I liked this problem because there are so many sub-problems that are interview problems in their own right-- (reverse linked list, find midpoint)

var Node = function(val) {
  this.val = val;
  this.next = null;

var a = new Node('r');
var b = a.next = new Node('a');
var c = b.next = new Node('c');
var d = c.next = new Node('e');
var e = d.next = new Node('c');
var f = e.next = new Node('a');
var g = f.next = new Node('r');

function checkOdd(linkedList) {
  var count = 0;
  var curr = linkedList;
  while (curr) {
    curr = curr.next;
  return count % 2 === 1;

function findMidPoint(linkedList) {
  var slow = linkedList;
  var fast = linkedList;

  while (fast && fast.next) {
    fast = fast.next.next;
    slow = slow.next

  if (checkOdd(linkedList)) {
    return slow.next
  } else {
    return slow;

function checkPalindrome(linkedList) {
  var midpoint = findMidPoint(linkedList);

  var firstHead = linkedList;
  var secondHead = reverseLL(midpoint);

  var flag = true;

  while (secondHead) {
    if (secondHead.val !== firstHead.val) {
      flag = false;
    secondHead = secondHead.next;
    firstHead = firstHead.next;
  return flag;

function reverseLL(root) {
  var prevNode = null;
  var currNode = root;
  var nextNode;
  while (currNode) {
    nextNode = currNode.next;
    currNode.next = prevNode;
    prevNode = currNode;
    currNode = nextNode;
  return prevNode;

console.log(checkPalindrome(a)); = true

2 Answers 2

  • checkOdd is unwarranted waste of time. You may deduce evenness of the list just by looking at fast: if it is null the list is even.

  • flag is a meaningless name. Failure to come up with a good name typically indicates that you'd be better off without such variable. Consider return false; as soon as a mismatch is found, and return true at the end of the function.

  • \$\begingroup\$ Yeah checkOdd is a waste of time now that I look at it more closely, thanks! \$\endgroup\$ Commented Dec 29, 2015 at 21:35
  • 1
    \$\begingroup\$ Failure to come up with a good name typically indicates that you'd be better off without such variable. Not necessarily, my variable naming is awful regardless of the functions they have. \$\endgroup\$
    – Mast
    Commented Dec 29, 2015 at 21:43

I agree that your code runs in \$O(n)\$ time and \$O(1)\$ space complexity, but the issue I have is that you do in-place modification of the list..... and thus a "processed" list is no longer the same list as it started as.

This makes the function useless for most practical purposes. A list that returns "true" on one call, will likely not return "true" on the second call.

A compromise solution would be to restore the list linking after reversing the second half, but your method will still not be reentrant - even though it would be more convenient to use.

In other words, you are using the list itself as a place to store temporary "state", and that list is not entirely contained in your function. This is as bad, or worse, than using global variables.

In a real interview I would challenge the requirement to have the \$O(1)\$ space, and suggest that \$O(n)\$ space is an acceptable compromise - or that a double-linked list is a better data structure to contain this problem.


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