nobody likes drama right before home time. The core moves:

• Find the k-th node ahead. No full group? Stop.
• Reverse exactly those k nodes with your spiciest three-pointer maneuver.
• Reconnect the reversed chunk back to main street (the rest of the list).

Bonus tip: a dummy node at the start saves you from annoying edge cases. Like a silent partner who never takes credit or bugs you for status updates.

🛠️ PseudoCode

• Set up a dummy node that points at head. Smile and wave at edge cases.
  

  ListNode dummy = new ListNode(0);
  dummy.next = head;

• Loop to find each k-group:
  

  ListNode kth = getKthNode(prev, k);

  If none left, escape the matrix.
• Do the reversal, using three-pointer gymnastics.
  

  ListNode prev = kth.next;
  ListNode curr = prevGroup.next;
  while (curr != groupNext)
    // swing pointers!

• Reconnect the group boundaries.
  

  prevGroup.next = kth;
  groupStart.next = groupNext;

• Move prev to end of reversed chunk, rinse, repeat

💻 The Code

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) { val = x; }
}
 public ListNode reverseKGroup(ListNode head, int k) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode prevGroup = dummy;
    while (true) {
        ListNode kth = getKthNode(prevGroup, k);
        if (kth == null) break;
        ListNode groupNext = kth.next;
        // Reverse group
        ListNode prev = groupNext, curr = prevGroup.next;
        while (curr != groupNext) {
            ListNode temp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = temp;
        }
        // Plug new head of group
        ListNode tmp = prevGroup.next;
        prevGroup.next = kth;
        prevGroup = tmp;
    }
    return dummy.next;
}
 private ListNode getKthNode(ListNode start, int k) {
    while (start != null && k > 0) {
        start = start.next;
        k--;
    }
    return start;
}

⚠️ Pitfalls, Traps & Runtime Smacks

• k = 1? You’re done—no reversal needed (but hey, thanks for playing).
• k > length? You’re also done, unless you enjoy empty loops.
• Smaller final group? Just don’t touch it. Seriously.
• Time/Space: Looks like O(n), smells like O(n), is O(n). Space? Constant. Unless you sneak in a stack, and your interviewer cries a single tear.

🧠 Interviewer Brain-Tattoo

“When your pointers go rogue, just remember: even seasoned devs have watched their lists disintegrate like Thanos snapped. Stay sharp, or bring backup coffee.”