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It was an easy round
Interview was taken by 3rd party agency and he asked coding question related to dsa
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I applied via Campus Placement and was interviewed before Sep 2023. There was 1 interview round.
Equation, ratio and proportion
I applied via Campus Placement and was interviewed before Nov 2023. There was 1 interview round.
Question were very basic from array and stacks
I applied via LinkedIn and was interviewed before Jul 2023. There were 3 interview rounds.
Basic DSA questions were asked.
I applied via Company Website and was interviewed before Jul 2023. There was 1 interview round.
Creating specific components for software development
Identify the requirements for the component
Design the component architecture
Implement the component using appropriate programming languages and tools
Test the component for functionality and performance
Integrate the component into the larger software system
I applied via Company Website and was interviewed before Feb 2023. There were 2 interview rounds.
Basic DSA Questions of array
I was interviewed in Feb 2021.
Round duration - 60 minutes
Round difficulty - Easy
In the first inerview round, I was asked two problems, that were based on Data Structures.
I realized a key point in this round : that sometimes, the interviewer is wrong, and you are right. Still, you should not argue with the interviewer, rather, just accept, and move on.
This happened in the first coding problem itself, when, I actually used a "map" in c++. The interviewer asked me, to use an "unordered map" Instead of map, saying, that using an unordered map would make the insertion cost O(1) instead of O(log(n)). I refused to this by saying, that using unordered map is risky, as in case of a lot of collisions, it could go O(n).
Still, the interviewer said, that I were wrong, and said that he had done some competitive programming himself, and he never faced a TLE verdict because of using unordered map.
And I knew very well, that an Anti-Hash-Test could be created, to enforce a TLE verdict. The interviewer went on to saying that "every competitive programmer in the world would never use a map, rather, always use unordered map".
I slightly smiled to this, but agreed, and later cleared that interview, because of the smooth second problem.
Your task is to determine if two given strings are anagrams of each other. Two strings are considered anagrams if you can rearrange the letters of one string to form the...
Determine if two strings are anagrams of each other by checking if they can be rearranged to form the other string.
Create a frequency map of characters for both strings and compare them.
Sort both strings and compare them.
Use a set to store characters and their counts for each string, then compare the sets.
Explanation: Sanyam wishes to distribute 'N' candies among 'K' friends. The friends are arranged based on Sanyam's order of likeness. He initially distributes candies s...
Distribute N candies among K friends in Sanyam's order of likeness, incrementing distribution by K each round.
Distribute candies starting from 1st friend, incrementing by K each round
If remaining candies are fewer than what a friend is supposed to receive, stop distribution
Output the number of candies each friend ends up with at the end of distribution
Round duration - 60 minutes
Round difficulty - Medium
I were asked several questions involving DBMS. I was given data of patients of entire country, and was asked to make a dataBase for the same.
Tip 1 : I did a lot of problems on Codeforces, Codechef and other online platforms
Tip 2 : Apart from normal CP, it is better to do some interview prep separately
Tip 3 : Must note - you need to have development skills as well to clear the projects and DBMS interview.
Tip 1 : Make a 1 page resume only
Tip 2 : It is good to have some projects and decent ranks on your resume
I applied via Job Fair and was interviewed before Nov 2023. There were 3 interview rounds.
3 coding question around prime number arrays and strings
Round 1 -> Coding interview -> ask 2 Leetcode problems
Round 2-> HR
based on 1 interview
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