Number of subarrays with GCD equal to 1

Given an array arr[], the task is to find the number of sub-arrays with a 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  

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 it’s equal to 1.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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

Javascript

<script>

// Javascript implementation of the approach

function __gcd(a, b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
}


// Function to return the required count
function cntSubArr(arr, n)
{
    // To store the final answer
    var ans = 0;

    for (var i = 0; i < n; i++) {

        // To store the GCD starting from
        // index 'i'
        var curr_gcd = 0;

        // Loop to find the gcd of each subarray
        // from arr[i] to arr[i...n-1]
        for (var 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
var arr = [1, 1, 1];
var n = arr.length;
document.write( cntSubArr(arr, n));

</script>

Output:

6

Time Complexity: O(N²log(max(arr[])) )

Auxiliary Space: O(1)