# Smallest Subarray with given GCD

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(n3), O(n2) 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(n2logn), O(n2) 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).

1. If we have any number equal to ‘k’ in the array then the answer is 1 as GCD of k is k. Return 1.
2. If there is no number which is divisible by k, then GCD doesn’t exist. Return -1.
3. 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 ` `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"` `<< ````" "````; ` `        ``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 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.

.

.

.

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.