# Minimum LCM and GCD possible among all possible sub-arrays

Given an array arr[] of N positive integers, the task is to find the minimum LCM and GCD between elements of all possible sub-array.

Examples:

Input: arr[] = {4, 4, 8}
Output: LCM = 4, GCD = 4
All possible sub-arrays are:
{4} -> LCM = 4, GCD = 4
{8} -> LCM = 8, GCD = 8
{4, 8} -> LCM = 8, GCD = 4

Input: arr[] = {2, 66, 14, 521}
Output: LCM = 2, GCD = 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We have to approach this problem greedily. It is obvious that when we decrease the number of elements then the LCM will become smaller and when we increase the number of elements then the GCD will become smaller. So we will take the smallest element of the array which is a single element and will be the required LCM. Now for GCD, minimum GCD will be the GCD of all the elements of the array.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return minimum GCD ` `// among all subarrays ` `int` `minGCD(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``int` `minGCD = 0; ` ` `  `    ``// Minimum GCD among all sub-arrays will be ` `    ``// the GCD of all the elements of the array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``minGCD = __gcd(minGCD, arr[i]); ` ` `  `    ``return` `minGCD; ` `} ` ` `  `// Function to return minimum LCM ` `// among all subarrays ` `int` `minLCM(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``int` `minLCM = arr; ` ` `  `    ``// Minimum LCM among all sub-arrays will be ` `    ``// the minimum element from the array ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``minLCM = min(minLCM, arr[i]); ` ` `  `    ``return` `minLCM; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 2, 66, 14, 521 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << ``"LCM = "` `<< minLCM(arr, n) ` `         ``<< ``", GCD = "` `<< minGCD(arr, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `// Function to return minimum GCD ` `// among all subarrays ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(a == ``0``)  ` `        ``return` `b;  ` `    ``return` `__gcd(b % a, a);  ` `}  ` `static` `int` `minGCD(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``int` `minGCD = ``0``; ` ` `  `    ``// Minimum GCD among all sub-arrays will be ` `    ``// the GCD of all the elements of the array ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``minGCD = __gcd(minGCD, arr[i]); ` ` `  `    ``return` `minGCD; ` `} ` ` `  `// Function to return minimum LCM ` `// among all subarrays ` `static` `int` `minLCM(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``int` `minLCM = arr[``0``]; ` ` `  `    ``// Minimum LCM among all sub-arrays will be ` `    ``// the minimum element from the array ` `    ``for` `(``int` `i = ``1``; i < n; i++) ` `        ``minLCM = Math.min(minLCM, arr[i]); ` ` `  `    ``return` `minLCM; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` `  `    ``int` `arr[] = { ``2``, ``66``, ``14``, ``521` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(``"LCM = "` `+ minLCM(arr, n) ` `        ``+ ``" GCD = "``+minGCD(arr, n)); ` ` `  `} ` `} ` `// This code is contributed by Code_Mech. `

 `# Python3 implementation of the approach  ` `from` `math ``import` `gcd ` ` `  `# Function to return minimum GCD  ` `# among all subarrays  ` `def` `minGCD(arr, n) : ` ` `  `    ``minGCD ``=` `0``; ` `     `  `    ``# Minimum GCD among all sub-arrays  ` `    ``# will be the GCD of all the elements ` `    ``# of the array ` `    ``for` `i ``in` `range``(n) : ` `        ``minGCD ``=` `gcd(minGCD, arr[i]); ` `         `  `    ``return` `minGCD;  ` ` `  `# Function to return minimum LCM  ` `# among all subarrays  ` `def` `minLCM(arr, n) : ` ` `  `    ``minLCM ``=` `arr[``0``];  ` ` `  `    ``# Minimum LCM among all sub-arrays   ` `    ``# will be the minimum element from  ` `    ``# the array  ` `    ``for` `i ``in` `range``(``1``, n) :  ` `        ``minLCM ``=` `min``(minLCM, arr[i]);  ` ` `  `    ``return` `minLCM;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``2``, ``66``, ``14``, ``521` `];  ` `    ``n ``=` `len``(arr);  ` ` `  `    ``print``(``"LCM = "``, minLCM(arr, n),  ` `          ``", GCD ="``, minGCD(arr, n));  ` ` `  `# This code is contributed by Ryuga `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Function to return minimum GCD ` `    ``// among all subarrays ` `    ``static` `int` `__gcd(``int` `a, ``int` `b)  ` `    ``{  ` `        ``if` `(a == 0)  ` `            ``return` `b;  ` `        ``return` `__gcd(b % a, a);  ` `    ``}  ` `    ``static` `int` `minGCD(``int` `[] arr, ``int` `n) ` `    ``{ ` `        ``int` `minGCD = 0; ` `     `  `        ``// Minimum GCD among all sub-arrays  ` `        ``// will be the GCD of all the  ` `        ``// elements of the array ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``minGCD = __gcd(minGCD, arr[i]); ` `     `  `        ``return` `minGCD; ` `    ``} ` `     `  `    ``// Function to return minimum LCM ` `    ``// among all subarrays ` `    ``static` `int` `minLCM(``int` `[] arr, ``int` `n) ` `    ``{ ` `     `  `        ``int` `minLCM = arr; ` `     `  `        ``// Minimum LCM among all sub-arrays ` `        ``// will be the minimum element from  ` `        ``// the array ` `        ``for` `(``int` `i = 1; i < n; i++) ` `            ``minLCM = Math.Min(minLCM, arr[i]); ` `     `  `        ``return` `minLCM; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `     `  `        ``int` `[] arr = { 2, 66, 14, 521 }; ` `        ``int` `n = arr.Length; ` `     `  `        ``Console.WriteLine(``"LCM = "` `+ minLCM(arr, n) +        ` `                          ``", GCD = "` `+ minGCD(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik. `

 ` `

Output:
```LCM = 2, GCD = 1
```

Time Complexity: O(N)

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