Check Palindrome Linked List - javascript

I intended to reverse the original LinkedList and compare it with reversed one. If it is identical return true. However, I keep getting the wrong results. Where I did wrong?
var isPalindrome = function(head) {
if(head == null || head.next == null) return true;
let reversedHead = reverse(head)
let count = 0
while (head != null && reversedHead != null) {
if(head.val != reversedHead.val) {
return false
}
head = head.next
reversedHead = reversedHead.next
}
if (head == null && reversedHead == null){
return true
} else {
return false
}
};
var reverse =(head)=> {
let prev = null
let next = new ListNode()
let curr = new ListNode()
curr = head
while(curr){
next = curr.next
curr.next = prev
prev = curr
curr = next
}
return prev
}

After you call:
let reversedHead = reverse(head)
You will have reversed the list, i.e. the head has become the tail, and a tail's next reference is null. Hence your loop will quickly end, as after:
head = head.next
... head will be null.
You need to rethink the algorithm. There are several ways to do it...
Instead of an inplace reversal, you could let reverse return a completely new linked list that will have the reversed order:
function reverse(head) { // Does not modify the given list. Returns a new one.
let node = head;
let reversedHead = null;
while (node) {
reversedHead = new ListNode(node.val, reversedHead);
node = node.next;
}
return reversedHead;
}
I assume here that the ListNode constructor accepts a second argument:
class ListNode {
constructor(val, next=null) {
this.val = val;
this.next = next;
}
}

Related

Trying to solve Lowest common ancestor

var lowestCommonAncestor = function(root, p, q) {
// return the path to the node
let path = []
const search = (node, target) => {
if (node === null) return false
path.push(node)
if (node === target) return true
const leftSearched = search(node.left, target)
if (leftSearched) return true
const rightSearched = search(node.right,target)
if (rightSearched) return true
path.pop()
}
search(root, p)
const pathP = path
path = []
search(root, q)
const pathQ = path
let result
while(pathP.length > 0 && pathQ.length > 0 && pathP[0] === pathQ[0]) {
result = pathP[0]
pathP.shift()
pathQ.shift()
}
return result
};
console.log(lowestCommonAncestor([3,5,1,6,2,0,8,null,null,7,4],5,1));
Iam getting following error message
const leftSearched = search(node.left, target)
^
TypeError: Cannot read property 'left' of undefined
Could someone help me to fix this issue
Leet Code, as some other code challenge sites, will transform the array input (actually the text input having JSON notation) into an instance of TreeNode, and will pass that as argument to the function with your solution code.
When you want to run the solution locally, you'll have to take care of this transformation yourself. For that purpose you could make use of the fromList function -- specifically for binary trees.
NB: you have a bug in your search function. if (node === target) should be if (node.val === target).
// LeetCode template:
function TreeNode(val) {
this.val = val;
this.left = this.right = null;
}
// Tool to convert array input to a tree:
function fromList(values) {
if (!values) return;
let it = (function* () {
for (let value of values) {
yield value == null ? null : new TreeNode(value);
}
while (true) yield null;
})();
let root = it.next().value;
let nextlevel = [root];
while (nextlevel.length) {
let level = nextlevel;
nextlevel = [];
for (let node of level) {
if (node) {
node.left = it.next().value;
node.right = it.next().value;
nextlevel.push(node.left, node.right);
}
}
}
return root;
}
// Your function
var lowestCommonAncestor = function(root, p, q) {
// return the path to the node
let path = []
const search = (node, target) => {
if (node === null) return false;
path.push(node);
if (node.val === target) return true;
const leftSearched = search(node.left, target);
if (leftSearched) return true;
const rightSearched = search(node.right,target);
if (rightSearched) return true;
path.pop();
}
search(root, p);
const pathP = path;
path = [];
search(root, q);
const pathQ = path;
let result;
while(pathP.length > 0 && pathQ.length > 0 && pathP[0] === pathQ[0]) {
result = pathP[0];
pathP.shift();
pathQ.shift();
}
return result;
};
// Running your solution on some input
let input = [3,5,1,6,2,0,8,null,null,7,4];
// Make the conversion that LeetCode would do
let root = fromList(input);
let lca = lowestCommonAncestor(root,5,1);
// For your own purposes, just print the value of that node:
console.log(lca.val); // 3

Implement the addInPos method inside the LinkedList prototype in Javascript

I need to implement the addInPos method inside the LinkedList prototype but I don't know what it is bad in my code...because the test no pass.
Implement the addInPos method inside the LinkedList prototype that must add an element in the indicated position. Both data will be provided as a parameter (pos, value). Where "pos" will be the position in which the "value" value should be added. In the event that the position in which the insertion is to be made is invalid, that is, it exceeds the size of the current list, it must return false.
If the node was added correctly return true.
Note: the zero position corresponds to the head of the LinkedList.
My code:
LinkedList.prototype.addInPos = function(pos, value) {
if(pos > this.size) {
return false;
}
const newNode = new Node(pos, value);
let current = this.head;
let previous;
if(pos === 0) {
newNode.next = current;
current.prev = newNode;
this.head = newNode;
} else {
for(let i = 0; i < pos; i++){
previous = current;
current = current.next;
}
newNode.next = current;
newNode.prev = previous;
current.prev = newNode;
previous.next = newNode;
}
}
Thanks.
With your structure ideally insertAt function would look like:
var insertAt = function(head, item, index) {
var curr = head;
var count = 0;
while (curr !== null) {
if (count === index) {
var newNode = { data: item, next: curr.next, prev: curr };
curr.next = newNode;
break;
}
count++;
curr = curr.next;
}
};
insertAt(head, 2, 3);
Let me know if this works.
Also look at this LinkedList class I have created, insertAt function is currently missing, but from this stack question, I am planning to add it in my class as well.
Class GitHub
NPM package - #dsinjs/linked-list
Complete documentation

Copy List with Random Pointer - Copy not point to -

I'm trying to solve the problem below:
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
Return a deep copy of the list.
The Linked List is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:
val: an integer representing Node.val
random_index: the index of the node (range from 0 to n-1) where random pointer points to, or null if it does not point to any node.
Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]
My idea was loop the head and create a new node with the same value and set it to my ans Node. It will be a copy because I'm creating a new one. Then, I loop again for the random pointer which looks in my ans Node the node with the value that I should point to. It works perfectly for the input above. However, if I have an input with duplicate values, the findeNode method might return an invalid Node. For example it fails for the input below:
[[3,null],[5,17],[4,null],[-9,6],[-10,3],[5,15],[0,11],[6,null],[-6,16],[3,16],[-6,11],[9,12],[-2,1],[-3,11],[-1,10],[2,11],[-3,null],[-9,7],[-2,4],[-8,null],[5,null]]
My output:
[[3,null],[5,3],[4,null],[-9,null],[-10,3],[5,15],[0,11],[6,null],[-6,13],[3,13],[-6,11],[9,12],[-2,1],[-3,11],[-1,8],[2,11],[-3,null],[-9,7],[-2,4],[-8,null],[5,null]]
Correct output:
[[3,null],[5,17],[4,null],[-9,6],[-10,3],[5,15],[0,11],[6,null],[-6,16],[3,16],[-6,11],[9,12],[-2,1],[-3,11],[-1,10],[2,11],[-3,null],[-9,7],[-2,4],[-8,null],[5,null]]
Node definition:
function Node(val, next, random) {
this.val = val;
this.next = next;
this.random = random;
};
My code:
var copyRandomList = function(head) {
if(!head) return head;
let ans = new Node(0);
let temp = ans;
let cur = head;
while(cur) {
const node = new Node(cur.val);
temp.next = node;
cur = cur.next;
temp = temp.next;
}
cur = head;
temp = ans.next;
while(cur) {
const randomNode = cur.random;
if(!randomNode) {
temp.random = null;
} else {
const node = findNode(ans, randomNode.val);
temp.random = node;
}
cur = cur.next;
temp = temp.next;
}
return ans.next;
};
const findNode = (list, val) => {
if(!val)
return null;
while(list) {
if(list.val === val)
return list;
list = list.next;
}
return null;
}

Remove elements from singly linked list (javascript)

I'm doing a CodeFights problem, trying to remove elements from a singly linked list that has the value k.
Below is what I have (l is the list and k is the value):
function removeKFromList(l, k) {
//figure out the head of the new list so we can reference it later
var head = l;
while (head.value === k){
head = head.next;
}
var node = head;
var temp = null;
while (node && node !== null) {
if (node.next.value === k){
temp = node.next.next;
node.next.next = null;
node.next = temp;
}
node = node.next;
console.log("+++", head)
}
console.log("---", head)
}
The CodeFight test cast is 3 -> 1 -> 2 -> 3 -> 4 -> 5. The final result would have been 1 -> 2 -> 4 -> 5. But my '---' console log keeps returning "Empty" (according to the CodeFights console).
My '+++' console log returns the correct head with element each loop.
I've been pulling my hair out over this, any idea what is missing here?
You need to return the list if you delete the first node.
Then you need a loop for taking the next not while the value is not found.
At last you need a check if the last node exist and if the value is found, then assign the next node to the last next property.
function removeNode(list, value) {
var node = list,
last;
if (node && node.value === value) {
return node.next;
}
while (node && node.value !== value) {
last = node,
node = node.next;
}
if (last && node.value === value) {
last.next = node.next;
}
return list;
}
var list = { value: 1, next: { value: 2, next: { value: 3, next: { value: 4, next: { value: 5, next: { value: 6, next: { value: 7, next: null } } } } } } };
list = removeNode(list, 5);
console.log(list)
list = removeNode(list, 1);
console.log(list)
list = removeNode(list, 7);
console.log(list)
.as-console-wrapper { max-height: 100% !important; top: 0; }
Try this:
function myRemove(l, k){
if(l == null){
return l;
}
while(l.value == k){
l = l.next;
}
thisNode = l;
nextNode = thisNode.next;
while(nextNode != null){
if(nextNode.value == k){
thisNode.next = nextNode.next;
// No more nodes, ie last node was to be removed
if(thisNode.next == null)
break;
}
thisNode = thisNode.next;
nextNode = thisNode.next;
}
return l;
}
This could be also done by recursion:
function removeKFromList({value, next}, k) {
if(value === k){
return next ? removeKFromList(next, k) : null;
} else {
return {
next : removeKFromList(next, k),
value
};
}
}

Doubly Linked List in javascript

I was building linked list in javascript.
I dont understand one part.
function Node(element) {
this.element = element;
this.next = null;
this.previous = null;
}
function LList() {
this.head = new Node("head");
this.find = find;
this.findLast = findLast;
this.remove = remove;
this.insert = insert;
this.display = display;
this.dispReverse = dispReverse;
}
function find(item) {
var currNode = this.head;
while(currNode.element != item) {
currNode = currNode.next;
}
return currNode;
}
function display(list) {
var currNode = this.head.next;
while (currNode != null) {
console.log(currNode.element);
currNode = currNode.next;
}
}
function insert(newElement, item) {
var newNode = new Node(newElement);
var current = this.find(item);
newNode.next = current.next;
newNode.previous = current;
current.next = newNode;
// Why I dont need this part?
// Since new node got inserted, on my thoughts,
// the next node of the current node should point the new node as a previous one
// current.next.previous = newNode;
}
function remove(item) {
var currNode = this.find(item);
if (currNode.next != null) {
currNode.previous.next = currNode.next;
currNode.next.previous = currNode.previous;
currNode.next = null;
currNode.previous = null;
}
}
function findLast() {
var currNode = this.head;
while (currNode.next != null) {
currNode = currNode.next;
}
return currNode;
}
function dispReverse() {
var currNode = this.head;
currNode = this.findLast();
while(currNode.previous != null) {
console.log(currNode.element);
currNode = currNode.previous;
}
}
var cities = new LList();
cities.insert("Conway", "head");
cities.insert("Russellville", "Conway");
cities.insert("Carlisle", "Russellville");
cities.insert("Alma", "Carlisle");
cities.display();
cities.remove("Carlisle");
cities.display();
cities.dispReverse();
/*
Output should look like this:
Conway
Russellville
Carlisle
Alma
Conway
Russellville
Alma
Alma
Russellville
Conway
*/
Problem is the insert function!
Let's say if I have A B C node already.
And I want to insert K after B.
Currently, the next and previous of B are C and A for each.
The previous element of C is B.
Once I put K after B,
A B K C
(1) the next element of K will be C
(2) the previous element of K will be B.
(3) The next element of B is K
(4) previous element of C is K.
On the code I wrote in Insert function, each line of the codes below should process the upper statements.
(1) newNode.next = current.next;
(2) newNode.previous = current;
(3) current.next = newNode;
(4) current.next.previous = newNode;
But when I run this whole code including (4), error has occurred.
I don understand why...
Without the (4) line of codes, it works.
Is there anyone who can help me understand this?
You need to do step 4 before step 3:
current.next.previous = newNode
current.next = newNode
As it is, the reference of current.next (C) is being set to newNode (K) before you lookup the "old" current.next's previous property ( current.next.previous when points to B). Your reference to the current node is changed as soon as you assign it a new value. This is why current.next.previous is actually returning newNode.previous instead of the node reference you expect.
Your insert logic seems wrong in the last line :
current.next = newNode;
current.next.previous = newNode;
This actually means
newNode.previous=newNode;
Since you are setting the value of current.next to newNode in the 3rd statement
It should be :
newNode.next.previous = newNode.

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