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

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Cvent
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4 rounds | 7 Coding problems

Interview preparation journey

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Journey
My journey started with mastering the basics, progressively diving into more complex concepts. I embraced challenges, sought help, and turned setbacks into learning opportunities. Continuous learning was key, as I stayed updated with industry trends. Interview preparation included problem-solving practice, mock interviews, and in-depth company research. Rejections didn't deter me; instead, they fuelled improvement. With a strong foundation, resilience, and dedicated preparation, I successfully cracked job interviews. My journey highlights the significance of learning, perseverance, and adaptability.
Application story
Cvent India visited our campus for on-campus hiring, setting a CGPA requirement of above 7. The recruitment process included two online test rounds followed by two interview rounds. This stringent criterion showcased the company's emphasis on academic excellence and technical competence in potential candidates.
Why selected/rejected for the role?
During the final round, an internet issue prevented me from activating my webcam. The interviewer kindly advised me to resolve the problem. It took approximately 25 minutes to switch to a different laptop with a stable connection. Despite the delay, the interviewer conducted the interview for a shortened duration of 30 minutes.
Preparation
Duration: 6 months
Topics: I prepared in-depth for the following topics: Object-Oriented Programming (OOPS), Operating Systems (OS), Database Management Systems (DBMS), Stack, Queue, Arrays, and Strings.
Tip
Tip

Tip 1: Engage in Regular Coding Practice Sessions.
Tip 2: Create Real-World Projects to Apply Knowledge Effectively.

Application process
Where: Campus
Eligibility: 7
Resume Tip
Resume tip

Tip 1: Enhance your resume with relevant experiences that demonstrate your skills.
Tip 2: Include impressive and impactful projects that showcase your abilities effectively.

Interview rounds

01
Round
Easy
Online Coding Interview
Duration60 mins
Interview date24 Aug 2023
Coding problem2

1. Find Minimum in Rotated Sorted Array

Moderate
10m average time
90% success
0/80
Asked in companies
OlaAmazonMorgan Stanley

You are given an array 'arr' of size 'n' having unique elements that has been sorted in ascending order and rotated between 1 and 'n' times which is unknown.


The rotation involves shifting every element to the right, with the last element moving to the first position. For example, if 'arr' = [1, 2, 3, 4] was rotated one time, it would become [4, 1, 2, 3].


Your task is to find the minimum number in this array.


Note :
All the elements in the array are distinct. 


Example :
Input: arr = [3,4,5,1,2]

Output: 1

Explanation: The original array was [1,2,3,4,5] and it was rotated 3 times.


Try solving now

2. Reverse Only Letters

Easy
15m average time
85% success
0/40
Asked in companies
AppleOYOPayPal

You are given a string, ‘S’. You need to reverse the string where characters that are not an alphabet stay in the same place, and the rest reverse their positions.

Eg: “a-bcd” becomes “d-cba”

Try solving now
02
Round
Easy
Online Coding Interview
Duration30 Minutes
Interview date24 Aug 2023
Coding problem2

1. Combination Sum

Easy
15m average time
85% success
0/40
Asked in companies
UberFacebookLinkedIn

You are given an array 'ARR' of 'N' distinct positive integers. You are also given a non-negative integer 'B'.


Your task is to return all unique combinations in the array whose sum equals 'B'. A number can be chosen any number of times from the array 'ARR'.


Elements in each combination must be in non-decreasing order.


For example:
Let the array 'ARR' be [1, 2, 3] and 'B' = 5. Then all possible valid combinations are-

(1, 1, 1, 1, 1)
(1, 1, 1, 2)
(1, 1, 3)
(1, 2, 2)
(2, 3)
Problem approach

int mod=1e9+7;
int dp [1001] [1001]; 
int solve(int a, int b,int cnt1,int cnt2,int mx, int mn) 
{
int len=a*cnt1+b*cnt2;
if (len>mx)
return 0;
if(dp [cnt1] [cnt2]!=-1)
return dp[cnt1] [cnt2];
int ans=(len>=mn)?1:0;
int ans1=solve (a,b, cnt1+1, cnt2,mx, mn)%mod;
int ans2=solve(a,b, cnt1, cnt2+1, mx,mn )%mod;
ans=(ans+ans2+ans1)%mod;
return dp [cnt1][cnt2]=ans;

}


int StartupNames (int minLength, int maxLength, int cntone, int cntTwo){

memset (dp,-1, sizeof (dp));
return solve (cntOne, cntTwo, 0, 0, maxLength, minLength) ;
}

Try solving now

2. Row Wave Form

Easy
15m average time
80% success
0/40
Asked in companies
AmazonVir SoftechCvent

You are given a 2D array with dimensions ‘N*M’. You need to read the array elements row-wise and return a linear array that stores the elements like a wave i.e the 1st-row elements are stored from left to right, 2nd-row elements are stored from right to left, and so on.

Try solving now
03
Round
Easy
Video Call
Duration60 minutes
Interview date24 Aug 2023
Coding problem2

1. Stack using queue

Moderate
25m average time
65% success
0/80
Asked in companies
DunzoOptumBig Basket

Implement a Stack Data Structure specifically to store integer data using two Queues.


There should be two data members, both being Queues to store the data internally. You may use the inbuilt Queue.


Implement the following public functions :

1. Constructor:
It initializes the data members(queues) as required.

2. push(data) :
This function should take one argument of type integer. It pushes the element into the stack and returns nothing.

3. pop() :
It pops the element from the top of the stack and, in turn, returns the element being popped or deleted. In case the stack is empty, it returns -1.

4. top :
It returns the element being kept at the top of the stack. In case the stack is empty, it returns -1.

5. size() :
It returns the size of the stack at any given instance of time.

6. isEmpty() :
It returns a boolean value indicating whether the stack is empty or not.
Operations Performed on the Stack:
Query-1(Denoted by an integer 1): Pushes an integer data to the stack. (push function)

Query-2(Denoted by an integer 2): Pops the data kept at the top of the stack and returns it to the caller. (pop function)

Query-3(Denoted by an integer 3): Fetches and returns the data being kept at the top of the stack but doesn't remove it, unlike the pop function. (top function)

Query-4(Denoted by an integer 4): Returns the current size of the stack. (size function)

Query-5(Denoted by an integer 5): Returns a boolean value denoting whether the stack is empty or not. (isEmpty function)
Example
Operations: 
1 5
1 10
2
3
4

Enqueue operation 1 5: We insert 5 at the back of the queue.
  Queue: [5]

Enqueue operation 1 10: We insert 10 at the back of the queue.
  Queue: [5, 10]

Dequeue operation 2: We remove the element from the front of the queue, which is 5, and print it.
  Output: 5
  Queue: [10]

Peek operation 3: We return the element present at the front of the queue, which is 10, without removing it.
  Output: 10
  Queue: [10]

IsEmpty operation 4: We check if the queue is empty.
  Output: False
  Queue: [10]
Problem approach

#include
#include
using namespace std;
queueq1;
queueq2;

void pushElement(int x,int size)
{
if(size==q1.size())
{
cout<<"Stack is Full"<>size;
while(n!=-1)
{
cout<<"Enter Your choice"<>n;
if(n==1)
{
int x;
cin>>x;
pushElement(x,size);
}
else if(n==2)
{
int temp=topElement();
if(temp==-1)
cout<<"Stack is empty"< else
cout<<"Top element is "< }
else if(n==3)
{
popElement();
}
else if(n==4)
{
printStack();
}
else break;
}
return 0;
}

Try solving now

2. Min Stack

Easy
0/40
Asked in companies
Natwest GroupPostmanPayPal

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

1. Push(num): Push the given number in the stack.
2. Pop: Remove and return the top element from the stack if present, else return -1.
3. Top: return the top element of the stack if present, else return -1.
4. getMin: Returns minimum element of the stack if present, else return -1.

For Example:

For the following input: 
1
5
1 1
1 2
4
2
3

For the first two operations, we will just insert 1 and then 2 into the stack which was empty earlier. So now the stack is => [2,1]
In the third operation, we need to return the minimum element of the stack, i.e., 1. So now the stack is => [2,1]
For the fourth operation, we need to pop the topmost element of the stack, i.e., 2. Now the stack is => [1]
In the fifth operation, we return the top element of the stack, i.e. 1 as it has one element. Now the stack is => [1]

So, the final output will be: 
1 2 1
Problem approach

class MinStack {
public:
stack>s;
MinStack() {

}

void push(int val) {
if(s.size()==0)
{
s.push({val,val});
}
else
{
auto x=s.top();
if(x.second {
s.push({val,x.second});
}
else
{
s.push({val,val});
}
}
}

void pop() {
if(s.size()==0)
return ;
else
s.pop();
}

int top() {
return s.top().first;
}

int getMin() {
return s.top().second;
}
};

Try solving now
04
Round
Easy
Video Call
Duration30 minutes
Interview date24 Aug 2023
Coding problem1

1. Product Of Array Except Self

Easy
26m average time
0/40
Asked in companies
FacebookDelhiveryIntuit

You have been given an integer array/list (ARR) of size N. You have to return an array/list PRODUCT such that PRODUCT[i] is equal to the product of all the elements of ARR except ARR[i]

 Note :
Each product can cross the integer limits, so we should take modulo of the operation. 

Take MOD = 10^9 + 7 to always stay in the limits.
Follow up :
Can you try solving the problem in O(1) space?
Problem approach

class Solution {
public:
vector productExceptSelf(vector& arr) {
int n=arr.size();
vectorleft(n,1);

for(int i=0;i=0;i--)
{
int k=temp;
temp*=arr[i];
if(i==0)
arr[i]=k;
else if(i==n-1)
arr[i]=left[i-1];
else{
arr[i]=k*left[i-1];
}
}
return arr;
}
};

Try solving now

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Skill covered: Programming

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