Amazon interview experience Real time questions & tips from candidates to crack your interview
Solutions & Support Engineer
Amazon
2 rounds | 3 Coding
problems
Selected Interview preparation journey
Preparation
Duration: 3 months
Topics: Data structures & algorithms, Computer networks, COA, Web development
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
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
Coding problem2
The timing was 4-6 pm
It was an online round ,questions were easy to hard
1. Rod cutting problem
Moderate
40m average time
75% success
0/80
Asked in companies
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.
Note:
1. The sizes will range from 1 to ‘N’ and will be integers.
2. The sum of the pieces cut should be equal to ‘N’.
3. Consider 1-based indexing.
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.
2. Coin Change(Finite Supply)
Hard
0/120
Asked in companies
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.
You have to find the number of ways to make the sum ‘V’ by selecting some(or all) coins from the array.
The answer can be very large. So, return the answer modulo 1000000007.
For Example :
‘N’ = 3, ‘coins’ = {1, 2, 3}, ‘freq’ = {1, 1, 3}, ‘V’ = 6
For the given example, we can make six by using the following coins:
{1, 2, 3}
{3. 3}
Hence, the answer is 2.
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).
Finally, as we have to find the total number of ways, so we will add these 2 possible choices, i.e count(coins, n, sum – coins[n-1] ) + count(coins, n-1, sum );
02
Round
Medium
Video Call
Duration60 minutes
Interview date7 Sep 2022
Coding problem1
Asked some DSA questions and one programming logic questions
1. Painting Fences
Moderate
25m average time
75% success
0/80
Asked in companies
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.
As the answer can be too large, return the answer modulo 10^9 + 7.
For Example:
Consider If N = 2, then there can be 4 different ways to color fences such that at most 2 adjacent fences have the same color-:
[ [0, 1],
[1, 0],
[1, 1],
[0, 0] ]
Hence, the answer is 4.
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
diff = k * (k-1)* (k-1)
//(k-1) choices for the first place
// k choices for the second place
//(k-1) choices for the third place
same = k * (k-1) * 2
// 2 is multiplied because consider two color R and B
// R R B or B R R
// B B R or R B B
c'' != c, (k-1) choices for it
Hence we deduce that,
total[i] = same[i] + diff[i]
same[i] = diff[i-1]
diff[i] = (diff[i-1] + diff[i-2]) * (k-1)
= total[i-1] * (k-1)
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Skill covered: Programming
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