thekhairul

// radius (largest row of stars) must be an odd number
function makeDiamondWithStars(radius) { // let's say radius is 9
    let starsOrder = []; // we list down the number of stars needed to put on each line
    for (let i = 1; i <= radius; i+=2) {
        starsOrder.push(i);
    }
    // after this loop starsOrder array is [1, 3, 5, 7, 9]
    starsOrder = starsOrder.concat(starsOrder.slice(0,-1).reverse()); // now starsOrder is [1, 3, 5, 7, 9, 7, 5, 3, 1]
    const midOfTarget = Math.floor(radius/2); // 4
    for (let j=0; j<starsOrder.length; j++) {

thekhairul

function sumOfFirstN(n) {
  return (n*(n+1)) / 2;
}

// Time complexity - O(1)
// Space complexity - O(1)

thekhairul

def twoSum(nums, target):
    dict = {}

    for i in range(len(nums)):
        if ((target-nums[i]) in dict):
            return [dict[target - nums[i]], i]

        dict[nums[i]] = i

    return []

thekhairul

def removeNthFromEnd(head,n):
# O(2n) || O(1)
    count = 0
    cur = head

    while cur:
        count += 1
        cur = cur.next

    m = count - n

thekhairul

def hasCycle(head):
    fast, slow = head, head
    while fast and fast.next:
        fast = fast.next.next
        slow = slow.next
        if fast == slow:
            return True
    return False

thekhairul

def isPalindrome(head):
    if head == None or head.next == None:
        return True

    # find mid
    mid, fast = head, head
    prevOfMid = None

    while fast and fast.next:
        prevOfMid = mid

thekhairul

def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
    if head == None or head.next == None:
        return head

    curNode = head
    prevNode = None

    while curNode:
        nextNode = curNode.next
        curNode.next = prevNode

thekhairul

def deleteMiddle(head):
    if head.next:
        slow = head
        fast = head
        beforeMid = None

        while fast and fast.next:
            fast = fast.next.next
            beforeMid = slow
            slow = slow.next

thekhairul

def middleNode(head):
    #find total length
    count = 1
    cur = head
    while cur.next:
        count += 1
        cur = cur.next
    #find mid
    mid = int(count/2 + 1)
    #find mid node

thekhairul

def isPowerOfTwo(n):
    left = 0
    right = 31
    result = False

    while left <= right:
        mid = (left + right) // 2

        if pow(2, mid) == n:
            result = True