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Salesforce· MTS II

Salesforce | SMTS | Hyderabad | Feb 2026 [offer]

Location
Hyderabad
Experience
9 years
Posted
9 Apr 2026
437 reads

Summary

I cleared the Salesforce interview process, solved the OA and DSA rounds, completed LLD and HLD rounds, and received an offer for an SMTS position in Hyderabad.

How it went

Position: LMTS YOE: 9+

Recruiter reached out on Linked, after discussion about interview availability, he shared Hackerrank OA link.

OA Q1. Min length subarray with k distinct integers Q2. Shortest cycle in DAG

I was able to solve both problems.

Round 1 (DSA) Q1: For every window of size k, we need the minimum, now among all those mins, we have to return the meximum Solved this problem.

Q2: This was cycle detection in a DAG related problem. For this, interview wanted to know my approach only, not the actual code so I explained him and this round finished here.

Round 2 (LLD) I was given a LLD problem for a stock trading app, the interview had already listed the functionalities and critereas to cover under the LLD.

Round 3 (HLD) I was asked to design distributed rate limiter. First, interviewer wanted to know which RL algorithm I would choose, we spent some time on this then went ahead with the design.

Round 4 (HM) This round was behavioral and LP related. I could have messed up a bit in terms of LP.

Verdict Hired but downlevelled to SMTS, HR didn't tell the reason behind it though.

Questions asked (3)

  1. 01

    Minimum Length Subarray with K Distinct Integers

    Data Structures & Algorithms

    Find the shortest (minimum length) contiguous subarray that contains exactly k distinct integers.

  2. 02

    Shortest Cycle in a DAG

    Data Structures & Algorithms

    Given a directed graph that is supposed to be a DAG, find the length of the shortest cycle present in the graph (if any).

  3. 03

    Maximum of Window Minimums (Meximum)

    Data Structures & Algorithms

    For an array and a window size k, compute the minimum element in each sliding window of size k. Among all these minima, return the maximum (sometimes referred to as the "meximum").

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