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# Cohesity Internship Interview Experience

• Difficulty Level : Medium
• Last Updated : 19 Aug, 2018

Cohesity came to our campus for 2 month internship.It has 1 online round and 2 F2F zoom round each of half hour.

Online round was on HackerRank having 2 question, We have to do it in 1:10 minutes.
Question 1 : Count maximum number of pair (a, b) possible from an array so that a is less than equal to 2*b. Also tell number of unpaired element left.(50 marks)
So for Input
1, 3, 5, 6, 7
ans = 2 1
as 1, 5 and 3, 6 are 2 pair and 7 is the only unpaired elelment left.
Question 2 : Given 2 string, say s1 and s2 find total number of substring such that s[i .. j]=s[k. . l] and j-i+1 is minimal. (20 Marks)
It seems hard, but is simple as j-i+1 should be minimal .One can use count sort to count number of char and then do summation as
sum+=count1[i]*count2[i];
where count1 and count2 are 26 length array giving number of times a char occurred in s1 and s2, respectively.

After round 1 only 5 Candidates were shortlisted for interview which had more than 57 marks.

For each Zoom round, We have only 30 min to understand the problem and give its pure C++ code.

Zoom Round 1:

I/P: 1-2-3-3-2-1
O/P: 2-3-2-1
Answer can be given in Recursive or Iterative approach.
Remember to take edge cases as List can be null or can have only 1 element.

Zoom Round 2: Find Diameter of a Binary tree.

Only 1 candidate was selected for the internship out of 5, me ðŸ˜‰ ..

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