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++ 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 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 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# 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 |
<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> |
6
Time Complexity: O(N2log(max(arr[])))
Auxiliary Space: O(1)