Flipkart Interview Experience for SDE-1 (Off-Campus) 2024

I don’t remember that I have applied for this job role maybe because I was applying for a lot of jobs…

I gave my OA on May 15th and then I got a mail on 28th may ig that I have got shortlisted for the interview. after around 10 days I got another mail that I have interview on June 10 so make sure you guys start preparing at least after you receive the mail about interview shortlist. this was the mistake I made I did’t prepare anything.

there will be 4 rounds 1 – OA, 2 – Technical interview 1, 3 – Technical interview 2, 4 – HR Interview

2 players are playing a game of batons. These players can each hold a maximum of x and y batons in their hands respectively.

The game ends when the sum of batons in the hands of both the players equals to target_value.

Players can throw or pick baton with the constraint that:

throw all the batons in there hand pick more batons equivalent to the number of batons in their hand currently

There is an unlimited supply of batons. Players get a chance to pick/throw the batons.

You need to find if at any point of time, the sum of number of batons in both the players’ hands reaches target_value.

3,5 ====== 4

(0,5) -> (3,2) -> (0,2) -> (2,0) -> (2,5) -> (3,4) -> (0,4) true

Q2. Flipkart employee cab system is as follows:

When an employee arrives, they will wait in line for the next available cab. You can get on a cab that departs at T1 time if you arrive at T2 time where T2 <= T1, and the cab is not full. Employee with the earliest arrival times get on the cab first.

You are given an integer array cab of length n, where cab[i] represents the departure time of the ith cab. You are also given an integer array employees of length m, where employees[j] represents the arrival time of the jth employee. All cab departure times are unique. All employee arrival times are unique.

You are given an integer capacity, which represents the maximum number of employees that can get on each cab.

When a cab arrives at the pickup point:

If capacity or fewer employees are waiting for a cab, they will all get on the cab.

Return the last(as late as possible) time you may arrive at the cab pickup point to catch a cab. You cannot arrive at the same time as another employee.

Input cabs = [10,20], employees = [2,17,18,19], capacity = 2

Output: 16

Input cabs = [20,30,10], employees = [19,13,26,4,25,11,21], capacity = 2

Output: 20

these were the questions I have got