# Minimum deletions required to make GCD of the array equal to 1

Given an array arr[] of N integers, the task is to find the minimum deletions required to make the GCD of the resulting array elements equal to 1. If it is impossible then print -1.

Examples:

Input: arr[] = {2, 4, 6, 3}
Output: 0
It is clear that GCD(2, 4, 6, 3) = 1
So, we do not need to delete any elements.

Input: arr[] = {8, 14, 16, 26}
Output: -1
No matter how many elements get deleted, the gcd will never be 1.

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

Approach: If the GCD of the initial array is 1 then we do not need to delete any of the elements and the result will be 0. If GCD is not 1 then the GCD can never be 1 no matter what elements we delete. Let’s say gcd be the gcd of the elements of the array and we delete k elements. Now, N – k elements are left and they still have gcd as their factor. So, it is impossible to get GCD of the array elements equal to 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum ` `// deletions required ` `int` `MinDeletion(``int` `a[], ``int` `n) ` `{ ` ` `  `    ``// To store the GCD of the array ` `    ``int` `gcd = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``gcd = __gcd(gcd, a[i]); ` ` `  `    ``// GCD cannot be 1 ` `    ``if` `(gcd > 1) ` `        ``return` `-1; ` ` `  `    ``// GCD of the elements is already 1 ` `    ``else` `        ``return` `0; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 3, 6, 12, 81, 9 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``cout << MinDeletion(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Recursive function to return gcd of a and b  ` `    ``static` `int` `__gcd(``int` `a, ``int` `b)  ` `    ``{  ` `        ``// Everything divides 0  ` `        ``if` `(a == ``0``)  ` `        ``return` `b;  ` `        ``if` `(b == ``0``)  ` `        ``return` `a;  ` `         `  `        ``// base case  ` `        ``if` `(a == b)  ` `            ``return` `a;  ` `         `  `        ``// a is greater  ` `        ``if` `(a > b)  ` `            ``return` `__gcd(a-b, b);  ` `        ``return` `__gcd(a, b-a);  ` `    ``}  ` ` `  `    ``// Function to return the minimum ` `    ``// deletions required ` `    ``static` `int` `MinDeletion(``int` `a[], ``int` `n) ` `    ``{ ` `     `  `        ``// To store the GCD of the array ` `        ``int` `gcd = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``gcd = __gcd(gcd, a[i]); ` `     `  `        ``// GCD cannot be 1 ` `        ``if` `(gcd > ``1``) ` `            ``return` `-``1``; ` `     `  `        ``// GCD of the elements is already 1 ` `        ``else` `            ``return` `0``; ` `    ``} ` ` `  ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `a[] = { ``3``, ``6``, ``12``, ``81``, ``9` `}; ` `        ``int` `n = a.length; ` ` `  `        ``System.out.print(MinDeletion(a, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python3 implementation of the approach  ` `from` `math ``import` `gcd ` ` `  `# Function to return the minimum  ` `# deletions required  ` `def` `MinDeletion(a, n) :  ` ` `  `    ``# To store the GCD of the array  ` `    ``__gcd ``=` `0``;  ` `    ``for` `i ``in` `range``(n) : ` `        ``__gcd ``=` `gcd(__gcd, a[i]);  ` ` `  `    ``# GCD cannot be 1  ` `    ``if` `(__gcd > ``1``) : ` `        ``return` `-``1``;  ` ` `  `    ``# GCD of the elements is already 1  ` `    ``else` `: ` `        ``return` `0``;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``a ``=` `[ ``3``, ``6``, ``12``, ``81``, ``9` `];  ` `    ``n ``=` `len``(a)  ` ` `  `    ``print``(MinDeletion(a, n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Recursive function to return gcd of a and b  ` `    ``static` `int` `__gcd(``int` `a, ``int` `b)  ` `    ``{  ` `        ``// Everything divides 0  ` `        ``if` `(a == 0)  ` `            ``return` `b;  ` `        ``if` `(b == 0)  ` `            ``return` `a;  ` `         `  `        ``// base case  ` `        ``if` `(a == b)  ` `            ``return` `a;  ` `         `  `        ``// a is greater  ` `        ``if` `(a > b)  ` `            ``return` `__gcd(a-b, b);  ` `        ``return` `__gcd(a, b-a);  ` `    ``}  ` ` `  `    ``// Function to return the minimum ` `    ``// deletions required ` `    ``static` `int` `MinDeletion(``int` `[]a, ``int` `n) ` `    ``{ ` `     `  `        ``// To store the GCD of the array ` `        ``int` `gcd = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``gcd = __gcd(gcd, a[i]); ` `     `  `        ``// GCD cannot be 1 ` `        ``if` `(gcd > 1) ` `            ``return` `-1; ` `     `  `        ``// GCD of the elements is already 1 ` `        ``else` `            ``return` `0; ` `    ``} ` ` `  ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `[]a = { 3, 6, 12, 81, 9 }; ` `        ``int` `n = a.Length; ` ` `  `        ``Console.WriteLine(MinDeletion(a, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:

```-1
```

Time Complexity: O(N)

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Improved By : AnkitRai01, vt_m