Count number of sub-sequences with GCD 1

Given an array of N numbers, the task is to count the number of subsequences which have gcd equal to 1.

Examples:

Input: a[] = {3, 4, 8, 16} 
Output: 7
The subsequences are:  
{3, 4}, {3, 8}, {3, 16}, {3, 4, 8},
{3, 4, 16}, {3, 8, 16}, {3, 4, 8, 16}

Input: a[] = {1, 2, 4}
Output: 4

A simple solution is to generate all subsequences or subsets. Foe every subsequence, check if its GCD is 1 or not. If 1, increment the result.

When we have values in array (say all smaller than 1000), we can optimize the above solution as we know that number of possible GCDs would be small. We modify the recursive subset generation algorithm where consider two cases for every element, we either include or exclude it. We keep track of current GCD and if we have already counted for this GCD, we return the count. So when we are considering a subset, some GCDs would appear again and again. Therefore the problem can be solved using Dynamic Programming. Given below are the steps to solve the above problem:

  • Start from every index and call the recursive function by taking the index element.
  • In the recursive function, we iterate till we reach N.
  • The two recursive calls will be based on either we take the index element or not.
  • The base case will be to return 1 if we have reached the end and the gcd till now is 1.
  • Two recursive calls will be func(ind+1, gcd(a[i], prevgcd)) and func(ind+1, prevgcd)
  • The overlapping subproblems can be avoided by using memoization technique.

Below is the implementation of the above approach:

C++

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// C++ program to find the number
// of subsequences with gcd 1
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Recursive function to calculate the number
// of subsequences with gcd 1 starting with 
// particular index
int func(int ind, int g, int dp[][MAX], int n, int a[])
{
  
    // Base case
    if (ind == n) {
        if (g == 1)
            return 1;
        else
            return 0;
    }
  
    // If already visited
    if (dp[ind][g] != -1)
        return dp[ind][g];
  
    // Either we take or we do not
    int ans = func(ind + 1, g, dp, n, a) 
              + func(ind + 1, gcd(a[ind], g), dp, n, a);
  
    // Return the answer
    return dp[ind][g] = ans;
}
  
// Function to return the number of subsequences
int countSubsequences(int a[], int n)
{
  
    // Hash table to memoize
    int dp[n][MAX];
    memset(dp, -1, sizeof dp);
  
    // Count the number of subsequences
    int count = 0;
  
    // Count for every subsequence
    for (int i = 0; i < n; i++)
        count += func(i + 1, a[i], dp, n, a);
  
    return count;
}
  
// Driver Code
int main()
{
    int a[] = { 3, 4, 8, 16 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countSubsequences(a, n);
    return 0;
}

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Java

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// Java program to find the number 
// of subsequences with gcd 1 
class GFG
{
      
static final int MAX = 1000
static int gcd(int a, int b) 
    if (a == 0
        return b; 
    return gcd(b % a, a); 
  
// Recursive function to calculate the number 
// of subsequences with gcd 1 starting with 
// particular index 
static int func(int ind, int g, int dp[][],
                int n, int a[]) 
  
    // Base case 
    if (ind == n) 
    
        if (g == 1
            return 1
        else
            return 0
    
  
    // If already visited 
    if (dp[ind][g] != -1
        return dp[ind][g]; 
  
    // Either we take or we do not 
    int ans = func(ind + 1, g, dp, n, a) 
            + func(ind + 1, gcd(a[ind], g), dp, n, a); 
  
    // Return the answer 
    return dp[ind][g] = ans; 
  
// Function to return the
// number of subsequences 
static int countSubsequences(int a[], int n) 
  
    // Hash table to memoize 
    int dp[][] = new int[n][MAX]; 
    for(int i = 0; i < n; i++)
        for(int j = 0; j < MAX; j++)
            dp[i][j] = -1;
  
    // Count the number of subsequences 
    int count = 0
  
    // Count for every subsequence 
    for (int i = 0; i < n; i++) 
        count += func(i + 1, a[i], dp, n, a); 
  
    return count; 
  
// Driver Code 
public static void main(String args[])
    int a[] = { 3, 4, 8, 16 }; 
    int n = a.length; 
    System.out.println(countSubsequences(a, n)); 
}
  
// This code is contributed by Arnab Kundu

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Python3

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# Python3 program to find the number 
# of subsequences with gcd 1 
  
MAX = 1000
  
def gcd(a, b): 
    if (a == 0):
        return
    return gcd(b % a, a)
  
# Recursive function to calculate the 
# number of subsequences with gcd 1 
# starting with particular index 
def func(ind, g, dp, n, a):
  
    # Base case 
    if (ind == n): 
        if (g == 1): 
            return 1
        else:
            return 0
  
    # If already visited 
    if (dp[ind][g] != -1): 
        return dp[ind][g] 
  
    # Either we take or we do not 
    ans = (func(ind + 1, g, dp, n, a) + 
           func(ind + 1, gcd(a[ind], g), 
                             dp, n, a)) 
  
    # Return the answer 
    dp[ind][g] = ans
    return dp[ind][g]
  
# Function to return the number 
# of subsequences 
def countSubsequences(a, n): 
  
    # Hash table to memoize 
    dp = [[-1 for i in range(MAX)] 
              for i in range(n)] 
  
    # Count the number of subsequences 
    count = 0
  
    # Count for every subsequence 
    for i in range(n): 
        count += func(i + 1, a[i], dp, n, a) 
  
    return count 
  
# Driver Code 
a = [3, 4, 8, 16
n = len(a)
print(countSubsequences(a, n)) 
  
# This code is contributed by mohit kumar 29

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C#

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// C# program to find the number 
// of subsequences with gcd 1 
using System;
  
class GFG
{
  
static int gcd(int a, int b) 
    if (a == 0) 
        return b; 
    return gcd(b % a, a); 
  
// Recursive function to calculate the number 
// of subsequences with gcd 1 starting with 
// particular index 
static int func(int ind, int g, int [][] dp,
                int n, int [] a) 
  
    // Base case 
    if (ind == n) 
    
        if (g == 1) 
            return 1; 
        else
            return 0; 
    
  
    // If already visited 
    if (dp[ind][g] != -1) 
        return dp[ind][g]; 
  
    // Either we take or we do not 
    int ans = func(ind + 1, g, dp, n, a) 
            + func(ind + 1, gcd(a[ind], g), dp, n, a); 
  
    // Return the answer 
    return dp[ind][g] = ans; 
  
// Function to return the
// number of subsequences 
static int countSubsequences(int [] a, int n) 
  
    // Hash table to memoize 
    int [][] dp = new int[n][]; 
    for(int i = 0; i < n; i++)
        for(int j = 0; j < 1000; j++)
            dp[i][j] = -1;
  
    // Count the number of subsequences 
    int count = 0; 
  
    // Count for every subsequence 
    for (int i = 0; i < n; i++) 
        count += func(i + 1, a[i], dp, n, a); 
  
    return count; 
  
// Driver Code 
public static void Main()
    int [] a = { 3, 4, 8, 16 }; 
    int n = 4; 
    int x = countSubsequences(a, n);
    Console.Write(x); 
}
  
// This code is contributed by 
// mohit kumar 29

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PHP

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<?php
// PHP program to find the number 
// of subsequences with gcd 1 
  
$GLOBALS['MAX'] = 1000;
  
function gcd($a, $b
    if ($a == 0) 
        return $b
    return gcd($b % $a, $a); 
  
// Recursive function to calculate the 
// number of subsequences with gcd 1 
// starting with particular index 
function func($ind, $g, $dp, $n, $a
  
    // Base case 
    if ($ind == $n)
    
        if ($g == 1) 
            return 1; 
        else
            return 0; 
    
  
    // If already visited 
    if ($dp[$ind][$g] != -1) 
        return $dp[$ind][$g]; 
  
    // Either we take or we do not 
    $ans = func($ind + 1, $g, $dp, $n, $a) + 
           func($ind + 1, gcd($a[$ind], $g), 
                              $dp, $n, $a); 
  
    // Return the answer 
    $dp[$ind][$g] = $ans
      
    return $dp[$ind][$g] ;
  
// Function to return the number
// of subsequences 
function countSubsequences($a, $n
  
    // Hash table to memoize 
    $dp = array(array()) ;
      
    for($i = 0 ; $i < $n ; $i++)
        for($j = 0; 
            $j < $GLOBALS['MAX']; $j++)
            $dp[$i][$j] = -1 ;
              
    // Count the number of subsequences 
    $count = 0; 
  
    // Count for every subsequence 
    for ($i = 0; $i < $n; $i++) 
        $count += func($i + 1, $a[$i],
                       $dp, $n, $a); 
  
    return $count
  
// Driver Code 
$a = array(3, 4, 8, 16); 
$n = sizeof($a) ;
  
echo countSubsequences($a, $n); 
  
// This code is contributed by Ryuga
?>

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Output:

7

Alternate Solution : Count number of subsets of a set with GCD equal to a given number



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Striver(underscore)79 at Codechef and codeforces D

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