Envié una solicitud electrónica. Acudí a una entrevista en Amazon (Seattle, WA) en may 2025
Entrevista
There were three 1hr long interviews shuffled, I was just given the interview link and slots. The interviews happened one after the other so in a span of 3.5hrs I gave three interviews. The rounds were as follows: 1hr coding interview, 1hr of behavioral / leadership principles and 1hr of low level design with leadership principles.
Preguntas de entrevista [1]
Pregunta 1
In coding round I was asked the Next Permutation question from LeetCode but I had to find 5th next permutation. Follow up was to find the greatest possible permutation from the given digits.
Waiting for interview since a month after my OA and their email saying that I cleared the OA. They replied me on the follow up mail that they are in the process of scheduling interviews but still I am doubtful as they closed the job opening for new candidates for that role.
Envié una solicitud electrónica. El proceso duró 3 semanas. Acudí a una entrevista en Amazon (Hyderabad) en feb 2025
Entrevista
1 session with HR, just verifying if I am interested, when I am free, etc.
3 technical/Bar-Raising interviews scheduled over span of a week
1 session with HR confirming my existing employment, notice period, etc.
Preguntas de entrevista [1]
Pregunta 1
Technical Round 1
Q1> Variation of Celebrity Problem
Given a square matrix of n, where matrix[i][j] = 1 indicates that employee i "acknowledges" employee j.
Determine the "ultimate employee" - Employee acknowledged by all, but does not acknowledge anyone else. If no ultimate decision maker, return -1.
Q2> Some kind of BFS problem.
Consider a graph of Amazon "Fulfilment Center" (Root) and Delivery Stations [Node]
FC Manager wants to know how many DS are there for which a package will have to go through X intermediate DSes
Technical Round 2
Q1> Variation of subarray with sum k.
Calculate subarrays where sum of elements of subarray is equal to length of subarray.
Q2> BFS type problem - Rotten Tomatoes
Find least "iterations" for all oranges to become rotten.
A rotten orange at index (i,j) can rot all of its neighbours (up, down, left, right. No horizontal). If it is impossible to rot every orange then simply return -1;
Behavioural Qs -
Have you ever gone above and beyond what your role demands?
Have you ever had to work on a very tight deadline?
Have you ever had to go out of your way to dive deep and learn something for the job?
