Razorpay interview experience Real time questions & tips from candidates to crack your interview

SDET-2

Razorpay

1 rounds | 2 Coding problems

Anonymous

Fresher | Dec 2020

Rejected

Interview preparation journey

Preparation

Duration: 4 months

Topics: Data Structures, Algorithms, System Design, Aptitude, OOPS

Tip

Tip 1 : Must do Previously asked Interview as well as Online Test Questions.
Tip 2 : Go through all the previous interview experiences from Codestudio and Leetcode.
Tip 3 : Do at-least 2 good projects and you must know every bit of them.

Application process

Where: Other

Eligibility: Above 7 CGPA

Resume tip

Tip 1 : Have at-least 2 good projects explained in short with all important points covered.
Tip 2 : Every skill must be mentioned.
Tip 3 : Focus on skills, projects and experiences more.

Interview rounds

Round

Easy

Video Call

Duration60 minutes

Interview date22 Dec 2020

Coding problem2

This was a technical interview round with questions on DSA.

1. Stock Buy and Sell

Hard

20m average time

80% success

0/120

Asked in companies

You have been given an array 'PRICES' consisting of 'N' integers where PRICES[i] denotes the price of a given stock on the i-th day. You are also given an integer 'K' denoting the number of possible transactions you can make.

Your task is to find the maximum profit in at most K transactions. A valid transaction involves buying a stock and then selling it.

Note

You can’t engage in multiple transactions simultaneously, i.e. you must sell the stock before rebuying it.

For Example

Input: N = 6 , PRICES = [3, 2, 6, 5, 0, 3] and K = 2.
Output: 7

Explanation : The optimal way to get maximum profit is to buy the stock on day 2(price = 2) and sell it on day 3(price = 6) and rebuy it on day 5(price = 0) and sell it on day 6(price = 3). The maximum profit will be (6 - 2) + (3 - 0) = 7.

Problem approach

The idea is to traverse the given list of prices and find a local minimum of every increasing sequence. We can gain maximum profit if we buy the shares at the starting of every increasing sequence (local minimum) and sell them at the end of the increasing sequence (local maximum).

Steps :
1. Find the local minima and store it as starting index. If not exists, return.

2. Find the local maxima. and store it as an ending index. If we reach the end, set the end as the ending index.

3. Update the solution and Increment count of buy-sell pairs.

4. Repeat the above steps till the end is not reached.

Try solving now

2. Trapping Rain Water

Moderate

15m average time

80% success

0/80

Asked in companies

You have been given a long type array/list 'arr’ of size 'n’.

It represents an elevation map wherein 'arr[i]’ denotes the elevation of the 'ith' bar.

Print the total amount of rainwater that can be trapped in these elevations.

Note :

The width of each bar is the same and is equal to 1.

Example:

Input: ‘n’ = 6, ‘arr’ = [3, 0, 0, 2, 0, 4].

Output: 10

Explanation: Refer to the image for better comprehension:

Note :

You don't need to print anything. It has already been taken care of. Just implement the given function.

Problem approach

Basic Insight :
An element of the array can store water if there are higher bars on the left and right. The amount of water to be stored in every element can be found out by finding the heights of bars on the left and right sides. The idea is to compute the amount of water that can be stored in every element of the array.

Algorithm :
1) Create two arrays left and right of size n. create a variable max_ = INT_MIN.

2) Run one loop from start to end. In each iteration update max_ as max_ = max(max_, arr[i]) and also assign left[i] = max_

3) Update max_ = INT_MIN.

4) Run another loop from end to start. In each iteration update max_ as max_ = max(max_, arr[i]) and also assign right[i] = max_

5) Traverse the array from start to end.

6) The amount of water that will be stored in this column is min(a,b) – array[i],(where a = left[i] and b = right[i]) add this value to total amount of water stored

7) Print the total amount of water stored.

Try solving now