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Monotype Interview Question for Senior Software Development Engineers


Country: India
Interview Type: In-Person
More Questions from This Interview



Comment hidden because of low score. Click to expand.
4
of 4 vote

A power of 2, when represented as binary number, is one bit 1 with all the remaining bits 0.

This code below flips the rightmost bit 1 of an integer. If the resulting number becomes zero, you know it had only one bit 1 before, thus a power of 2.

int power2(int a)
{
	a = a & (a-1);

	return a == 0;
}

- Victor December 05, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Its nice little trick

- Stuart December 06, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

The code above is incorrect, since it doesn't handle an important edge case [of 0 and negative numbers].
Handling of such edge cases is important for these kind of interview questions. Here's correct version:

bool power2(int a)
{
	return (a > 0) && ((a & (a-1)) == 0);
}

- 0xF4 December 14, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

A power of 2 is never a negative number.
2^1 = 2
2^0 = 1
2^(-1) = 1/2
...
2^(-inf) = 0

- Anonymous December 16, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

You're thinking of a multiple of 2. A power of two, in binary, will be represented by 00000010, 00000100, 00001000, 00010000, etc. Thus you need to count the # of bits. If it's more than one (you can stop when you get to 2), it's not a power of 2. Or if the 1's place is 1. So it would look like this:

if(a&1) return false;
for(i = 0; i < 32; i++) {
	if(a = a>>1  & 1)count ++;
	if (count>1) return false;
}
if count = 1 return true;
else return false;

- kdr213 August 10, 2016 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

int powerTobeFound = 2147483648;
    if ((powerTobeFound & (~powerTobeFound+1)) == powerTobeFound) {
        NSLog(@"Power of TWO");
    }

- deepak.hebbar December 09, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

return !a && a & (a - 1)

- Aman June 20, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>

using namespace std;

int main()
{

    int powerOfTwo;

     char exit = 'n';

    while(  exit=='n')
    {
        cout << "Enter number :";
        cin>> powerOfTwo;

        if(powerOfTwo==0)
        {
            cout << "Number is 0 only.";
        }
        if(!(powerOfTwo &(powerOfTwo-1)))
        {
            cout<<"Number is Power of 2..."<<endl;
        }
        else{
            cout<<"Number is not power of 2..."<<endl;
        }

        cout << "Do you want to exit ? Y/N" << endl;
        cin >>exit;
    }

    return 0;
}

- hilleybilley January 23, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>

main()
{
int a;
int o;
printf("enter # \n");
scanf("%d", &a);
o=-a;

if((a&o)==a) // get two's complement of the number and & with the given number gives
printf(" YES \n"); //the same number if its power of 2
else
printf(" NO \n");
}

- justin March 24, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>

main()
{
    int a;
    int o;
    printf("enter # \n");
    scanf("%d", &a);
    o=-a;

    if((a&o)==a)
        printf(" YES \n");
    else
        printf(" NO \n");
}

- justin March 24, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

The most effective algorithm to check for the number is a power of 2 or not we will follow the below implementation:

Implementation:

return x && (!(x&(x - 1)));

which will return either false for the number is not a power of 2 or will return true if it is.

- swapnilkant11 June 13, 2019 | Flag Reply


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