Ho presentato la mia candidatura online. Ho sostenuto un colloquio presso Amazon (Seattle, WA) nel mese di mag 2025
Colloquio
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.
Domande di colloquio [1]
Domanda 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.
Ho sostenuto un colloquio presso Amazon (Washington, DC)
Colloquio
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.
Ho presentato la mia candidatura online. La procedura ha richiesto 3 settimane. Ho sostenuto un colloquio presso Amazon (Hyderabad) nel mese di feb 2025
Colloquio
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.
Domande di colloquio [1]
Domanda 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?
