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)