Python program to Find the Jumbo GCD subarray

Given an array, write a program to find the maximum GCD among all the subarray of the size >= 2 of the given array.

Examples:

Input list: [2, 3, 4, 4, 4]
Output: 4

Input list: [3, 7, 2, 9, 18, 5, 1, 13 ]
Output: 9

Approach:

• Import the math module for python
• Introduce a variable(say, V1) to store the gcd of each element of the list while looping through the list.
• Iterate through the elements of the array or list using a loop.
• At each iteration call the math.gcd() function.
• Store the outcome of the math.gcd() function to another variable (say, V2) at each iteration.
• Now compare V1 and V2. If V2 is greater than V1, set V1 equal to V2 else pass.
• Let the loop run through and print out the final value of V1.

Below you can find the implementation of the above-mentioned approach:
Examples 1:

Python3

import math

# input list
List = [2, 3, 4, 4, 4 ]

max1 = 0
for i in range(len(List)-1):

  # use math.gcd() function
  gcd1 = math.gcd(List[i], List[i + 1])
  if(gcd1>max1):
    max1 = gcd1

# print max1
# as the result
print(max1)

Output:

4

Explanation: For the given array one of the subarrays having maximum gcd is[3, 5] which has gcd 4.

Time Complexity:  O(n * log(min(a, b))), as it iterates through the input list once of size n, and gcd() method takes log(min(a,b)) time. 
Auxiliary Space: O(log(n)), as it uses recursion and the maximum depth of recursion is log(n).

Example 2:

Python3

import math

# input list
List = [3, 7, 2, 9, 18, 5, 1, 13 ]

max1 = 0
for i in range(len(List)-1):

  # use math.gcd() function
  gcd1 = math.gcd(List[i], List[i + 1])
  if(gcd1>max1):
    max1 = gcd1

# print max1
# as the result
print(max1)

Output:

9

Explanation:
For the given array one of the subarrays having maximum gcd is[4, 5] which has gcd 9.

Time Complexity:  O(n * log(min(a, b))), as it iterates through the input list once of size n, and gcd() method takes log(min(a,b)) time. 
Auxiliary Space: O(log(n)), as it uses recursion and the maximum depth of recursion is log(n).