# Number of subarrays with GCD equal to 1

Given an array arr[], the task is to find the number of sub-arrays with GCD value equal to 1.

Examples:

Input: arr[] = {1, 1, 1}
Output: 6
All the subarrays of the given array
will have GCD equal to 1.

Input: arr[] = {2, 2, 2}
Output: 0

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

Approach: The key observation is that if the GCD of all the elements of the sub-array arr[l…r] is known then the GCD of all the elements of the sub-array arr[l…r+1] can be obtained by simply taking the GCD of the previous sub-array with arr[r + 1].
Thus, for every index i, keep iterating forward and compute the GCD from index i to j and check if its 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 required count ` `int` `cntSubArr(``int``* arr, ``int` `n) ` `{ ` `    ``// To store the final answer ` `    ``int` `ans = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// To store the GCD starting from ` `        ``// index 'i' ` `        ``int` `curr_gcd = 0; ` ` `  `        ``// Loop to find the gcd of each subarray ` `        ``// from arr[i] to arr[i...n-1] ` `        ``for` `(``int` `j = i; j < n; j++) { ` `            ``curr_gcd = __gcd(curr_gcd, arr[j]); ` ` `  `            ``// Increment the count if curr_gcd = 1 ` `            ``ans += (curr_gcd == 1); ` `        ``} ` `    ``} ` ` `  `    ``// Return the final answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 1, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << cntSubArr(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Function to return the required count ` `static` `int` `cntSubArr(``int` `[]arr, ``int` `n) ` `{ ` `    ``// To store the final answer ` `    ``int` `ans = ``0``; ` ` `  `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// To store the GCD starting from ` `        ``// index 'i' ` `        ``int` `curr_gcd = ``0``; ` ` `  `        ``// Loop to find the gcd of each subarray ` `        ``// from arr[i] to arr[i...n-1] ` `        ``for` `(``int` `j = i; j < n; j++)  ` `        ``{ ` `            ``curr_gcd = __gcd(curr_gcd, arr[j]); ` ` `  `            ``// Increment the count if curr_gcd = 1 ` `            ``ans += (curr_gcd == ``1``) ? ``1` `: ``0``; ` `        ``} ` `    ``} ` ` `  `    ``// Return the final answer ` `    ``return` `ans; ` `} ` ` `  `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(b == ``0``)  ` `        ``return` `a;  ` `    ``return` `__gcd(b, a % b);      ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args)  ` `{ ` `    ``int` `arr[] = { ``1``, ``1``, ``1` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(cntSubArr(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` `from` `math ``import` `gcd ` ` `  `# Function to return the required count  ` `def` `cntSubArr(arr, n) : ` ` `  `    ``# To store the final answer  ` `    ``ans ``=` `0``;  ` ` `  `    ``for` `i ``in` `range``(n) : ` ` `  `        ``# To store the GCD starting from  ` `        ``# index 'i'  ` `        ``curr_gcd ``=` `0``;  ` ` `  `        ``# Loop to find the gcd of each subarray  ` `        ``# from arr[i] to arr[i...n-1]  ` `        ``for` `j ``in` `range``(i, n) : ` `            ``curr_gcd ``=` `gcd(curr_gcd, arr[j]);  ` ` `  `            ``# Increment the count if curr_gcd = 1  ` `            ``ans ``+``=` `(curr_gcd ``=``=` `1``); ` ` `  `    ``# Return the final answer  ` `    ``return` `ans;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``arr ``=` `[ ``1``, ``1``, ``1` `];  ` `    ``n ``=` `len``(arr);  ` ` `  `    ``print``(cntSubArr(arr, n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `                     `  `class` `GFG  ` `{ ` ` `  `// Function to return the required count ` `static` `int` `cntSubArr(``int` `[]arr, ``int` `n) ` `{ ` `    ``// To store the final answer ` `    ``int` `ans = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// To store the GCD starting from ` `        ``// index 'i' ` `        ``int` `curr_gcd = 0; ` ` `  `        ``// Loop to find the gcd of each subarray ` `        ``// from arr[i] to arr[i...n-1] ` `        ``for` `(``int` `j = i; j < n; j++)  ` `        ``{ ` `            ``curr_gcd = __gcd(curr_gcd, arr[j]); ` ` `  `            ``// Increment the count if curr_gcd = 1 ` `            ``ans += (curr_gcd == 1) ? 1 : 0; ` `        ``} ` `    ``} ` ` `  `    ``// Return the final answer ` `    ``return` `ans; ` `} ` ` `  `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(b == 0)  ` `        ``return` `a;  ` `    ``return` `__gcd(b, a % b);      ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args)  ` `{ ` `    ``int` `[]arr = { 1, 1, 1 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(cntSubArr(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```6
```

Time Complexity: O(N2)

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Improved By : AnkitRai01, Rajput-Ji