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Backend Developer

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1 rounds | 2 Coding problems

Interview preparation journey

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Journey
I already knew Java and Spring Boot, so I just brushed up on my concepts and was prepared. However, I made sure not to fail on the hands-on coding, so I practiced coding extensively—about 3–4 hours a day. This might seem little, but I was also working at the time.
Application story
I applied on Naukri.com. An agency called me, took basic details, and then asked some relevant questions, such as my current location and whether I would be willing to relocate. After this, they scheduled my first round.
Why selected/rejected for the role?
I did not show up for the next round because I got the offer letter from somewhere else, so ideally, I dropped this job.
Preparation
Duration: 0.2 Months
Topics: Java, Spring, Spring Boot, DSA, Dynamic Programming (DP), Core Java
Tip
Tip

Tip 1: Solve a sufficient number of DSA questions.

Tip 2: Brush up on everything you already know.

Tip 3: Practice all concepts thoroughly.

Application process
Where: Naukri
Eligibility: 2+ year experience, (Salary Package: 11 LPA)
Resume Tip
Resume tip

Tip 1: Should be organized and not messy.

Tip 2: Convey clearly what you know.

Interview rounds

01
Round
Medium
Online Coding Test
Duration60 Minutes
Interview date13 Jan 2026
Coding problem2

It was a primarily coding-based round.

1. Koko Eating Bananas

Moderate
25m average time
70% success
0/80
Asked in companies
AtlassianTennrUnthinkable Solutions LLP

A monkey is given ‘n’ piles of bananas, where the 'ith' pile has ‘a[i]’ bananas. An integer ‘h’ is also given, which denotes the time (in hours) in which all the bananas should be eaten.


Each hour, the monkey chooses a non-empty pile of bananas and eats ‘m’ bananas. If the pile contains less than ‘m’ bananas, then the monkey consumes all the bananas and won’t eat any more bananas in that hour.


Find the minimum number of bananas ‘m’ to eat per hour so that the monkey can eat all the bananas within ‘h’ hours.


Example:

Input: ‘n’ = 4, ‘a’ =  [3, 6, 2, 8] , ‘h’ = 7

Output: 3

Explanation: If ‘m’ = 3, then 
The time taken to empty the 1st pile is 1 hour.
The time taken to empty the 2nd pile is 2 hour.
The time taken to empty the 3rd pile is 1 hour.
The time taken to empty the 4th pile is 3 hour.
Therefore a total of 7 hours is taken. It can be shown that if the rate of eating bananas is reduced, they can’t be eaten in 7 hours.
Problem approach
class Solution {
   public int minEatingSpeed(int[] piles, int h) {
       int max = 0;
       for (int pile : piles) {
           if (pile > max) {
               max = pile;
           }
       }
       int left = 1;
       int right = max;
       while (left < right) {
           int mid = left + (right - left) / 2;
           int hours = 0;
           for (int pile : piles) {
               hours += (pile + mid - 1) / mid; // ceiling division
           }
           if (hours <= h) {
               right = mid;
           } else {
               left = mid + 1;
           }
       }
       return left;
   }
}
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2. Minimized Maximum of Products Distributed to Any Store

Moderate
0/80

You are an operations manager for a company with n specialty retail stores. There are m different product types, and you are given a 0-indexed integer array quantities, where quantities[i] represents the number of products available for the i-th product type.


Your task is to distribute all products to the n retail stores, adhering to the following rules:

  A single store can only be stocked with products of at most one type. It can, however, be given any amount of that product type.

  After the distribution, you want to ensure the workload is balanced. Let x be the maximum number of products given to any single store. Your goal is to make x as small as possible.


Return the minimum possible value of x.

Problem approach

I first sort the product quantities in descending order to prioritize distributing larger quantities. Then, I use binary search to determine the smallest possible value of the maximum number of products per store (x) such that all products can be distributed among the given number of stores, n. For each candidate value of x during the binary search, I calculate how many stores are required by dividing each product’s quantity by x and summing the total number of stores needed. Based on whether the required number of stores is less than or equal to n, I adjust the binary search bounds accordingly to find the optimal value of x.

Try solving now

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Skill covered: Programming

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