## Bloomberg LP Interview Question for Software Engineers

Country: United States
Interview Type: In-Person

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``````def pow(x, n):
if n == 0:
return 1
neg = False
if n < 0:
neg = True
n = -n
pow = 1
while n > 1:
if n % 2 == 1:
pow *= x
n /= 2      # x^n = (x ^ 2) ^ (n / 2)
x *= x
return 1/(pow * x ) if neg else pow * x``````

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double power(int x, int n)
{
if (n < 0) return 1/pow(x,n);
else return pow(x,n);
}

double pow (int x, int n)
{
if (n == 0) return 1;

double v = power(x, n >> 1);

if (x % 2 == 0) return v*v;
else return v*v*x;
}

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@ aonecoding :
How your case handles fractional power?

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public static double pow(int x, int y) {
if(y == 0) return 1;
if(x == 0) return 0;
if(y < 0) return 1/pow(x, Math.abs(y));
String[] ind = Integer.toString(y).split(".");
int indd = Integer.toString(y).indexOf(".");
if(ind.length > 1 && Integer.parseInt(ind) > 0){
int denom = 1;
while(indd++ < Integer.toString(y).length()-1) denom *= 10;
return pow(x,Integer.parseInt(ind + ind))/pow(x,denom);
} else {
return x*pow(x, y-1);
}
}

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``````public int power(int number, int power) { int powerNumber=1;while(power !=0) { powerNumber = powerNumber * number;
power--;}``````

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{}

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{{{ int r =4; int n = 20; long double r=1; for (int i = 1; i <= n; ++i) { r*=x; } std::cout << "V=" << r; std::cout <<", STD=" << std::pow(x, n) << std::endl;
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iint pow(int x, int y){
if(y == 0){
return 1;
}
if(y == 1){
return x;
}
int temp = pow(x, y/2);
if(y%2 ==0){
temp = temp * temp;
}else{
temp = x * temp * temp;
}
return temp;
}

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0
of 0 vote

int pow(int x, int y){
if(y == 0){
return 1;
}
if(y == 1){
return x;
}
int temp = pow(x, y/2);
if(y%2 ==0){
temp = temp * temp;
}else{
temp = x * temp * temp;
}
return temp;
}

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of 0 vote

``````public static int power(int num, int pow) {
int res=1;
while(pow >= 1) {
res= num *	res;
pow --;
}``````

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of 0 vote

int pow(int x, int n)
{
if(n == 0)
return 1;

return x * pow(x, n-1);
}

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of 0 vote

``````int pow(int x, int n)
{
return n == 0 ? 1 : x * pow(x, n-1);
}``````

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0
of 0 vote

``````int pow(int x, int n)
{
return n == 0 ? 1 : x * pow(x, n-1);``````

}

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