Round 1 (Coding and Problem solving round held on bluejeans)
http://collabedit.com/ was used for code sharing , the video and screen share were on throughout the interview

Question 1: Given an array of stock prices with index representing the days, return the maximum profit . The user can buy only once and sell only once. The buy and sell date cannot be same.
It is a similar leetcode question:
https://leetcode.com/problems/best-time-to-buy-and-sell-stock/

def maxProfit(self, prices: List[int]) -> int:
minCostPrice,maxSellPrice=float(‘inf’),float(‘-inf’)
if(len(prices)==0):
return 0
for i in range(len(prices)):
if(minCostPrice>prices[i]):
minCostPrice=prices[i]
maxSellPrice=max(maxSellPrice,prices[i]-minCostPrice)

return maxSellPrice

Question 2: Construct a stack with functions
Push: Push an Element in a stack(i.e. at the top)
Pop: Pop an element (i.e. remove an element from top)
Min_value_in_stack: returns minimum value in stack at any point
Similar question on leetcode: https://leetcode.com/problems/min-stack/

def __init__(self):
self.stck=[]
# print(self.stck)
def push(self, x: int) -> None:
if(len(self.stck)==0):
self.stck.append([x,x])
# print(self.stck)
else:
self.stck.append([x,min(self.stck[-1][1],x)])
# print(self.stck)
def pop(self) -> None:
if(len(self.stck)>0):
self.stck.pop()
# print(self.stck)

def top(self) -> int:
if self.stck:
return self.stck[-1][0]
# print(self.stck)
def getMin(self) -> int:
return self.stck[-1][1]

Round 2(Coding and Data Structures Round)
Held on Bluejeans video call and code in http://collabedit.com/
Question : What is Binary Search Tree(BST)? (Answer: Specify the conditions)
Question:Why this example is not a BST ?(Answer: 3 cannot be in the right side of root node 5)
5

3 7

2 4 38
Question:What are some applications of BST? (Answer: Application related to searching as time complexity is O(logn), the inorder traversal gives a sorted order (so can store values to be in sorted order)
Question: Give Order Traversals of a tree

Pre: root,left,right
In: left,root,right
Post: left,right,root
def preorder(root):
if(root==None):
return
print(root.val)
self.preorder(root.left)
self.preorder(root.right)
def inorder(root):
if(root==None):
return
self.inorder(root.left)
print(root.val)
self.inorder(root.right)
def postorder(root):
if(root==None):
return
self.postorder(root.left)
self.postorder(root.right)
print(root.val)

Question : Give an array having numbers from 1 to n , where 2 numbers are missing. Find the two missing numbers.
Answer:
Sorted: Apply linear search or binary serach
Not sorted : Try applying average and sum formula
Snippet for the equations:

a:1 to n
ideal_sum=n*(n+1)/2
curr_sum=sum(a)
diff=ideal_sum-curr_sum
x+y=diff 
1<=x,y<=n
ideal_sum-x-y/n=(n+1)/2-x/n-y/n

Question : Given a linked list with duplicate elements , sum up the duplicates and return the end result (do not use extra space like map to store frequency)
1->2->2->3->3->3->9->None
1->4->9->9->Null
1->4->18->NULL
Snippet of Code:

new_val=0
while(ptr.next!=None):

if(ptr.next.val==ptr.val):
new_val+=ptr->next.val

else: 
if(new_val!=0):
intial_pos.val=new_val
intial_pos.next=ptr.next
new_val=ptr.next.val

new_val=0
ptr=ptr.next