Cognzant Technology Solutions Interview Question for Android Engineers

Country: India
Interview Type: Written Test

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2
of 2 vote

``````public static long calculate(int n){
int sum = 0;
for(int i=1, factorial=1, sign=1; i <= n; i++, factorial*=i, sign*=-1)
sum += sign*factorial;
return sum;
}``````

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1
of 1 vote

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``````

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0
of 0 vote

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)
}``````

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1
of 1 vote

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0
of 0 vote

``````'use strict';

function compute(n) {
if (n <= 1) return 1;
let factorial = 1;
let result = 0;

for (let i = 1; i <= n; i++) {
factorial = factorial * i;
if (i % 2 == 0) {
result = result - factorial;
}
else {
result = result + factorial;
}
}

return result;
}``````

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0
of 0 vote

Swift 2.2. Doesn't account for overflow.

``````func sumOfFactorialsTo(n: Int)-> Int {
var result = 0
var factorial = 1
for i in 1 ... n {
factorial *= i
result += (i % 2 == 0 ? -1 : 1) * factorial

}
return result

}``````

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0
of 0 vote

``````// 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

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0
of 0 vote

``````public static int getFactorialSum(int n){

if(n == 1) return 1;

int factorial = 1;
int cummlativeSum = 0;
for(int j = 1; j <=n; j++){
factorial = factorial * j;
if(j % 2 != 0) cummlativeSum = cummlativeSum - factorial;
cummlativeSum = cummlativeSum + factorial;
}
return cummlativeSum;
}``````

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0
of 0 vote

Calculates the new Factorial in every iteration

``````public int Calculate(int n)
{
int total = 0;
int fact = -1;

for (int i=1; i <=n; i++)
{
fact *= -i;
total += fact;
}

}``````

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