## Facebook Interview Question for Software Engineers

• 1
of 1 vote

Country: United States
Interview Type: In-Person

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1
of 1 vote

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``````public static int divide(int dividend, int divisor) {

if(divisor == 0) throw new RuntimeException();

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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0
of 0 vote

``````private static void Divide(int a, int b, out int quotient, out int reminder)
{
quotient = 0;
reminder = 0;
int aCopy = a;
while (aCopy >= b)
{
quotient++;
aCopy -= b;
}

reminder = aCopy;
}``````

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0
of 0 vote

``````private static void Divide(int a, int b, out int quotient, out int reminder)
{
quotient = 0;
reminder = 0;
int aCopy = a;
while (aCopy >= b)
{
quotient++;
aCopy -= b;
}

reminder = aCopy;
}``````

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0
of 0 vote

As the function returns an int we suppose we are trimming the decimals and we just only return the int part, e.g. 3/2 = 1.5, result = 1

now we have that a/b = c thus giving us a=c*b, then we need to find a value where c times b is equal to a, so we iterate over the possibles values of c

``````function divide(a,b) {
i=0;
while(b*i<=a){
i++;
}
console.log(i-1);
}``````

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