Engineer 1 Interview Questions

1,033 engineer 1 interview questions shared by candidates

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?
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Software Development Engineer 1

Interviewed at Amazon

3.5
Jul 15, 2025

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?

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