Queue Using Stacks

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Queue Using Stacks

Question: How would you use stacks to implement a queue?

Answer: So how could we go about doing this? What if we used two stacks and used one as the incoming stack and the other as the outgoing stack? A queue is a FIFO structure, so when we enqueue something, we need to make sure that it does not get popped off before something that was already there. Similarly, when we dequeue something, we have to make sure that elements that were inserted earlier get ejected first.

What’s better than some code!

Stack in;
Stack out;
void enqueue( int value ) {
while( !out.isEmpty() ) {
in.push( out.pop() );
}
in.push( value );
}

int dequeue() {
while( !in.isEmpty() ) {
out.push( in.pop() );
}
return out.pop();
}

number of bit set

Question: Write a function that determines the number of bits set to 1 in the binary representation of an integer.

Answer: Going by the brute force approach, it would be easy to come up with the following solution. If the least significant bit of a number is 1, it will return a result of 1 when we AND it with 1. Using this logic, we can shift bits in the number to the right and keep testing if the least significant bit is 1. When there are no 1′s left in the number, it will become zero. Here’s some code to illustrate this simple logic.

int numOnesInInteger( int num )
{
int numOnes = 0;
while( num != 0 )
{
if( ( num & 1 ) == 1 ) {
numOnes++;
}
num = num >> 1;
}
return numOnes;
}
Think we can do better? The above algorithm runs in O(n) where n in the number of bits in the integer. Maybe we can make this a tad better. But this would need some real creative bit manipulation skills. So let’s think, what happens to a number when it we AND it with (number – 1). Let’s take 7 (0111) as an example.

7 & 6 = 0111 & 0110 = 0110 = 6
6 & 5 = 0110 & 0101 = 0100 = 4
4 & 3 = 0100 & 0011 = 0000 = 0

Do you see a pattern yet? Essentially, when you AND a number with (number – 1), the last 1 bit in the number gets set to zero. So if we continue to set every 1 bit in the number to zero, the number will eventually become zero. If we use this logic to count the number of 1s, we have an algorithm that runs in O(m) where m is the number of bits set to 1 in the number. Here’s some code.

int numOnesInInteger( int num )
{
int numOnes = 0;
while( num != 0 ){
num = num & (num – 1);
numOnes++;
}
return numOnes;
}

bit swaps required to convert integer A to integer B

Question: How would you find the number of bit swaps required to convert integer A to integer B?

Answer: Gut instinct might suggest that you go through every bit of A and every bit of B while simultaneously updating a count of the bits that are different. But how about a cooler way to solve this?

A key to solving a lot of bit manipulation questions is the use of the XOR functionality.

int bit_swaps_required( int a, int b ) {
unsigned int count = 0;
for( int c = a ^ b; c != 0; c = c >> 1 ) {
count += c & 1;
}
return count;
}
Simple right?

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puzzel

How Strong is an Egg?

Question: You have two identical eggs. Standing in front of a 100 floor building, you wonder what is the maximum number of floors from which the egg can be dropped without breaking it. What is the minimum number of tries needed to find out the solution?

Answer: The easiest way to do this would be to start from the first floor and drop the egg. If it doesn’t break, move on to the next floor. If it does break, then we know the maximum floor the egg will survive is 0. If we continue this process, we will easily find out the maximum floors the egg will survive with just one egg. So the maximum number of tries is 100 that is when the egg survives even at the 100th floor.

Can we do better? Of course we can. Let’s start at the second floor. If the egg breaks, then we can use the second egg to go back to the first floor and try again. If it does not break, then we can go ahead and try on the 4th floor (in multiples of 2). If it ever breaks, say at floor x, then we know it survived floor x-2. That leaves us with just floor x-1 to try with the second egg. So what is the maximum number of tries possible? It occurs when the egg survives 98 or 99 floors. It will take 50 tries to reach floor 100 and one more egg to try on the 99th floor so the total is 51 tries. Wow, that is almost half of what we had last time.

Can we do even better? Yes we can (Bob, the builder). What if we try at intervals of 3? Applying the same logic as the previous case, we need a max of 35 tries to find out the information (33 tries to reach 99th floor and 2 more on 97th and 98th floor).

Interval – Maximum tries
1 – 100
2 – 51
3 – 35
4 – 29
5 – 25
6 – 21
7 – 20
8 – 19
9 – 19
10 – 19
11 – 19
12 – 19
13 – 19
14 – 20
15 – 20
16 – 21

So picking any one of the intervals with 19 maximum tries would be fine.

Update: Thanks to RiderOfGiraffes for this solution.

Instead of taking equal intervals, we can increase the number of floors by one less than the previous increment. For example, let’s first try at floor 14. If it breaks, then we need 13 more tries to find the solution. If it doesn’t break, then we should try floor 27 (14 + 13). If it breaks, we need 12 more tries to find the solution. So the initial 2 tries plus the additional 12 tries would still be 14 tries in total. If it doesn’t break, we can try 39 (27 + 12) and so on. Using 14 as the initial floor, we can reach up to floor 105 (14 + 13 + 12 + … + 1) before we need more than 14 tries. Since we only need to cover 100 floors, 14 tries is sufficient to find the solution.

Therefore, 14 is the least number of tries to find out the solution.

If you have any questions, please feel free to send me an email at support@mytechinterviews.com. If you have any interview questions which you feel would benefit others, I would love to hear about it.

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