## Cognzant Technology Solutions Interview Question

Android Engineers**Country:**India

**Interview Type:**Written Test

Observation:

As there are only 3 elements of series it's hard to assume anything. But if the assumption of deducting even factorial from the previous odd factorial.

here is the solution with O(1) space and O(n) time

```
def fancyFactorial(n):
n1fact = 1
result = 1
if n <= 1:
return n1fact
i = 2
while i <= n:
n1fact = n1fact * i
if i % 2 == 0:
result = result - n1fact
else:
result = result + n1fact
i += 1
return result
```

Swift

```
///func factorial(n:Int, result:Int = 1)->Int {
/// if n < 1 { return result }
/// return factorial(n - 1, result: result * n)
///}
///func nfactorial(n:Int, result:Int = 0)->Int {
/// if n < 1 { return result }
/// return nfactorial(n - 1, result: factorial(n) + result)
///}
```

nfactorial(5) // 153

// I did read the problem wrong. Need new glasses. Thanks!

```
func factorial(n:Int, result:Int = 1)->Int {
if n < 1 { return result }
return factorial(n - 1, result: result * n)
}
func nfactorial(n:Int, val:Int = 1, result:Int = 0)->Int {
if val > n { return result }
let f = factorial(val)
let m = val % 2 == 0 ? (result - f) : (result + f)
return nfactorial(n, val:val + 1, result:m)
}
```

```
// return 1! - 2! + 3! - 4!... n!
public static int calculate(int n) {
int []factorial = new int[n];
factorial[0] = 1; //1!
int result = factorial[0];
for(int i=2;i<=n;i++) { // i = 1
factorial [i-1] = factorial[i - 2] * i;
if(n % 2 == 0) {
result -= factorial[i-1];
} else {
result += factorial[i-1];
}
}
return result;
}
```

Time Complexity : O(n) for iterations, O(1) - constant access for array

Space Complexity : Additional space of size n

- Roman August 16, 2016