Number of subarrays with GCD = 1 | Segment tree
Given an array arr[], the task is to find the count of sub-arrays with GCD equal to 1.
Examples:
Input: arr[] = {1, 1, 1}
Output: 6
Every single subarray of the given array has GCD
of 1 and there are a total of 6 subarrays.
Input: arr[] = {2, 2, 2}
Output: 0
Approach: This problem can be solved in O(NlogN) using segment-tree data structure. The segment that will be built can be used to answer range-gcd queries.
Let’s understand the algorithm now. Use the two-pointer technique to solve this problem. Let’s make a few observations before discussing the algorithm.
- Let’s say G is the GCD of the subarray arr[l…r] and G1 is the GCD of the subarray arr[l+1…r]. G smaller than or equal to G1 always.
- Let’s say for the given L1, R1 is the first index such that GCD of the range [L, R] is 1 then for any L2 greater than or equal to L1, R2 will also be greater than or equal to R1.
After the above observation, two-pointer technique makes perfect sense i.e. if the length
of the smallest R is known for an index L then for an index L + 1, the search needs to be started from R on-wards.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define maxLen 30 // Array to store segment-tree int seg[3 * maxLen]; // Function to build segment-tree to // answer range GCD queries int build( int l, int r, int in, int * arr) { // Base-case if (l == r) return seg[in] = arr[l]; // Mid element of the range int mid = (l + r) / 2; // Merging the result of left and right sub-tree return seg[in] = __gcd(build(l, mid, 2 * in + 1, arr), build(mid + 1, r, 2 * in + 2, arr)); } // Function to perform range GCD queries int query( int l, int r, int l1, int r1, int in) { // Base-cases if (l1 <= l and r <= r1) return seg[in]; if (l > r1 or r < l1) return 0; // Mid-element int mid = (l + r) / 2; // Calling left and right child return __gcd(query(l, mid, l1, r1, 2 * in + 1), query(mid + 1, r, l1, r1, 2 * in + 2)); } // Function to find the required count int findCnt( int * arr, int n) { // Building the segment tree build(0, n - 1, 0, arr); // Two pointer variables int i = 0, j = 0; // To store the final answer int ans = 0; // Looping while (i < n) { // Incrementing j till we don't get // a gcd value of 1 while (j < n and query(0, n - 1, i, j, 0) != 1) j++; // Updating the final answer ans += (n - j); // Increment i i++; // Update j j = max(j, i); } // Returning the final answer return ans; } // Driver code int main() { int arr[] = { 1, 1, 1, 1 }; int n = sizeof (arr) / sizeof ( int ); cout << findCnt(arr, n); return 0; } |
Java
// Java implementation of the above approach class GFG { static int maxLen = 30 ; // Array to store segment-tree static int []seg = new int [ 3 * maxLen]; // Function to build segment-tree to // answer range GCD queries static int build( int l, int r, int in, int [] arr) { // Base-case if (l == r) return seg[in] = arr[l]; // Mid element of the range int mid = (l + r) / 2 ; // Merging the result of left and right sub-tree return seg[in] = __gcd(build(l, mid, 2 * in + 1 , arr), build(mid + 1 , r, 2 * in + 2 , arr)); } // Function to perform range GCD queries static int query( int l, int r, int l1, int r1, int in) { // Base-cases if (l1 <= l && r <= r1) return seg[in]; if (l > r1 || r < l1) return 0 ; // Mid-element int mid = (l + r) / 2 ; // Calling left and right child return __gcd(query(l, mid, l1, r1, 2 * in + 1 ), query(mid + 1 , r, l1, r1, 2 * in + 2 )); } // Function to find the required count static int findCnt( int [] arr, int n) { // Building the segment tree build( 0 , n - 1 , 0 , arr); // Two pointer variables int i = 0 , j = 0 ; // To store the final answer int ans = 0 ; // Looping while (i < n) { // Incrementing j till we don't get // a gcd value of 1 while (j < n && query( 0 , n - 1 , i, j, 0 ) != 1 ) j++; // Updating the final answer ans += (n - j); // Increment i i++; // Update j j = Math.max(j, i); } // Returning the final answer return ans; } static int __gcd( int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver code public static void main(String []args) { int arr[] = { 1 , 1 , 1 , 1 }; int n = arr.length; System.out.println(findCnt(arr, n)); } } // This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the above approach from math import gcd maxLen = 30 ; # Array to store segment-tree seg = [ 0 ] * ( 3 * maxLen); # Function to build segment-tree to # answer range GCD queries def build(l, r, i, arr) : # Base-case if (l = = r) : seg[i] = arr[l]; return seg[i]; # Mid element of the range mid = (l + r) / / 2 ; # Merging the result of left and right sub-tree seg[i] = gcd(build(l, mid, 2 * i + 1 , arr), build(mid + 1 , r, 2 * i + 2 , arr)); return seg[i]; # Function to perform range GCD queries def query(l, r, l1, r1, i) : # Base-cases if (l1 < = l and r < = r1) : return seg[i]; if (l > r1 or r < l1) : return 0 ; # Mid-element mid = (l + r) / / 2 ; # Calling left and right child return gcd(query(l, mid, l1, r1, 2 * i + 1 ), query(mid + 1 , r, l1, r1, 2 * i + 2 )); # Function to find the required count def findCnt(arr, n) : # Building the segment tree build( 0 , n - 1 , 0 , arr); # Two pointer variables i = 0 ; j = 0 ; # To store the final answer ans = 0 ; # Looping while (i < n) : # Incrementing j till we don't get # a gcd value of 1 while (j < n and query( 0 , n - 1 , i, j, 0 ) ! = 1 ) : j + = 1 ; # Updating the final answer ans + = (n - j); # Increment i i + = 1 ; # Update j j = max (j, i); # Returning the final answer return ans; # Driver code if __name__ = = "__main__" : arr = [ 1 , 1 , 1 , 1 ]; n = len (arr); print (findCnt(arr, n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the above approach using System; class GFG { static int maxLen = 30; // Array to store segment-tree static int []seg = new int [3 * maxLen]; // Function to build segment-tree to // answer range GCD queries static int build( int l, int r, int iN, int [] arr) { // Base-case if (l == r) return seg[iN] = arr[l]; // Mid element of the range int mid = (l + r) / 2; // Merging the result of left and right sub-tree return seg[iN] = __gcd(build(l, mid, 2 * iN + 1, arr), build(mid + 1, r, 2 * iN + 2, arr)); } // Function to perform range GCD queries static int query( int l, int r, int l1, int r1, int iN) { // Base-cases if (l1 <= l && r <= r1) return seg[iN]; if (l > r1 || r < l1) return 0; // Mid-element int mid = (l + r) / 2; // Calling left and right child return __gcd(query(l, mid, l1, r1, 2 * iN + 1), query(mid + 1, r, l1, r1, 2 * iN + 2)); } // Function to find the required count static int findCnt( int [] arr, int n) { // Building the segment tree build(0, n - 1, 0, arr); // Two pointer variables int i = 0, j = 0; // To store the final answer int ans = 0; // Looping while (i < n) { // Incrementing j till we don't get // a gcd value of 1 while (j < n && query(0, n - 1, i, j, 0) != 1) j++; // Updating the final answer ans += (n - j); // Increment i i++; // Update j j = Math.Max(j, i); } // Returning the final answer return ans; } static int __gcd( int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver code public static void Main(String []args) { int []arr = { 1, 1, 1, 1 }; int n = arr.Length; Console.WriteLine(findCnt(arr, n)); } } // This code is contributed by PrinciRaj1992 |
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