# 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

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

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++ implementation of the approach #include 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 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 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# 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

Output:
10

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Improved By : AnkitRai01, princiraj1992