# 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