Count number of sub-sequences with GCD 1

Given an array of N numbers, the task is to count the number of subsequences that 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. For every subsequence, check if its GCD is 1 or not. If 1, increment the result.

When we have values in the 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 was considering two cases for every element, we either include or exclude it. We keep track of the 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 whether 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 the memoization technique.

Below is the implementation of the above approach:  

C++

// 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;
}

Java

// 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

Python3

# Python3 program to find the number
# of subsequences with gcd 1

MAX = 1000

def 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
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

C#

// 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

PHP

<?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
?>

Javascript

<script>

// JavaScript program to find the number
// of subsequences with gcd 1
var 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
        var 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
    function countSubsequences(a , n) {

        // Hash table to memoize
        var dp = Array(n).fill().map(()=>Array(MAX).fill(0));
        for (i = 0; i < n; i++)
            for (j = 0; j < MAX; j++)
                dp[i][j] = -1;

        // Count the number of subsequences
        var 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

        var a = [ 3, 4, 8, 16 ];
        var n = a.length;
        document.write(countSubsequences(a, n));

// This code contributed by Rajput-Ji

</script>

Output

7

Complexity Analysis:

• Time Complexity: O(logN*N²), as we are recursively calling the function two times but we are using memorization and the GCD function will cost O(logN). Where N is the number of elements in the array.
• Auxiliary Space: O(N*1000), as we are using extra space for the DP matrix. Where N is the number of elements in the array.

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

Dynamic programming approach to this problem without memoization:

Basically, the approach will be making a 2d matrix in which i coordinate will be the position of elements of the given array and the j coordinate will be numbers from 0 to 100 ie. gcd can vary from 0 to 100 if array elements are not enough large.  we will iterate on the given array and the 2d matrix will store information that till ith position that how many subsequences are there having gcd vary from 1 to 100. later on, we will add dp[i][1] to get all subsequences having gcd as 1.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// This function calculates
// gcd of two number
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// This function will return total
// subsequences
int countSubsequences(int arr[], int n)
{

   // Declare a dp 2d array
   long long int dp[n][101] = {0};

   // Iterate i from 0 to n - 1
   for(int i = 0; i < n; i++)
   {

       dp[i][arr[i]] = 1;

       // Iterate j from i - 1 to 0
       for(int j = i - 1; j >= 0; j--)
       {
           if(arr[j] < arr[i])
           {

              // Iterate k from 0 to 100
              for(int k = 0; k <= 100; k++)
              {

                 // Find gcd of two number
                 int GCD = gcd(arr[i], k);

                 // dp[i][GCD] is summation of
                 // dp[i][GCD] and dp[j][k]
                 dp[i][GCD] = dp[i][GCD] +
                                    dp[j][k];
              }
           }
       }
   }

    // Add all elements of dp[i][1]
    long long int sum = 0;
    for(int i = 0; i < n; i++)
    {
       sum=(sum + dp[i][1]);
    }

    // Return sum
    return sum;
}

// Driver code
int main()
{
    int a[] = { 3, 4, 8, 16 };
    int n = sizeof(a) / sizeof(a[0]);

    // Function Call
    cout << countSubsequences(a, n);
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;

class GFG{

// This function calculates
// gcd of two number
static int gcd(int a, int b)
{
    if (b == 0)
        return a;

    return gcd(b, a % b);
}

// This function will return total
// subsequences
static long countSubsequences(int arr[],
                              int n)
{

    // Declare a dp 2d array
    long  dp[][] = new long[n][101];

    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < 101; j++)
        {
            dp[i][j] = 0;
        }
    }

    // Iterate i from 0 to n - 1
    for(int i = 0; i < n; i++)
    {

        dp[i][arr[i]] = 1;

        // Iterate j from i - 1 to 0
        for(int j = i - 1; j >= 0; j--)
        {
            if (arr[j] < arr[i])
            {

                // Iterate k from 0 to 100
                for(int k = 0; k <= 100; k++)
                {

                    // Find gcd of two number
                    int GCD = gcd(arr[i], k);

                    // dp[i][GCD] is summation of
                    // dp[i][GCD] and dp[j][k]
                    dp[i][GCD] = dp[i][GCD] +
                                 dp[j][k];
                }
            }
        }
    }

    // Add all elements of dp[i][1]
    long sum = 0;
    for(int i = 0; i < n; i++)
    {
        sum = (sum + dp[i][1]);
    }

    // Return sum
    return sum;
}

// Driver code
public static void main(String args[])
{
    int a[] = { 3, 4, 8, 16 };
    int n = a.length;

    // Function Call
    System.out.println(countSubsequences(a, n));
}
}

// This code is contributed by bolliranadheer

Python3

# Python3 program for the
# above approach

# This function calculates
# gcd of two number
def gcd(a, b):
    if (b == 0):
        return a;
    return gcd(b, a % b);

# This function will return total
# subsequences
def countSubsequences(arr, n):

    # Declare a dp 2d array
    dp = [[0 for i in range(101)] for j in range(n)]

    # Iterate i from 0 to n - 1
    for i in range(n):
        dp[i][arr[i]] = 1;

        # Iterate j from i - 1 to 0
        for j in range(i - 1, -1, -1):
            if (arr[j] < arr[i]):

                # Iterate k from 0 to 100
                for k in range(101):

                    # Find gcd of two number
                    GCD = gcd(arr[i], k);

                    # dp[i][GCD] is summation of
                    # dp[i][GCD] and dp[j][k]
                    dp[i][GCD] = dp[i][GCD] + dp[j][k];

    # Add all elements of dp[i][1]
    sum = 0;
    for i in range(n):
        sum = (sum + dp[i][1]);

    # Return sum
    return sum;

# Driver code
if __name__=='__main__':

    a = [ 3, 4, 8, 16 ]
    n = len(a)

    # Function Call
    print(countSubsequences(a,n))

# This code is contributed by Pratham76

C#

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG{

// This function calculates
// gcd of two number
static int gcd(int a, int b)
{
    if (b == 0)
        return a;

    return gcd(b, a % b);
}

// This function will return total
// subsequences
static long countSubsequences(int []arr,
                              int n)
{

    // Declare a dp 2d array
    long  [,]dp = new long[n, 101];

    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < 101; j++)
        {
            dp[i, j] = 0;
        }
    }

    // Iterate i from 0 to n - 1
    for(int i = 0; i < n; i++)
    {

        dp[i, arr[i]] = 1;

        // Iterate j from i - 1 to 0
        for(int j = i - 1; j >= 0; j--)
        {
            if (arr[j] < arr[i])
            {

                // Iterate k from 0 to 100
                for(int k = 0; k <= 100; k++)
                {

                    // Find gcd of two number
                    int GCD = gcd(arr[i], k);

                    // dp[i,GCD] is summation of
                    // dp[i,GCD] and dp[j,k]
                    dp[i, GCD] = dp[i, GCD] +
                                 dp[j, k];
                }
            }
        }
    }

    // Add all elements of dp[i,1]
    long sum = 0;
    for(int i = 0; i < n; i++)
    {
        sum = (sum + dp[i, 1]);
    }

    // Return sum
    return sum;
}

// Driver code
public static void Main(string []args)
{
    int []a = { 3, 4, 8, 16 };
    int n = a.Length;

    // Function Call
    Console.WriteLine(countSubsequences(a, n));
}
}

// This code is contributed by rutvik_56

Javascript

<script>
// javascript program for the
// above approach

    // This function calculates
    // gcd of two number
    function gcd(a , b) {
        if (b == 0)
            return a;

        return gcd(b, a % b);
    }

    // This function will return total
    // subsequences
    function countSubsequences(arr , n) {

        // Declare a dp 2d array
        var dp = Array(n).fill().map(()=>Array(101).fill(0));

        for (i = 0; i < n; i++) {
            for (j = 0; j < 101; j++) {
                dp[i][j] = 0;
            }
        }

        // Iterate i from 0 to n - 1
        for (i = 0; i < n; i++) {

            dp[i][arr[i]] = 1;

            // Iterate j from i - 1 to 0
            for (j = i - 1; j >= 0; j--) {
                if (arr[j] < arr[i]) {

                    // Iterate k from 0 to 100
                    for (k = 0; k <= 100; k++) {

                        // Find gcd of two number
                        var GCD = gcd(arr[i], k);

                        // dp[i][GCD] is summation of
                        // dp[i][GCD] and dp[j][k]
                        dp[i][GCD] = dp[i][GCD] + dp[j][k];
                    }
                }
            }
        }

        // Add all elements of dp[i][1]
        var sum = 0;
        for (i = 0; i < n; i++) {
            sum = (sum + dp[i][1]);
        }

        // Return sum
        return sum;
    }

    // Driver code

        var a = [ 3, 4, 8, 16 ];
        var n = a.length;

        // Function Call
        document.write(countSubsequences(a, n));

// This code contributed by aashish1995
</script>

Output

7

Complexity Analysis:

• Time Complexity: O(100*logN*N²), as we are using nested loops and the GCD function will cost O(logN). Where N is the number of elements in the array.
• Auxiliary Space: O(N*100), as we are using extra space for the DP matrix. Where N is the number of elements in the array.