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Sprinklr Interview Experience (On-Campus) 2022

Last Updated : 29 Aug, 2022
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Each and Every Round was eliminatory.

Round 1 (Online coding test, Platform: HackerEarth): Total 3 Questions. Those who solved at least 2 Questions out of 3 were shortlisted for interviews.

Verdict: 60 students were shortlisted out of a batch size of 370.

Now the chaos begins!!!!!!!!

Round-2 (Online Interview, Platform: Google meet). : Interview lasted for almost an hour. There were two interviewers. Both of them were friendly. In the beginning, they introduced themselves followed by a quick introduction from my side.

Q1. Given a piece of code, find its time complexity

C++




int cnt=0;
  
for(int i=n;i>0;i=i/2)
{
 for(int j=1;j<=i;j++)
   cnt++;
    
}


Ans: Time complexity = O(n). Standard series : cnt= n + n/2 + n/4 + …….. + 1 = 2n -1.

Q2. Arrange time complexities in ascending order.

  • O(n)
  • O(n^2)
  • O(n^(log2))
  • O(2^(logn))

Ans: c<a<b<d

Q3. https://www.geeksforgeeks.org/square-root-of-an-integer/

Q4: Design a queue such that it supports all the following three operations.

  • Enqueue: Must be done in O(1)
  • Copy queue: make a new copy of a queue. Must be done in O(1)
  • Print queue: Must be done in O(n).

The question was intentionally kept vague so that I ask clarification questions.

Ans. The interviewer didn’t ask me to code. He just wanted me to discuss my approach and think out loud. I started by explaining my straightforward linked-list approach where equeue can be done in O(1) and print can be done in O(n) but copy can’t be done in O(1). He gave some of the bottlenecks in the form of hints and also agreed that the data structure which I identified that is Linked-list is correct. After brainstorming for a couple of minutes, I arrived at a solution where we can assume it as a Tree of Linked-list. Whenever we want to copy our queue, we just add a new branch. He was satisfied with my solution.

Verdict: I was selected for Round 3. A total of 40 people were selected out of 60.

Round -3 (Offline, mode: pen-paper): The interviewer was Rude. He looked at my resume and asked me to explain projects. Then He gave me a piece of paper and asked for medium difficulty level DSA Ques.

Question: There are n players in a game. each player has some power, given in the form of an array. The maximum power a player can have is 100. You have ‘K’ spells. you can use these spells to increase the power of any player by the following formula: power += [power%10 + no of spells used]/10. Your task is to find the maximum sum of the power of all players after using spells.

sample: power=[77,89] and K=4 
Ans = 78 + 90 = 168. ( use 3 spells for player 1 and 1 spell for player 2).

My approach: I solved it using the sorting technique with the time complexity of O(nlogn). He asked me to optimize it. It took a couple of minutes but I was not able to figure out how to optimize it, so I asked for the hint which he denied. Then I tried to think out loud and told what was going through my head. He was busy doing some stuff on his laptop and completely ignored (That was Rude). He was basically multitasking and it was quite evident that he was not at all interested in interviewing me.

After a few minutes, He was like leave it, you won’t be able to solve it (that was frightening) :(.

Question: (Designed to Reject a candidate ): Given n coins where the probability of getting head on coin-1 is p1, on coin-2 is p2, on coin-3, is p3,…….. on coin-n is pn respectively. Find the probability of getting exactly K heads when all n coins are tossed.

He gave this question and again started working on his laptop, completely ignorant of what I was writing on paper or what I was saying.

I was not able to solve it.

Verdict: Got rejected (Obviously). only 20 people were called for Round 4.

Final Verdict: 5 people received an offer

They took a total of 6 interviews of each candidate who received an offer and for some even 7 interviews. Quite hectic. (However, it was mentioned that there will be only 4 interviews in their Pre-placement talk)

Pro Tip: Don’t Apply in Sprinklr. 🙂


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