Find original numbers from gcd() every pair in Python


Suppose we have an array A where GCD of every possible pair of elements of another array is given, we have to find the original numbers which are used to compute the given GCD array.

So, if the input is like A = [6, 1, 1, 13], then the output will be [13, 6] as gcd(13, 13) is 13, gcd(13, 6) is 1, gcd(6, 13) is 1, gcd(6, 6) is 6

To solve this, we will follow these steps −

  • n := size of A

  • sort array A in descending order

  • occurrence := an array of size A[0] and fill with 0

  • for i in range 0 to n, do

    • occurrence[A[i]] := occurrence[A[i]] + 1

  • size := integer of square root of n

  • res := an array of size same as size of A and fill with 0

  • l := 0

  • for i in range 0 to n, do

    • if occurrence[A[i]] > 0, then

      • res[l] := A[i]

      • occurrence[res[l]] := occurrence[res[l]] - 1

      • l := l + 1

      • for j in range 0 to l, do

        • if i is not same as j, then

          • g := gcd(A[i], res[j])

          • occurrence[g] := occurrence[g] - 2

  • return res[from index 0 to size]

Example

Let us see the following implementation to get better understanding −

 Live Demo

from math import sqrt, gcd
def get_actual_array(A):
   n = len(A)
   A.sort(reverse = True)
   occurrence = [0 for i in range(A[0] + 1)]
   for i in range(n):
      occurrence[A[i]] += 1
   size = int(sqrt(n))
   res = [0 for i in range(len(A))]
   l = 0
   for i in range(n):
      if (occurrence[A[i]] > 0):
         res[l] = A[i]
         occurrence[res[l]] -= 1
         l += 1
         for j in range(l):
            if (i != j):
               g = gcd(A[i], res[j])
               occurrence[g] -= 2
   return res[:size]
A = [6, 1, 1, 13]
print(get_actual_array(A))

Input

[6, 1, 1, 13]

Output

[13, 6]

Updated on: 25-Aug-2020

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