Zoho Interview Experience | 2021
Hi friends,
Myself GopalaKrishnan. I’m going to share my On-Campus Zoho Interview Experience. Zoho came to my college (GCE, Salem) this year (2021) . There were 4 rounds. Zoho Corporation is an Indian software development company.
Level – 1 ( 2 hours)
10-15 Aptitude Questions
20-30 C programming Questions and Flow Charts
Level – 2 (30 MINUTES)
3 – C Programming Output Questions
2 – Flow Chart Questions
Level -3 (3 HOURS)
This is Programming round. 6 questions were asked.
1.Remove the given element in a array.
Input : [1, 5, 6, 5, 2,4]
Output: [ 1, 6, 2 ,4]
int rem = 5, j =0;
int a[] = {1, 5, 6, 5, 2,4};
for(int i =0; i<a.length; i++)
{
if(a[i] != rem)
{
a[j++] = a[i];
}
}
for(int i =0; i<j ; i++)
{
System.out.print(a[i]+"");
}
2.Remove the duplicate element in the array
For this question I used variable ‘len’ to store the initial size of the array. I checked the every element of the array (from 0 to len) with other element of the array. If it match, move that matched element to last of the array and reduce the size of the l. Finally print the array elements from o to len.
3.Rotate the array elements according to the given input (1.left rotate 2.right rotate)for the given number of times (n)
Consider array as [15,21,32,42,27,51]
- If the user select option 2 and (num of rotation) n =2, then we have to right rotate , for that first I reverse the whole array (passing the argument as i=0,j=Length-1)[51,27,42,32,21,15]
- Then I reverse the array (from first n element , n will no of rotation(n = 2)[27,51,42,32,21,15]
- Then I reverse the remaining element (from n index to last)[27,51,15,21,32,42]
4. Sort the given array according to its 1’s count in binary.
Input: [ 4, 0, 3]
Output[ 3, 4 0 ]
Explanation: 3( 0 1 1) has 2 set bits and 4 ( 1 0 0 ) has 1 set bits and 0 ( 000) has no set bits.
Here for sorting I use normal sorting algorithm with O(n^2) time complexity. For getting 1’s count for each number I used function called onesCount(). In that function , For the given number I make a AND operation with 1 if the result is 1 then I increase the count value and then I right shift the given number by 1. Refer below code.
onesCount(int n)
{
int count = 0;
while (n) {
count += n & 1;
n >>= 1;
}
return count;
}
6. Printing the elements of the matrix from the given index
5 6 15 16 25 4 7 14 17 24 3 8 13 18 23 2 9 12 19 22 1 10 11 20 21 input :4 0 output:1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 input:4 1 output:10 9 8 7 6 15 14 13 12 11 20 19 18 17 16 25 24 23 22 21 1 2 3 4 5
Technical & HR (15-20 Minutes)
1.Some Technical questions from java (like interface vs abstract class)
2.Some Questions from Project
3.Puzzle ( a car have 4 tyre + 1 spare tyre each tyre can travel a maximum distance of 20Km what is the maximum distance the car can travel ?
Solved this puzzle Answer:25 Km we have to swap the tyre for every 5km)
Finally I accepted the Offer 🙂
