# Smallest subarray with GCD as 1 | Segment Tree

Given an array arr[], the task is to find the smallest sub-arrays with GCD equal to 1. If there is no such sub-array then print -1.

Examples:

Input: arr[] = {2, 6, 3}
Output: 3
{2, 6, 3} is the only sub-array with GCD = 1.

Input: arr[] = {2, 2, 2}
Output: -1

## 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++

 // 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 length int findLen(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 = INT_MAX;        // 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++;            if (j == n)             break;            // Updating the final answer         ans = min((j - i + 1), ans);            // Incrementing i         i++;            // Updating j         j = max(j, i);     }        // Returning the final answer     if (ans == INT_MAX)         return -1;     else         return ans; }    // Driver code int main() {     int arr[] = { 2, 2, 2 };     int n = sizeof(arr) / sizeof(int);        cout << findLen(arr, n);        return 0; }

## Java

 // Java implementation of the 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 length static int findLen(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 = Integer.MAX_VALUE;        // 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++;            if (j == n)             break;            // Updating the final answer         ans = Math.min((j - i + 1), ans);            // Incrementing i         i++;            // Updating j         j = Math.max(j, i);     }        // Returning the final answer     if (ans == Integer.MAX_VALUE)         return -1;     else         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[] = { 2, 2, 2 };     int n = arr.length;        System.out.println(findLen(arr, n)); } }    // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation of the approach from math import gcd as __gcd    maxLen = 30    # Array to store segment-tree seg = [0 for i in range(3 * maxLen)]    # Function to build segment-tree to # answer range GCD queries def build(l, r, inn, arr):            # Base-case     if (l == r):         seg[inn] = arr[l]         return seg[inn]        # Mid element of the range     mid = (l + r) // 2        # Merging the result of      # left and right sub-tree     seg[inn] = __gcd(build(l, mid,                            2 * inn + 1, arr),                       build(mid + 1, r,                             2 * inn + 2, arr))        return seg[inn]    # Function to perform range GCD queries def query(l, r, l1, r1, inn):            # Base-cases     if (l1 <= l and r <= r1):         return seg[inn]     if (l > r1 or r < l1):         return 0        # Mid-element     mid = (l + r) // 2        # Calling left and right child     x=__gcd(query(l, mid, l1, r1,                   2 * inn + 1),              query(mid + 1, r, l1, r1,                   2 * inn + 2))     return x    # Function to find the required length def findLen(arr, n):            # Buildinng the segment tree     build(0, n - 1, 0, arr)        # Two pointer variables     i = 0     j = 0        # To store the finnal answer     ans = 10**9        # Loopinng     while (i < n):            # Incrementinng 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            if (j == n):             break;            # Updatinng the finnal answer         ans = minn((j - i + 1), ans)            # Incrementinng i         i += 1            # Updatinng j         j = max(j, i)        # Returninng the finnal answer     if (ans == 10**9):         return -1     else:         return ans    # Driver code arr = [2, 2, 2] n = len(arr)    print(findLen(arr, n))    # This code is contributed by Mohit Kumar

## C#

 // C# implementation of the 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 ind, int[] arr)     {         // Base-case         if (l == r)             return seg[ind] = arr[l];                // Mid element of the range         int mid = (l + r) / 2;                // Merging the result of left and right sub-tree         return seg[ind] = __gcd(build(l, mid, 2 * ind + 1, arr),                                 build(mid + 1, r, 2 * ind + 2, arr));     }            // Function to perform range GCD queries     static int query(int l, int r,                       int l1, int r1, int ind)     {         // Base-cases         if (l1 <= l && r <= r1)             return seg[ind];         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 * ind + 1),                      query(mid + 1, r, l1, r1, 2 * ind + 2));     }            // Function to find the required length     static int findLen(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 = int.MaxValue;                // 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++;                    if (j == n)                 break;                    // Updating the final answer             ans = Math.Min((j - i + 1), ans);                    // Incrementing i             i++;                    // Updating j             j = Math.Max(j, i);         }                // Returning the final answer         if (ans == int.MaxValue)             return -1;         else             return ans;     }            static int __gcd(int a, int b)      {          return b == 0 ? a : __gcd(b, a % b);          }             // Driver code     public static void Main()     {         int []arr = { 2, 2, 2 };         int n = arr.Length;                Console.WriteLine(findLen(arr, n));     } }    // This code is contributed by kanugargng

Output:

-1

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