Facebook Interview Question for Software Engineers


Country: United States
Interview Type: In-Person




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2
of 2 vote

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public static int divide(int dividend, int divisor) {
        //if(divisor == 0) throw new Exception();
        int sign = 1;
        //figure out the sign of the result
        if((dividend < 0 && divisor > 0)
                || (dividend > 0 && divisor < 0)) {
            sign = -1;
        }
        if(dividend < 0) dividend = -dividend;
        if(divisor < 0) divisor = -divisor;
        int n = 1; //get the range of result [n, 2n], where n is doubled every round
        while(dividend > (n << 1) * divisor) {
            n <<= 1;
        }
        //now it's known that the result is between [n, 2n]
        //so in the dividend has a sum of more than n and less than 2n divisors in total
        //to figure out exactly how many divisors sum up to the dividend,
        // break down the problem to result = n + divide(dividend - n * divisor, divisor)
        // next round the dividend becomes (dividend - n * divisor) with n added to the result.
        // proceed to figure out the range [x, 2x] of result for the new dividend, where x can be n/2, n/4, n/8...
        // whenever the x is found, add x to the result and deduce x * divisor from the dividend
        // util n is 0 or dividend is 0

        //If written in math, the formula looks like dividend = ([T/F]* n + [T/F]* n/2 + [T/F]* n/ 4 + [T/F]* n/ 8...) * divisor
        //The quotient will be ([T/F]* n + [T/F]* n/2 + [T/F]* n/ 4 + [T/F]* n/ 8...), [T/F] depends on (dividend - n * divisor) >= 0
        int result = 0;
        while(n > 0 && dividend > 0) {
            if(dividend - n * divisor >= 0) {
                dividend -= n * divisor;
                result += n;
            }
            n >>= 1;
        }
        return result * sign;
    }

- aonecoding July 15, 2017 | Flag Reply
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1
of 1 vote

a divided by b:

// a/b
        int a = 41;
        int b = 5;
        int rem = a;
        int ans =0;
        while(rem>=b){
            rem-=b;
            ans++;
        }
        System.out.println("ans:"+ans+"|rem="+rem);

- krupen.ghetiya July 22, 2017 | Flag Reply
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0
of 0 vote

Why make it more complex, this is clean, neat... and works.

def div_wo(a,b){
  panic( b == 0 , 'Terrible parameter!')
  printf('%d/%d%n', a,b)
  sign_change = false  
  if ( a < 0  ){
    sign_change = true   
    a = -a 
  } 
  if ( b < 0 ){
    sign_change ^= true
    b = -b
  } 
  res = 0 
  while ( a >= b ){
     res += 1
     a -= b 
  } 
  sign_change ? -res : res 
}
println( div_wo(0,1) )
println( div_wo(1,1) )
println( div_wo(1,2) )
println( div_wo(2,1) )
println( div_wo(2,-1) )
println( div_wo(-2,-1) )
println( div_wo(-2,1) )
println( div_wo(3,2) )

- NoOne July 16, 2017 | Flag Reply
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0
of 0 vote

@NoOne: it's then just linear runtime O(dividend/divisor) whereas aone did actually place a logarithm in there (O(lg(dividend/divisor)).

1. terminology:
quotient = dividend / divisor

2. assumptions, constraints:
a) integer values (including signs)
b) positive and negative
c) since python, integers are 'unbound' no need to check overflow, but with Java or c++ you should check overflow, actually!

3. solution
1) turn the signs around until only positive integers remain
2) now do a 'high school division' on paper kind of thing in base 2, e.g.
1001001101 : 101 = (1 + 4 + 8 + 64 + 512) / 5
1111234567
1) 1001 / 101 = 1 remainder 100
2) 1000 / 101 = 1 remainder 11
3) 110 / 101 = 1 remainder 1
4) 11 / 101 = 0 remainder 11
5) 111 / 101 = 1 remainder 10
6) 100 / 101 = 0 remainder 100
7) 1001 / 100 = 1 remainder 100
= 1110101 Remainder 100

def division(dividend, divisor):
  # division by zero case
  if divisor == 0: raise ZeroDivisionError()

  # turn signs to get rid of -dividend, -divisor
  if dividend < 0: return -division(-dividend, divisor)  
  if divisor < 0: return -division(dividend, -divisor)

  # sdiv = divisor * 2^n, sdiv > divisor and sdiv <= divisor * 2
  sdiv = divisor
  while sdiv < dividend:
    sdiv <<= 1 

  # remove divisior * 2^i from dividend until divisor * 2^0
  # therefore the quotient is multiplied by two in each iteration
  quotient = 0
  while sdiv >= divisor:
    quotient <<= 1 
    if dividend >= sdiv:
      dividend -= sdiv
      quotient |= 1
    sdiv >>= 1

  return quotient

print(division(981, 12) == 81)
print(division(11, 12) == 0)
print(division(13, 13) == 1)
print(division(99191, 15) == 6612)
print(division(-1, 1) == -1)
print(division(1, -2) == 0)
print(division(2, -1) == -2)
print(division(-9, -3) == 3)

- ChrisK July 17, 2017 | Flag Reply
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0
of 0 vote

inp1 divided by inp2:

int inp1 = 32;
        int inp2 = -8;
        int a = Math.abs(inp1);
        int b = Math.abs(inp2);
        int rem = a;
        int ans =0;
        if(b>0 && a>0){
            while(rem>=b){
                rem-=b;
                ans++;
            }
        }
        if((inp1<0 && inp2>0) ||(inp1>0 && inp2<0) ){
            ans=-ans;
        }
        System.out.println("ans:"+ans);

- krupen.ghetiya July 22, 2017 | Flag Reply
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0
of 0 vote

public static void main(String args[]) throws Exception {
		System.out.println(divideWHTSign(300, 30));
	}

	public static int divideWHTSign(int a, int b) {
		int quotient = 0;
		while (a >= b) {
			quotient++;
			a -= b;
		}
		return quotient;
	}

- koustav.adorable July 23, 2017 | Flag Reply


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