## Facebook Interview Question for Software Engineers

Country: United States
Interview Type: In-Person

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3
of 3 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;
}``````

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1
of 1 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) )``````

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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);``````

Comment hidden because of low score. Click to expand.
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)``````

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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);``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````public static void divideNumbers(int a , int b){
int result = 0;
int remainder = 0;

while(a-b > 0){
result++;
remainder = a-b;
a = remainder;
}

System.out.println("Result = "+ result);
System.out.println("Remainder = "+ remainder);
}``````

Comment hidden because of low score. Click to expand.
-1
of 1 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;
}``````

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