## Siemens Interview Question

Java Developers**Country:**India

**Interview Type:**Phone Interview

+1 I think the answer at stackoverflow is the closest answer. However, it's still not O(1) as claimed, because you need to iterate the array to calculate the actual sum and the squared sum of the array.

If the sum of all numbers and the sum of squares of all numbers of the array are not given, this question cannot be solved in less than linear time.

The code for the stackoverflow link:

```
#include <bits/stdc++.h>
using namespace std;
int main () {
int n;
scanf("%d",&n);
int Nsum = 0;
int NSsum = 0;
Nsum = (n*(n+1))/2;
NSsum = n*(n+1)*(2*n+1)/6;
vector < int > v (n+1, 0);
int sum = 0;
int Ssum = 0;
for (int i=0; i<n-2; i++) {
scanf("%d",&v[i]);
sum = sum + v[i];
Ssum = Ssum + v[i]*v[i];
}
int aplusb = Nsum - sum;
int asqplusbsq = NSsum - Ssum;
int twoab = aplusb * aplusb - asqplusbsq;
int aminusb = int(sqrt(asqplusbsq - twoab));
int x = (aplusb + aminusb)/2;
int y = (aplusb - aminusb)/2;
cout << x << "x " << y << "y " << endl;
return 0;
}
```

Sum of the missing numbers = (100*101)/2 - Add all numbers

Product of the missing numbers = product of 100 numbers/ product of actual numbers

Use above both to find out missing numbers

He already mentioned, that we should not traverse the complete Array. Then how can you find the sum of complete Array

@prashantreddy: Isn't it obvious you have to look at at least Omega(n) numbers? i.e. the restriction is bogus, and made up by the person asking the question.

As @Sylla pointed out 100! is a very large number and won't fit in any basic primitive data type. You can argue that you can use a language that supports big numbers like Python or Java, but then I don't think that's the intention. @dn suggested the link at stackoverlow and that solution seems more reasonable. Using sum of numbers and sum of square of numbers, which both can be calculated with Gaussian formulas.

@oOZz: Even if you use Python or something like BigInteger in Java, you have to consider how those operations are implemented behind the covers. They are not O(1). Behind the covers, a single oversized integer has to be stored as a (possibly large) array of ints, longs, bytes, or other primitives (implementation-dependent). To compute a sum or product, the array representing the number has to be iterated over. There's no wand that a computer can waive to magically take huge numbers and multiply them together in one step. The language / library has to apply an algorithm that iterates over and operates on the underlying array of primitive types.

The cost of addition can be expected to be O(k) to add two k-digit numbers. Multiplication is at least O(k log k), but can be more depending on the implementation (is O(k^2) in the naive implementation).

It will not work in practice - 100! = 9.3326215443944152681699238856267e+157. Compared with Sum(1,100) = 5050 the difference is too big to solve the equation with acceptable precision.

@Prashant: Code is pretty simple. Let's do sum mathematics first...

Let a and b are the missing numbers.

{

Sum of 100 num = Sum of 98 num + a +b

=> a + b = (Sum of 100 num ) - (Sum of 98 num ) = x suppose

Product of 100 num = Product of 98 num * a *b

=> a*b = Product of 100 num /Product of 98 num = y suppose

x and y are known. We need to find a and b

a+b = x -----1

a*b = y .......2

(a-b)^2 = (a+b)^2 - 4ab

From 1 and 2

=> a-b = sqrt(x^2 - 4y) = z suppose

a + b = x

=> a = (x+z)/2

b = x - b

}

I dont think any special coding skills neede to write a program for this. :)

Finding Sum and product of given numbers takes O(n) time complexity and yes I dont think it is possible with lesser time complexity

I agree ..it looks pretty simple but i am not able to implement this in java.. I am not sure how to find out a and b after computing sum and product of missing number

```
public static void main(String[] args) {
int arr[]={2,1,4,6,5,8,9,10};
int finalSum = 10*(10+1)/2 ;
int finalProduct = 1*2*3*4*5*6*7*8*9*10;
// a and b are missing number
int a = 0,b = 0;
// product if missing number
int t1 = finalProduct/(1*2*4*5*6*8*9*10);
// sum of missing number
int t2 = finalSum - (1+2+4+5+6+8+9+10);
// Find out a and b ????
```

}

This question is nearly impossible to solve in less than O(n)..

-Solving n! and etc. etc. is really hectic task.

-Still we can generalize questions like this with "k" missing numbers by using an extra arrays.

-Using one extra Arrays you would take O(n) time to find all the missing "K" numbers.

LOGIC:

1.)Initialize all the elements of extra array to zero.

2.)Traverse the given array form the beginning.

3.)Replace zero with one in the extra array at the index of the element found.

eg: if 57 is found in the given array then replace - extra[57] with 1.

4.)At the end traverse the extra array and note down the index values of the places having 0.

stackoverflow.com/questions/10218791/find-two-missing-numbers

- dn May 22, 2013