Solutions & Support Engineer
Amazon
2 rounds | 3 Coding problems
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Interview preparation journey
Preparation
Duration: 3 months
Topics: Data structures & algorithms, Computer networks, COA, Web development
Tip

Tip 1 : Practice DSA a lot
Tip 2 : Have good knowledge about your projects and don't copy them.
Tip 3 : Revise core subjects before the interview

Application process
Where: Campus
Eligibility:
Resume tip

Tip 1 : Write only those things that you know well.
Tip 2 : Do not exaggerate your skills
Tip 3 : Include your projects at least two minimum

Interview rounds
01
Round
Easy
Online Coding Interview
Duration90 minutes
Interview date15 Jul 2022
Problems2
Rod cutting problem

#### Given a rod of length ‘N’ units. The rod can be cut into different sizes and each size has a cost associated with it. Determine the maximum cost obtained by cutting the rod and selling its pieces.

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Problem approach

Method 1 : A naive solution to this problem is to generate all configurations of different pieces and find the highest-priced configuration. This solution is exponential in terms of time complexity. Let us see how this problem possesses both important properties of a Dynamic Programming (DP) Problem and can efficiently be solved using Dynamic Programming.

Coin Change(Finite Supply)

#### You are given an array of integers ‘coins’ denoting the denomination of coins and another array of integers ‘freq’ denoting the number of coins of each denomination.

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Problem approach

We have 2 choices for a coin of a particular denomination, either i) to include, or ii) to exclude.
If we are at coins[n-1], we can take as many instances of that coin ( unbounded inclusion ) i.e count(coins, n, sum – coins[n-1] ); then we move to coins[n-2].
After moving to coins[n-2], we can’t move back and can’t make choices for coins[n-1] i.e count(coins, n-1, sum).
Fin...

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02
Round
Medium
Video Call
Duration60 minutes
Interview date7 Sep 2022
Problems1
Painting Fences

#### You are given ‘N’ fences. Your task is to find the total number of ways to paint fences using 2 colors only such that at most 2 adjacent fences are painted with the same color.

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Problem approach

diff = no of ways when color of last
two posts is different
same = no of ways when color of last
two posts is same
total ways = diff + same

for n = 1
diff = k, same = 0
total = k

for n = 2
diff = k * (k-1) //k choices for
first post, k-1 for next
same = k //k choices for common
color of two posts
total = k + k * (k-1)

for n = 3

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