InterviewBit academy

general
38

Can you tell us some of the examples and topic that were asked in the interview?

1 Like
39

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.

3 Likes
40

Wrote the code, where??, Is the interview on Google docs??

41

No, they have their own interface for that.

42

did you use recursion for the 3rd question

43

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";
44

Whatever, i haven’t received the mail yet, I’m losing hope now :frowning:

45

same dude idk what is happening.

46

How many did you solve

47

They have interviews till 26 th, so don’t worry you may get an email soon. Otherwise why not ping them directly and ask if results are still pending for some students?

1 Like
48

Maybe they’re going through the applications in the order they came, i had applied for the course just before the deadline
I’ll ping them by the end of the day when i reach home. Thanks

1 Like
49

Possible, I had applied around a month before or so.

1 Like
50

Hope so
(10 chars)

51

3 and a half

52

You probably get in, you should ping them too. :slight_smile:

53

I checked the website, it says, “Congrats, you have cleared the entrance test. As a next step please select a slot of 30 mins for personal interview”, but I didn’t receive any mail, Lol, I was worried, anyways, thanks for the support, good luck to all of you.

1 Like
54

What happens when the array is [1, 2, 2, 2, 2, 2], where will the upper bound pointer point at ??

1 Like
55

Didn’t know that there is also a case of direct selection .

56

You didn’t specify the number N, but lets take both 1 and 2
1-
lower_bound points to 0 and upper_bound points to 1. So, 1 - 0 = 1, and there only 1 instance of 1
2-
lower bound points to 1 and upper bound points to the end of the array that is 6, so 6 - 1 = 5 and there are 5 instances of 2.

1 Like
57

I meant n to be 2, anyways good solution, congrats on your selection!
Hopefully I’ll see you in IBAcademy then

1 Like