Question

Microsoft | OA Coding Questions and Solutions | Set - 36 | 2022

Answer 1

Given A & B find count of all num which falls between A & B(both inclusive), where num = x * (x+1)
A = 6
B = 20
Answer = 3

 

2 * 3 = 6
3 * 4 = 12
4 * 5 = 20

 

A>=6,12, 20<=B

Answer 2

I'm not sure how to solve it if the input is negative, but if you assume the number is positive the answer would be easy
first, get the min factor of the first number but satisfy the condition x*(x+1) :
the factors of 6 are 1,2,3,6 so we take 2

 

            for (int i = 1; i < a; i++)
                if (a % i == 0 && i*(i+1)==a){
                    Console.Write( i );
                    break;
                }

 

then find the max factor of the second number but satisfy the condition x*(x+1)
the factors of 20 are 1,2,4,5,10,20 so take 5

 

      for (int i = b; i>=1; i--)
                if (b % i == 0 && i*(i-1)==b){
                    Console.Write( i );
                    break;
                }

 

the answer is 5-2=3

 

you can find factors of the number:

 

         for (int i = 1; i <=a; i++)
                if (a % i == 0)
                    Console.Write( i );

 

not sure about my answer, but I hope it helps.

Answer 3

it doesn't matter for negative numbers, since product of two consecutive numbers is always >=0.
On this idea , if B<0 simple return 0

 

val = sqrt(B)
now if(val*(val+1)) <= B then that's your upper bound else do val--
similarly find limits for A and all numbers between val1, val2 will define your range.
just return val1 - val2 as your answer.