Adobe Interview Question for Backend Developers


Country: India
Interview Type: Written Test




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First make a sieve of 10^5 according to constraints given.Also make a hash of given primes.Now take a variable, gcd_val and then find gcd of all elements. Now divide the array elements by this gcd_val. Now find each elements factorization, and check whether each elements factor is available in the hash or not. If any element's factor of given array is not in the hash,then print as "not possible", else print "possible".
Time complexity - O(nlogn)
Space complexity - O(n+k)
as per constraints - 1<=arr[i],n,k<=10^5

Example:1 -arr[]={ 7,14,21 }
primes[]={2,3}
gcd_val = 7
after dividing by gcd_val = 7, arr[]={1,2,3}

Example:2 arr[]={2,4,6}
primes[]={3}
gcd_val = 2
after dividing by gcd_val=2, arr[]={1,2,3}
here 2 is not in our hash, so "not possible".

** If anyone find anything wrong with my approach, feel free to correct me...:) **

- pawan kumar July 31, 2016 | Flag Reply
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bool check(int a[])
{
int a = a[0];
int b = a[1];
int gcd = gcd(a, b);
int prime = 0;
if(!isPrime(gcd))
{
prime = Prime(gcd);
}
else
{
prime = gcd;
}

for(int i = 2 ; i < a.length - 1; i++){
if(a[i] % prime != 0){
return false;
}
}
return true;
}

- hprem991 July 31, 2016 | Flag Reply
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Take each number and divide it by every prime until you can, you will find a base number. If this base number is equal for all the numbers in the array then it is possible make all the numbers of the array equal, otherwise not. find the basic code below. I will soon upload an optimized version on
minimalcodes.wordpress.com

int baseNumber;
    vector<int> numbers, primes;	//vectors containing primes and numbers
    bool start = true, possible = true;
    for(int i = 0; i < numbers.size(); i++){
        int number = numbers.at(i);
        for(int j = 0; j < primes.size(); j++){
            while(number % primes.at(j) == 0){ //&& number != 1
                number = number / primes.at(j);
            }
            if(number == 1){
                break;
            }
        }

        if(start){
            start = false;
            baseNumber = number;
        } else if (baseNumber != number){
            possible = false;
            break;
        }
    }

    cout << "My  method : " << possible << endl;

- manangandhi7 August 01, 2016 | Flag Reply
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public class ArrayTest {


	public static void main(String[] args) {

		int[] arr = {3,15};
		int[] primes = {5};

		System.out.println(checkEquality(arr, primes));
	}

	public static boolean checkEquality(int[] arr,int[] primes){
		int max = arr[0];
		for(int i = 1;i< arr.length;i++){
			if(max < arr[i]){
				max = arr[i];
			}
		}
		//Iterating over elements array
		for (int i = 0; i < arr.length; i++) {
			if(arr[i] == max){
				continue;
			}
			//Iterating over primes array
			for (int j = 0; j < primes.length; j++) {
				int num = arr[i];
				boolean cont = false;
				for (int k = 1; k < 10; k++) {
					num *= primes[j];
					if(num == max){
						cont = true;
						break;
					}else if(num < max){
						continue;
					}else{
						if(j == primes.length -1){
							return false;
						}
						break;
					}
				}
				if(cont){
					break;
				}
			}
		}
		return true;
	}


}

- ultikhopdi August 02, 2016 | Flag Reply
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Step1. Find LCM of all numbers in the array O(n)
Step2. For each number a[i]
- Divide LCM by a[i]
- Use each input prime number to divide the result to remove all factors of input prime numbers (can use modulo to check divisibility)
- If left over number is not 1, return false;
Step 3: Return true

- Raj August 20, 2016 | Flag Reply
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test

- anonymous August 20, 2016 | Flag Reply


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