Given an array arr[] of n numbers and an integer k, find length of the minimum sub-array with gcd equals to k.

Example:

Input: arr[] = {6, 9, 7, 10, 12, 24, 36, 27}, K = 3 Output: 2 Explanation: GCD of subarray {6,9} is 3. GCD of subarray {24,36,27} is also 3, but {6,9} is the smallest

Note: Time complexity analysis of below approaches assume that numbers are fixed size and finding GCD of two elements take constant time.

**Method 1**

Find GCD of all subarrays and keep track of the minimum length subarray with gcd k. Time Complexity of this is O(n^{3}), O(n^{2}) for each subarray and O(n) for finding gcd of a subarray.

**Method 2**

Find GCD of all subarrays using segment tree based approach discussed here. Time complexity of this solution is O(n^{2}logn), O(n^{2}) for each subarray and O(logn) for finding GCD of subarray using segment tree.

**Method 3**

The idea is to use Segment Tree and Binary Search to achieve time complexity O(n (logn)^{2}).

- If we have any number equal to ‘k’ in the array then the answer is 1 as GCD of k is k. Return 1.
- If there is no number which is divisible by k, then GCD doesn’t exist. Return -1.
- If none of the above cases is true, the length of minimum subarray is either greater than 1 or GCD doesn’t exist. In this case, we follow following steps.
- Build segment tree so that we can quicky find GCD of any subarray using the approach discussed here
- After building Segment Tree, we consider every index as starting point and do binary search for ending point such that the subarray between these two points has GCD k

Following is C++ implementation of above idea.

// C++ Program to find GCD of a number in a given Range // using segment Trees #include <bits/stdc++.h> using namespace std; // To store segment tree int *st; // Function to find gcd of 2 numbers. int gcd(int a, int b) { if (a < b) swap(a, b); if (b==0) return a; return gcd(b, a%b); } /* A recursive function to get gcd of given range of array indexes. The following are parameters for this function. st --> Pointer to segment tree si --> Index of current node in the segment tree. Initially 0 is passed as root is always at index 0 ss & se --> Starting and ending indexes of the segment represented by current node, i.e., st[index] qs & qe --> Starting and ending indexes of query range */ int findGcd(int ss, int se, int qs, int qe, int si) { if (ss>qe || se < qs) return 0; if (qs<=ss && qe>=se) return st[si]; int mid = ss+(se-ss)/2; return gcd(findGcd(ss, mid, qs, qe, si*2+1), findGcd(mid+1, se, qs, qe, si*2+2)); } //Finding The gcd of given Range int findRangeGcd(int ss, int se, int arr[], int n) { if (ss<0 || se > n-1 || ss>se) { cout << "Invalid Arguments" << "\n"; return -1; } return findGcd(0, n-1, ss, se, 0); } // A recursive function that constructs Segment Tree for // array[ss..se]. si is index of current node in segment // tree st int constructST(int arr[], int ss, int se, int si) { if (ss==se) { st[si] = arr[ss]; return st[si]; } int mid = ss+(se-ss)/2; st[si] = gcd(constructST(arr, ss, mid, si*2+1), constructST(arr, mid+1, se, si*2+2)); return st[si]; } /* Function to construct segment tree from given array. This function allocates memory for segment tree and calls constructSTUtil() to fill the allocated memory */ int *constructSegmentTree(int arr[], int n) { int height = (int)(ceil(log2(n))); int size = 2*(int)pow(2, height)-1; st = new int[size]; constructST(arr, 0, n-1, 0); return st; } // Returns size of smallest subarray of arr[0..n-1] // with GCD equal to k. int findSmallestSubarr(int arr[], int n, int k) { // To check if a multiple of k exists. bool found = false; // Find if k or its multiple is present for (int i=0; i<n; i++) { // If k is present, then subarray size is 1. if (arr[i] == k) return 1; // Break the loop to indicate presence of a // multiple of k. if (arr[i] % k == 0) found = true; } // If there was no multiple of k in arr[], then // we can't get k as GCD. if (found == false) return -1; // If there is a multiple of k in arr[], build // segment tree from given array constructSegmentTree(arr, n); // Initialize result int res = n+1; // Now consider every element as starting point // and search for ending point using Binary Search for (int i=0; i<n-1; i++) { // a[i] cannot be a starting point, if it is // not a multiple of k. if (arr[i] % k != 0) continue; // Initialize indexes for binary search of closest // ending point to i with GCD of subarray as k. int low = i+1; int high = n-1; // Initialize closest ending point for i. int closest = 0; // Binary Search for closest ending point // with GCD equal to k. while (true) { // Find middle point and GCD of subarray // arr[i..mid] int mid = low + (high-low)/2; int gcd = findRangeGcd(i, mid, arr, n); // If GCD is more than k, look further if (gcd > k) low = mid; // If GCD is k, store this point and look for // a closer point else if (gcd == k) { high = mid; closest = mid; break; } // If GCD is less than, look closer else high = mid; // If termination condition reached, set // closest if (abs(high-low) <= 1) { if (findRangeGcd(i, low, arr, n) == k) closest = low; else if (findRangeGcd(i, high, arr, n)==k) closest = high; break; } } if (closest != 0) res = min(res, closest - i + 1); } // If res was not changed by loop, return -1, // else return its value. return (res == n+1) ? -1 : res; } // Driver program to test above functions int main() { int n = 8; int k = 3; int arr[] = {6, 9, 7, 10, 12, 24, 36, 27}; cout << "Size of smallest sub-array with given" << " size is " << findSmallestSubarr(arr, n, k); return 0; }

**Output:**

2

**Example:**

arr[] = {6, 9, 7, 10, 12, 24, 36, 27}, K = 3 // Initial value of minLen is equal to size // of array minLen = 8 No element is equal to k so result is either greater than 1 or doesn't exist.

**First index**

- GCD of subarray from 1 to 5 is 1.
- GCD < k
- GCD of subarray from 1 to 3 is 1.
- GCD < k
- GCD of subarray from 1 to 2 is 3
- minLen = minimum(8, 2) = 2

**Second Index**

- GCD of subarray from 2 to 5 is 1
- GCD < k
- GCD of subarray from 2 to 4 is 1
- GCD < k
- GCD of subarray from 6 to 8 is 3
- minLen = minimum(2, 3) = 2.

**.**

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**Sixth Index**

- GCD of subarray from 6 to 7 is 12
- GCD > k
- GCD of subarray from 6 to 8 is 3
- minLen = minimum(2, 3) = 2

**Time Complexity:** O(n (logn)^{2}), O(n) for traversing to each index, O(logn) for each subarray and O(logn) for GCD of each subarray.

This article is contributed by **Nikhil Tekwani.** If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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