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Find original Array from given Array of GCD of prefix

by admin
3 years ago
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Given an array B[] of size N, the duty is to print an array A[] such that for each ith aspect B[i] is gcd of the primary i components of A[] i.e. Bi = gcd (A1, A2, …., Ai) and if no such array A[] exists, print −1.

Examples:

Enter: B = {4, 2} 
Output: 4 2
Clarification: One attainable reply is [4, 2] as a result of 
B may be generated as follows: B=[gcd(4), gcd(4, 2)]=[4, 2].

Enter: B = {2, 6, 8, 10} 
Output: -1
Clarification: No array exists which satisfies the given situation.

Method: The issue may be solved based mostly on the next remark: 

  • We all know that  Bi = gcd(A1, A2,  . . .  , Ai) and Bi+1 = gcd(A1, A2, . . ., Ai, Ai+1) = gcd(gcd(A1, A2, . . ., Ai), Ai+1) = gcd(Bi, Ai+1). This fashion, we will write Bi+1 = gcd(Bi , Ai+1).
  • The implication of that is that Bi+1 have to be an element of Bi , Since gcd of two numbers is divisor of each numbers. Therefore, situation Bi+1 divides Bi ought to maintain for all 1 ≤ i <N.
  • So if the given array has any such i the place Bi+1 doesn’t divide Bi , no such A can exist.
  • The given array B is a sound candidate for A as Bi+1 = gcd(Bi, Ai+1), however we’ve got Ai+1 = Bi+1 . Since Bi+1 divide Bi , gcd(Bi, Bi+1) = Bi+1. So the given array B satisfies our situation and may be printed as array A.

Observe the beneath steps to unravel the issue:

  • Initialize a boolean variable flag = true.
  • Iterate on the given array and verify the next:
    • If the subsequent aspect shouldn’t be an element of the present aspect:
      • Set flag = false.
      • Terminate the loop.
  • If the flag is true:
    • Print the given array B[].
    • else print -1.

Under is the implementation of the above strategy.

Java

  

class GFG {

  

    

    

    static void findOriginal(int arr[], int n)

    {

  

        

        boolean flag = true;

  

        for (int i = 0; i < n - 1; i++) {

  

            

            

            if (arr[i] % arr[i + 1] != 0) {

                flag = false;

                break;

            }

        }

  

        if (flag == false)

            System.out.println(-1);

        else {

            for (int val : arr) {

                System.out.print(val + " ");

            }

        }

    }

  

    

    public static void major(String[] args)

    {

        

        int B[] = { 4, 2 };

        int N = B.size;

  

        

        findOriginal(B, N);

    }

}

Time Complexity: O(N) for traversing the given array.
Auxiliary Area: O(1) as fixed house is used.



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