## Samsung Interview Question

Developer Program Engineers**Country:**India

**Interview Type:**Written Test

If my understanding is correct, following is the way to solve this.

Taking a simple example of 2 numbers spread over 3 place.

array A = {1,2}; n=3

say the number is: xyz

I see that repetition is allowed, as per the example.

when n=3, we have to take into account n=2 and n=1 based on what I see from the example given in the question.

x can have 1 or 2; y also can have 1 or 2; z also can have 1 or 2.

So the possibilities of all types of 3 digit numbers is: 2*2*2 = 8

Same way, possibilities of all types of 2 digit numbers is 2*2=4

Same way possibilities of all types of 1 digit number is 2

In this case answer would be 14.

Going by this, following snippet would reap the result.

numOfPlaces = 1

while(numOfPlaces<=n) {

result = result + array.length ^ numOfPlaces;

numOfPlaces++;

}

Note: I have not tested the while loop. but conceptually, I believe I have given a fair answer based on the question.

Correct me if wrong.

Question is not clear. Can you please rephrase the question and edit the question. When n > 0, doesn't it imply that it can be greater than 4?? One more doubt I have is given example for n=5

- prudent_programmer March 05, 20181,11,111,1111,11111,12341 , the numbers 222, 2222, 333, 3333, .... are also valid right?