Yes, they’re emailing people one by one
Hey, Can you share your approach for the last problem?
On which test (A or B),Did you solve all 5 ?
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did you got email yet?
Test A(19 Oct).
When did you received mail?
No not yet.
No mail yet, but my roomate has received mail from InterviewBit
I also got the mail today.
I was able to solve only 3.
No not yet. btw there’s one more website providing this 6 months course(coding ninja). Y’all should check it out.
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When did you got mail? like rn or in the morning? I also solved 3 but havent received anything yet.
i got in morning around 10:30
Bro I’ve solved 4 and haven’t received any mail yet
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Just had my interview, they asked 4 very basic questions and I answered all of them, the interview only lasted for 20 mins. 
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Can you tell us some of the examples and topic that were asked in the interview?
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Sure.
Q1.
Interviewer(I) : Given a natural number N, find the numbers of bits required to store it.
Me(M) : \left \lfloor log_{2}(N) \right \rfloor + 1
Q2.
I : You are given 5 Apples, 3 Oranges and 6 Bananas, find the number of ways to arrange them in a single row?
M: \frac{14!}{5! \ . \ 3! \ . \ 6!}
Q3.
I : There is a person and an animal, the person can pet the animal only if either the name of the person is a sub-sequence of animal’s name or vice versa. You have tell if the person can pet the animal.
M: Wrote the code to do that in O(N) time and O(1) space, the interviewer seemed satisfied.
Q4.
I: Given a sorted array and a number N, find the number of occurrences of that number in the array in constant space.
M: subtract the lower_bound pointer for that number on that array from the upper_bound. Told him how these bounds can be implemented using Binary Search. Interviewer seemed satisfied.
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Wrote the code, where??, Is the interview on Google docs??
No, they have their own interface for that.
did you use recursion for the 3rd question
Nope iteration. Let’s a be the larger length string of the two(a and b are the input strings). Swap a and b if a is not longer in length than b.
Then this code should work. This is just a pseudo-code-
int length_a = a.length(), length_b = b.length();
if(length_a < length_b)
swap(a, b);
int ptr_a = 0, ptr_b = 0;
// find if b is a sub-sequence of a
while(ptr_a < length_a && ptr_b < length_b) {
if(a[ptr_a] == b[ptr_b])
++ptr_b;
++ptr_a;
}
// if all chars in b matches
if(ptr_b == length_b)
cout << "yes, can pet";
else
cout << "no, cant pet";