# Largest subsequence having GCD greater than 1

Given an array arr[], find the largest subsequence such that GCD of all those subsequence are greater than 1.
Examples:

Input: 3, 6, 2, 5, 4
Output: 3
Explanation: There are only three elements(6,
2, 4) having GCD greater than 1 i.e., 2. So the
largest subsequence will be 3

Input: 10, 15, 7, 25, 9, 35
Output: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach(Method 1)

Simple approach is to generate all the subsequence one by one and then find the GCD of all such generated set. Problem of this approach is that it grows exponentially in 2N

Iterative Approach(Method 2)

If we observe then we will found that to make gcd greater than 1, all such elements must contain comman factor greater than 1 which evenly divides all these values. So in order to get that factor we will iterate from 2 to Maximum element of array and then check for divisibility.

## C++

 // Simple C++ program to find length of // the largest subsequence with GCD greater // than 1. #include    using namespace std;    // Returns length of the largest subsequence // with GCD more than 1. int largestGCDSubsequence(int arr[], int n) {     int ans = 0;        // Finding the Maximum value in arr[]     int maxele = *max_element(arr, arr+n);        // Iterate from 2 to maximum possible     // divisor of all give values     for (int i=2; i<=maxele; ++i)     {         int count = 0;         for (int j=0; j

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## Java

 // Efficient Java program to find length of // the largest subsequence with GCD greater // than 1. import java.util.Arrays;    class GFG { // Returns length of the largest subsequence // with GCD more than 1. static int largestGCDSubsequence(int arr[], int n) {     int ans = 0;         // Finding the Maximum value in arr[]     int maxele = Arrays.stream(arr).max().getAsInt();;         // Iterate from 2 to maximum possible     // divisor of all give values     for (int i=2; i<=maxele; ++i)     {         int count = 0;         for (int j=0; j

## Python3

 # Simple Python 3 program to find length of # the largest subsequence with GCD greater # than 1.    # Returns length of the largest subsequence # with GCD more than 1. def largestGCDSubsequence(arr, n):     ans = 0        # Finding the Maximum value in arr[]     maxele = max(arr)        # Iterate from 2 to maximum possible     # divisor of all give values     for i in range(2, maxele + 1):         count = 0         for j in range(n):                            # If we found divisor,             # increment count             if (arr[j] % i == 0):                 count += 1         ans = max(ans, count)        return ans    # Driver code if __name__ == '__main__':     arr = [3, 6, 2, 5, 4]     size = len(arr)     print(largestGCDSubsequence(arr, size))    # This code is contributed by Rajput-Ji

## C#

 // Efficient C# program to find length of // the largest subsequence with GCD greater // than 1. using System; using System.Linq; public class GFG { // Returns length of the largest subsequence // with GCD more than 1. static int largestGCDSubsequence(int []arr, int n) {     int ans = 0;          // Finding the Maximum value in arr[]     int maxele = arr.Max();          // Iterate from 2 to maximum possible     // divisor of all give values     for (int i=2; i<=maxele; ++i)     {         int count = 0;         for (int j=0; j

## PHP



Output:

3

Time Complexity: O(n * max(arr[i])) where n is size of array.
Auxiliary Space: O(1)

Best Approach(Method 3)

An efficient approach is to use prime factorization method with the help of Sieve of Eratosthenes. First of all we will find the smallest prime divisor of all elements by pre-computed sieve. After that we will mark all the prime divisor of every element of arr[] by factorizing it with the help of pre-computed prime[] array.
Now we have all the marked primes occurring in all the array elements. The last step is to find the maximum count of all such prime factors.

## C++

 // Efficient C++ program to find length of // the largest subsequence with GCD greater // than 1. #include    using namespace std;    #define MAX 100001    // prime[] for storing smallest prime divisor of element // count[] for storing the number of times a particular // divisor occurs in a subsequence int prime[MAX], countdiv[MAX];    // Simple sieve to find smallest prime factors of numbers // smaller than MAX void SieveOfEratosthenes() {     for (int i = 2; i * i <= MAX; ++i)     {         if (!prime[i])             for (int j = i * 2; j <= MAX; j += i)                 prime[j] = i;     }        // Prime number will have same divisor     for (int i = 1; i < MAX; ++i)         if (!prime[i])             prime[i] = i; }    // Returns length of the largest subsequence // with GCD more than 1. int largestGCDSubsequence(int arr[], int n) {     int ans = 0;     for (int i=0; i < n; ++i)     {         int element = arr[i];            // Fetch total unique prime divisor of element         while (element > 1)         {             int div = prime[element];                // Increment count[] of Every unique divisor             // we get till now             ++countdiv[div];                // Find maximum frequency of divisor             ans = max(ans, countdiv[div]);                while (element % div==0)                 element /= div;         }     }        return ans; }    // Driver code int main() {     // Pre-compute smallest divisor of all numbers     SieveOfEratosthenes();        int arr[] = {10, 15, 7, 25, 9, 35};     int size = sizeof(arr) / sizeof(arr[0]);        cout << largestGCDSubsequence(arr, size);     return 0; }

## Java

 // Efficient Java program to find length of // the largest subsequence with GCD greater // than 1.    class GFG { static int MAX = 100001;    // prime[] for storing smallest prime divisor // of element count[] for storing the number // of times a particular divisor occurs  // in a subsequence static int[] prime = new int[MAX + 1]; static int[] countdiv = new int[MAX + 1];    // Simple sieve to find smallest prime  // factors of numbers smaller than MAX static void SieveOfEratosthenes() {     for (int i = 2; i * i <= MAX; ++i)     {         if (prime[i] == 0)             for (int j = i * 2; j <= MAX; j += i)                 prime[j] = i;     }        // Prime number will have same divisor     for (int i = 1; i < MAX; ++i)         if (prime[i] == 0)             prime[i] = i; }    // Returns length of the largest subsequence // with GCD more than 1. static int largestGCDSubsequence(int arr[], int n) {     int ans = 0;     for (int i = 0; i < n; ++i)     {         int element = arr[i];            // Fetch total unique prime divisor of element         while (element > 1)         {             int div = prime[element];                // Increment count[] of Every unique divisor             // we get till now             ++countdiv[div];                // Find maximum frequency of divisor             ans = Math.max(ans, countdiv[div]);                while (element % div == 0)                 element /= div;         }     }     return ans; }    // Driver code public static void main (String[] args) {     // Pre-compute smallest divisor of all numbers     SieveOfEratosthenes();        int arr[] = {10, 15, 7, 25, 9, 35};     int size = arr.length;        System.out.println(largestGCDSubsequence(arr, size)); } }    // This code is contributed by mits

## Python3

 # Efficient Python3 program to find length  # of the largest subsequence with GCD  # greater than 1. import math as mt    MAX = 100001    # prime[] for storing smallest # prime divisor of element # count[] for storing the number # of times a particular divisor  # occurs in a subsequence prime = [0 for i in range(MAX + 1)] countdiv = [0 for i in range(MAX + 1)]    # Simple sieve to find smallest prime  # factors of numbers smaller than MAX def SieveOfEratosthenes():        for i in range(2, mt.ceil(mt.sqrt(MAX + 1))):                if (prime[i] == 0):             for j in range(i * 2, MAX + 1, i):                 prime[j] = i            # Prime number will have same divisor     for i in range(1, MAX):         if (prime[i] == 0):             prime[i] = i    # Returns length of the largest  # subsequence with GCD more than 1. def largestGCDSubsequence(arr, n):        ans = 0     for i in range(n):            element = arr[i]            # Fetch total unique prime         # divisor of element         while (element > 1):                div = prime[element]                # Increment count[] of Every              # unique divisor we get till now             countdiv[div] += 1                # Find maximum frequency of divisor             ans = max(ans, countdiv[div])                while (element % div == 0):                 element = element // div                return ans    # Driver code    # Pre-compute smallest divisor # of all numbers SieveOfEratosthenes()    arr= [10, 15, 7, 25, 9, 35] size = len(arr)  print(largestGCDSubsequence(arr, size))    # This code is contributed  # by Mohit kumar 29

## C#

 // Efficient C# program to find length of  // the largest subsequence with GCD greater  // than 1.  using System;    class GFG {        static int MAX=100001;     // prime[] for storing smallest  // prime divisor of element count[] // for storing the number of times  // a particular divisor occurs in a subsequence  static int[] prime = new int[MAX + 1]; static int[] countdiv = new int[MAX + 1];     // Simple sieve to find smallest prime  //  factors of numbers smaller than MAX  static void SieveOfEratosthenes()  {      for (int i = 2; i * i <= MAX; ++i)      {          if (prime[i] == 0)              for (int j = i * 2; j <= MAX; j += i)                  prime[j] = i;      }         // Prime number will have same divisor      for (int i = 1; i < MAX; ++i)          if (prime[i] == 0)              prime[i] = i;  }     // Returns length of the largest subsequence  // with GCD more than 1.  static int largestGCDSubsequence(int []arr, int n)  {      int ans = 0;      for (int i = 0; i < n; ++i)      {          int element = arr[i];             // Fetch total unique prime divisor of element          while (element > 1)          {              int div = prime[element];                 // Increment count[] of Every unique divisor              // we get till now              ++countdiv[div];                 // Find maximum frequency of divisor              ans = Math.Max(ans, countdiv[div]);                 while (element % div==0)                  element /= div;          }      }      return ans;  }     // Driver code  public static void Main()  {     // Pre-compute smallest      // divisor of all numbers      SieveOfEratosthenes();         int []arr = {10, 15, 7, 25, 9, 35};      int size = arr.Length;         Console.WriteLine(largestGCDSubsequence(arr, size));  }  }    // This code is contributed by mits

Output:

4

Time complexity: O( n*log(max(arr[i])) ) + MAX*log(log(MAX))
Auxiliary space: O(MAX)