Adobe Interview Question for Software Engineer / Developers


Country: India
Interview Type: In-Person




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0
of 0 vote

Forgot to mention A and B are integers..... so function will be something like
int result(int a,int b) so for a=5,b=3 it returns 2.

- mangal.govind April 11, 2012 | Flag Reply
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0
of 0 votes

{{a.increment(-b) }}

- roshan April 11, 2012 | Flag
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bool isnotEqual(int a, int b)
{
            int i=0;
            loop(a)
            {
                        increment(1); //increment i
            }
           loop(b)
           {
                         increment(-1); //decrease i
            }
            if( i)  
                     return true;
            else
                        return false;
}

int result(a,b)
{
      diff =0;
     while(isnotEqual(a,b))
      {
                        increment(1) //a++;
                        diff++;
       }
       return diff;
}

it will work correctly only if a<b

- Mridul Malpani April 11, 2012 | Flag Reply
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0
of 0 vote

My Bad.
increament(a) means increase 'a' by 1 not increase 'b' by a times if b.increament(a)

@roshan:negation of a number was not allowed.
@mridul: when you do increament(-1) it will result in 0 not in -2,just think......

- mangal.govind April 12, 2012 | Flag Reply
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0
of 2 vote

We can solve this using the approach that first take zero and decrement it to b times. Now to make the number negative we can use the logic of the way integer implemented. That suppose if integer is 8 bit long then its range is defined as -128 to 127 and it is cyclic; means when we increment 0 to 255 times it gives -1 and not 255. So using this logic Algo will be

int num =0;
loop(b)
{
 loop(INT_MAX)
  {
   increment(num);
  }
}
loop(a)
 {
   increment(num);
 }

result=num;

Code for the same {loop is general for loop and increment operator, otherwise you can implement you own Loop(a) and increment(a)  :) }

int main()
{
 int i,j,l=0,a=8,b=5;
 for(i=0;i<b;i++)
  {
   for(j=0;j<2147483647;j++)
        l++;
 l++;
for(j=0;j<2147483647;j++)
        l++;
 }  

printf("\n%d",l); 
  for(i=0;i<a;i++)
    l++;  
   
  printf("\n%d",l);
 return 0; 
}

- Anonymous April 12, 2012 | Flag Reply
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0
of 0 votes

in this solution here i have taken integer as 32 bit long and while looping in for loop i can compare j to (INT_MAX/2)-1 times otherwise it will go in infinite loop.... so one additional L++ required after the loop..... and one additional loop for other half rotation.....

- mangal.govind April 12, 2012 | Flag
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0
of 0 votes

what does increment(num) do? increament num by 1 or increament num by num. where num is an integer.

- gautamvermaroshan May 10, 2012 | Flag
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0
of 0 vote

int result(int a, int b)
{
int i,j=0;
for(i=a,j;i>b;i--,j++);
return j;
}

it'll work only if a>b

- anonymus April 12, 2012 | Flag Reply
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0
of 0 votes

we don't have ++ and -- operators so we can't use those, in my solution above if you see i have given the logic first, using loop(n) and increment(n) but to check if logic works we had to use inbuilt for loop to implement loop and increment .... otherwise had to implement loop and increment separately... din't use it for logic implementation.... and it works for all if a > or < or = b

But what you are doing here is just making use of ++ and -- operator to implement the a-b ,c heck pls....

- mangal.govind April 12, 2012 | Flag
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0
of 0 vote

loop(B) //loop B times
{
A.increment(255); // Converts A to -(A-1)
loop(256) // converts -(A-1) into (A-1)
{
A.increment(1)
}
}

- Chetan Toshniwal April 13, 2012 | Flag Reply
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0
of 0 vote

loop(B) //loop B times
{
A.increment(255); // Converts A to -(A-1)
loop(256) // converts -(A-1) into (A-1)
{
A.increment(1)
}
}

- Chetan Toshniwal April 13, 2012 | Flag Reply
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0
of 0 vote

Isn't computing a-b the same as computing -b+a? Why not initialize a value to -b and increment it a times?

int val = -b;
increment(a);

- Anonymous April 29, 2012 | Flag Reply
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0
of 0 vote

#include<iostream.h>
#define Increment(count) count++
#define Loop(count) for(int k=0;k<count;k++)


void Decrement(int &A)
{
        int i,j;
        i=0;j=0;
        Loop(A)
        {
                j=i;
                Increment(i);
        }
        A=j;
        cout<<"A="<<A<<" i="<<i<<" j="<<j<<endl;
}

void main()
{
        int A=10,B=6;
        Loop(B)
        {
                Decrement(A);
        }
        cout<<"A="<<A<<endl;
}

- himanshugupta2103 August 31, 2012 | Flag Reply


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