# Largest subarray with GCD one

• Difficulty Level : Easy
• Last Updated : 07 Jul, 2022

There is an array with n elements. Find length of the largest subarray having GCD equal to 1. If no subarray with GCD 1, then print -1.

Examples :

```Input  : 1 3 5
Output : 3

Input : 2 4 6
Output :-1 ```

A simple solution is to consider every subarray and find its GCD and keep track of largest subarray with GCD one. Finally return length of the largest subarray with GCD 1.

An efficient solution is based on fact that if any two elements have GCD equals to one, then whole array has GCD one. So the output is either -1 or length of array.

## C++

 `// C++ program, to find length of the largest``// subarray with GCD equals to 1.``#include``using` `namespace` `std;` `int` `findLargest(``int` `arr[], ``int` `n)``{``    ``/*If gcd of any subarray is 1 then gcd of``     ``any number with the sub array will be 1.``     ``so if we are getting any subarray with``     ``gcd 1, then maximum number of element of``      ``the subarray will be equal to the number``      ``of elements of the array. Else it will be -1.*/``    ``int` `gcd = arr[0];``    ``for` `(``int` `i=1; i

## Java

 `// Java program, to find length of the``// largest subarray with GCD equals to 1.``class` `GFG {``    ` `    ``static` `int` `___gcd(``int` `a, ``int` `b)``    ``{``        ` `        ``// Everything divides 0``        ``if` `(a == ``0` `|| b == ``0``)``            ``return` `0``;``    ` `        ``// base case``        ``if` `(a == b)``            ``return` `a;``    ` `        ``// a is greater``        ``if` `(a > b)``            ``return` `___gcd(a - b, b);``            ` `        ``return` `___gcd(a, b - a);``    ``}``    ` `    ``static` `int` `findLargest(``int` `arr[],``                                ``int` `n)``    ``{``        ` `        ``/*If gcd of any subarray is 1``        ``then gcd of any number with the``        ``sub array will be 1. so if we``        ``are getting any subarray with``        ``gcd 1, then maximum number of``        ``element of the subarray will``        ``be equal to the number of ``        ``elements of the array. Else``        ``it will be -1.*/``        ``int` `gcd = arr[``0``];``        ` `        ``for` `(``int` `i = ``1``; i < n; i++)``            ``gcd = ___gcd(gcd, arr[i]);``    ` `        ``return` `(gcd == ``1``)? n : -``1``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = {``1``, ``3``, ``5``, ``7``};``        ``int` `n = arr.length;``        ` `        ``System.out.print(``"Length of the "``                   ``+ ``"largest subarray = "``                   ``+ findLargest(arr, n));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python program, to find``# length of the largest``# subarray with GCD equals to 1.` `def` `___gcd(a,b):` `    ``# Everything divides 0``    ``if` `(a ``=``=` `0` `or` `b ``=``=` `0``):``        ``return` `0`` ` `    ``# base case``    ``if` `(a ``=``=` `b):``        ``return` `a`` ` `    ``# a is greater``    ``if` `(a > b):``        ``return` `___gcd(a``-``b, b)``    ``return` `___gcd(a, b``-``a)``    ` `def` `findLargest(arr, n):` `    ``'''If gcd of any subarray is 1 then gcd of``     ``any number with the sub array will be 1.``     ``so if we are getting any subarray with``     ``gcd 1, then maximum number of element of``      ``the subarray will be equal to the number``      ``of elements of the array. Else it will be -1.'''``    ``gcd ``=` `arr[``0``]``    ``for` `i ``in` `range``(``1``,n):``        ``gcd ``=` `___gcd(gcd, arr[i])`` ` `    ``return` `n ``if` `(gcd ``=``=` `1``) ``else` `-``1``    ` `# Driver code``arr``=``[``1``, ``3``, ``5``, ``7``]``n``=``len``(arr)` `print``(``"Length of the largest subarray = "``,``         ``findLargest(arr, n))` `# This code is contributed``# by Anant Agarwal.`

## C#

 `// C# program, to find length of the``// largest subarray with GCD equals to 1.``using` `System;` `class` `GFG {``    ` `    ``static` `int` `___gcd(``int` `a, ``int` `b)``    ``{``        ` `        ``// Everything divides 0``        ``if` `(a == 0 || b == 0)``            ``return` `0;``    ` `        ``// base case``        ``if` `(a == b)``            ``return` `a;``    ` `        ``// a is greater``        ``if` `(a > b)``            ``return` `___gcd(a - b, b);``            ` `        ``return` `___gcd(a, b - a);``    ``}``    ` `    ``static` `int` `findLargest(``int` `[]arr,``                           ``int` `n)``    ``{``        ` `        ``// If gcd of any subarray is 1``        ``// then gcd of any number with the``        ``// sub array will be 1. so if we``        ``// are getting any subarray with``        ``// gcd 1, then maximum number of``        ``// element of the subarray will``        ``// be equal to the number of``        ``// elements of the array. Else``        ``// it will be -1.``        ``int` `gcd = arr[0];``        ` `        ``for` `(``int` `i = 1; i < n; i++)``            ``gcd = ___gcd(gcd, arr[i]);``    ` `        ``return` `(gcd == 1)? n : -1;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = {1, 3, 5, 7};``        ``int` `n = arr.Length;``        ` `        ``Console.Write(``"Length of the "``                       ``+ ``"largest subarray = "``                       ``+ findLargest(arr, n));``    ``}``}` `// This code is contributed by Nitin Mittal.`

## PHP

 ` ``\$b``)``        ``return` `___gcd(``\$a` `- ``\$b``, ``\$b``);``        ` `    ``return` `___gcd(``\$a``, ``\$b` `- ``\$a``);``}` `function` `findLargest(``\$arr``, ``\$n``)``{``    ` `    ``/*If gcd of any subarray is 1``    ``then gcd of any number with the``    ``sub array will be 1. so if we``    ``are getting any subarray with``    ``gcd 1, then maximum number of``    ``element of the subarray will``    ``be equal to the number of``    ``elements of the array. Else``    ``it will be -1.*/``    ``\$gcd` `= ``\$arr``[0];``    ` `    ``for` `(``\$i` `= 1; ``\$i` `< ``\$n``; ``\$i``++)``        ``\$gcd` `= ___gcd(``\$gcd``, ``\$arr``[``\$i``]);` `    ``return` `(``\$gcd` `== 1)? ``\$n` `: -1;``}` `// Driver code``\$arr` `= ``array``(1, 3, 5, 7);``\$n` `= ``count``(``\$arr``);` `echo` `"Length of the "` `.``     ``"largest subarray = "` `.``      ``findLargest(``\$arr``, ``\$n``);` `// This code is contributed by Sam007``?>`

## Javascript

 ``

Output

`Length of the largest subarray = 4`

Time Complexity: O(log(min(n)))
Auxiliary Space: O(1)

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