# Smallest subsequence having GCD equal to GCD of given array

Given an array arr[] of size N, the task is to find the smallest subsequence of the given array whose GCD of the subsequence is equal to the GCD of the given array. If more than one such subsequence exists, then print anyone of them.

Examples:

Input: arr[] = {4, 6, 12}
Output: 4 6
Explanation: Smallest subsequence having gcd equal to gcd of the given array(= 2) is {4, 6}. Therefore, the required output is {4, 6}

Input: arr[] = {6, 12, 18, 24}
Output: 6

Naive Approach: The simplest approach to solve this problem is to generate all possible subsequences of the given array and calculate GCD of each subsequence. Print the smallest subsequence having gcd equal to gcd of the given array.

Time Complexity: O(2N * N * log X), where X is the maximum element of the given array.
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach the idea is based on the following observations:

Length of the smallest subsequence having gcd equal to gcd of the given array must be either 1 or 2.

Consider, gcd of the given array is Y.

If arr[idx] = Y, then length of the smallest subsequence must be 1 for some value of idx.

Otherwise, length of the smallest subsequence must be 2.

If gcd of all possible pairs of the given array is greater than Y, then gcd of the given array must be greater than Y, which is not possible.
Therefore, at least one pair exists in the given array whose gcd is equal to Y.

Follow the steps below to solve the problem:

• Initialize a variable gcdArr to store GCD of the array.
• Traverse the given array and check if any array element is equal to gcdArr or not. If found to be true, print that element.
• Otherwise, print a pair from the given array whose gcd is equal to gcdArr.

Below is the implementation of the above approach.

## C++

 `// C++ program to implement` `// the above approach`   `#include ` `using` `namespace` `std;`   `// Function to print` `// the smallest subsequence` `// that satisfies the condition` `void` `printSmallSub(``int` `arr[], ``int` `N)` `{` `    ``// Stores gcd of the array.` `    ``int` `gcdArr = 0;`   `    ``// Traverse the given array` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// Update gcdArr` `        ``gcdArr = __gcd(gcdArr, arr[i]);` `    ``}`   `    ``// Traverse the given array.` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// If current element` `        ``// equal to gcd of array.` `        ``if` `(arr[i] == gcdArr) {` `            ``cout << arr[i] << ``" "``;` `            ``return``;` `        ``}` `    ``}`   `    ``// Generate all possible pairs.` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``for` `(``int` `j = i + 1; j < N;` `             ``j++) {`   `            ``// If gcd of current pair` `            ``// equal to gcdArr` `            ``if` `(__gcd(arr[i], arr[j])` `                ``== gcdArr) {`   `                ``// Print current pair` `                ``// of the array` `                ``cout << arr[i] << ``" "` `<< arr[j];` `                ``return``;` `            ``}` `        ``}` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 4, 6, 12 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``printSmallSub(arr, N);` `}`

## Java

 `// Java program to implement` `// the above approach` `import` `java.io.*;`   `class` `GFG{`   `// Function to calculate gcd` `// of two numbers` `static` `int` `gcd(``int` `a, ``int` `b)` `{` `    ``if` `(b == ``0``)` `        ``return` `a;` `        `  `    ``return` `gcd(b, a % b);` `}`   `// Function to print` `// the smallest subsequence` `// that satisfies the condition` `static` `void` `printSmallSub(``int``[] arr, ``int` `N)` `{` `    `  `    ``// Stores gcd of the array.` `    ``int` `gcdArr = ``0``;`   `    ``// Traverse the given array` `    ``for``(``int` `i = ``0``; i < N; i++)` `    ``{` `        `  `        ``// Update gcdArr` `        ``gcdArr = gcd(gcdArr, arr[i]);` `    ``}`   `    ``// Traverse the given array.` `    ``for``(``int` `i = ``0``; i < N; i++)` `    ``{` `        `  `        ``// If current element` `        ``// equal to gcd of array.` `        ``if` `(arr[i] == gcdArr)` `        ``{` `            ``System.out.print(arr[i] + ``" "``);` `            ``return``;` `        ``}` `    ``}`   `    ``// Generate all possible pairs.` `    ``for``(``int` `i = ``0``; i < N; i++)` `    ``{` `        ``for``(``int` `j = i + ``1``; j < N; j++) ` `        ``{` `            `  `            ``// If gcd of current pair` `            ``// equal to gcdArr` `            ``if` `(gcd(arr[i], arr[j]) == gcdArr) ` `            ``{` `                `  `                ``// Print current pair` `                ``// of the array` `                ``System.out.print(arr[i] + ``" "` `+ ` `                                 ``arr[j]);` `                ``return``;` `            ``}` `        ``}` `    ``}` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `arr[] = { ``4``, ``6``, ``12` `};` `    ``int` `N = arr.length;`   `    ``printSmallSub(arr, N);` `}` `}`   `// This code is contributed by akhilsaini`

## Python3

 `# Python3 program to implement` `# the above approach` `import` `math`   `# Function to print the ` `# smallest subsequence` `# that satisfies the condition` `def` `printSmallSub(arr, N):` `    `  `    ``# Stores gcd of the array.` `    ``gcdArr ``=` `0`   `    ``# Traverse the given array` `    ``for` `i ``in` `range``(``0``, N):` `        `  `        ``# Update gcdArr` `        ``gcdArr ``=` `math.gcd(gcdArr, arr[i])`   `    ``# Traverse the given array.` `    ``for` `i ``in` `range``(``0``, N):` `        `  `        ``# If current element` `        ``# equal to gcd of array.` `        ``if` `(arr[i] ``=``=` `gcdArr):` `            ``print``(arr[i], end ``=` `" "``)` `            ``return`   `        ``# Generate all possible pairs.` `        ``for` `i ``in` `range``(``0``, N):` `            ``for` `j ``in` `range``(i ``+` `1``, N):` `                `  `                ``# If gcd of current pair` `                ``# equal to gcdArr` `                ``if` `(math.gcd(arr[i], ` `                             ``arr[j]) ``=``=` `gcdArr):`   `                    ``# Print current pair` `                    ``# of the array` `                    ``print``(arr[i], end ``=` `" "``)` `                    ``print``(arr[j], end ``=` `" "``)` `                    ``return`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``arr ``=` `[ ``4``, ``6``, ``12` `]` `    ``N ``=` `len``(arr)` `    `  `    ``printSmallSub(arr, N)`   `# This code is contributed by akhilsaini`

## C#

 `// C# program to implement` `// the above approach` `using` `System;`   `class` `GFG{`   `// Function to calculate gcd` `// of two numbers` `static` `int` `gcd(``int` `a, ``int` `b)` `{` `    ``if` `(b == 0)` `        ``return` `a;` `        `  `    ``return` `gcd(b, a % b);` `}`   `// Function to print the ` `// smallest subsequence` `// that satisfies the condition` `static` `void` `printSmallSub(``int``[] arr, ``int` `N)` `{` `    `  `    ``// Stores gcd of the array.` `    ``int` `gcdArr = 0;`   `    ``// Traverse the given array` `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        `  `        ``// Update gcdArr` `        ``gcdArr = gcd(gcdArr, arr[i]);` `    ``}`   `    ``// Traverse the given array.` `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        `  `        ``// If current element` `        ``// equal to gcd of array.` `        ``if` `(arr[i] == gcdArr) ` `        ``{` `            ``Console.Write(arr[i] + ``" "``);` `            ``return``;` `        ``}` `    ``}`   `    ``// Generate all possible pairs.` `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        ``for``(``int` `j = i + 1; j < N; j++)` `        ``{` `            `  `            ``// If gcd of current pair` `            ``// equal to gcdArr` `            ``if` `(gcd(arr[i], arr[j]) == gcdArr)` `            ``{` `                `  `                ``// Print current pair` `                ``// of the array` `                ``Console.Write(arr[i] + ``" "` `+` `                              ``arr[j]);` `                ``return``;` `            ``}` `        ``}` `    ``}` `}`   `// Driver Code` `public` `static` `void` `Main()` `{`   `    ``int``[] arr = { 4, 6, 12 };` `    ``int` `N = arr.Length;`   `    ``printSmallSub(arr, N);` `}` `}`   `// This code is contributed by akhilsaini`

Output:

```4 6

```

Time Complexity: (N2 * log X), where X is the maximum element of the given array.
Auxiliary Space: O(1)

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Improved By : akhilsaini