Open In App
Related Articles

Ways to multiply n elements with an associative operation

Improve Article
Improve
Save Article
Save
Like Article
Like

Given a number n, find the number of ways to multiply n elements with an associative operation.

Examples : 

Input : 2
Output : 2
For a and b there are two ways to multiply them.
1. (a * b)        
2. (b * a)

Input : 3
Output : 12

Explanation(Example 2) : 

For a, b and c there are 12 ways to multiply them.
1.  ((a * b) * c)     2.  (a * (b * c))
3.  ((a * c) * b)     4.  (a * (c * b))
5.  ((b * a) * c)     6.  (b * (a * c))
7.  ((b * c) * a)     8.  (b * (c * a))
9.  ((c * a) * b)     10.  (c * (a * b))
11.  ((c * b) * a)    12.  (c * (b * a))

Approach: First, we try to find out the recurrence relation. From above examples, we can see h(1) = 1, h(2) = 2, h(3) = 12 . Now, for n elements there will be n – 1 multiplications and n – 1 parentheses. And, (a1, a2, …, an ) can be obtained from (a1, a2, …, a(n – 1)) in exactly one of the two ways : 

  1. Take a multiplication (a1, a2, …, a(n – 1))(which has n – 2 multiplications and n – 2 parentheses) and insert the nth element ‘an’ on either side of either factor in one of the n – 2 multiplications. Thus, for each scheme for n – 1 numbers gives 2 * 2 * (n – 2) = 4 * (n – 2) schemes for n numbers in this way.
  2. Take a multiplication scheme for (a1, a2, .., a(n-1)) and multiply on left or right by (‘an’). Thus, for each scheme for n – 1 numbers gives two schemes for n numbers in this way.

So after adding above two, we get, h(n) = (4 * n – 8 + 2) * h(n – 1), h(n) = (4 * n – 6) * h(n – 1). This recurrence relation with same initial value is satisfied by the pseudo-Catalan number. Hence, h(n) = (2 * n – 2)! / (n – 1)! 

C++




// C++ code to find number of ways to multiply n
// elements with an associative operation
# include <bits/stdc++.h>
using namespace std;
 
// Function to find the required factorial
int fact(int n)
{
    if (n == 0 || n == 1)   
        return 1 ;
 
    int ans = 1;  
    for (int i = 1 ; i <= n; i++)   
        ans = ans * i ;
 
    return ans ;
}
 
// Function to find nCr
int nCr(int n, int r)
{
    int Nr = n , Dr = 1 , ans = 1;
    for (int i = 1 ; i <= r ; i++ ) {
        ans = ( ans * Nr ) / ( Dr ) ;
        Nr-- ;
        Dr++ ;
    }
    return ans ;
}
 
// function to find the number of ways
int solve ( int n )
{
    int N = 2*n - 2 ;
    int R = n - 1 ;   
    return nCr (N, R) * fact(n - 1) ;
}
 
// Driver code
int main()
{
    int n = 6 ;
    cout << solve (n) ;   
    return 0 ;
}


Java




// Java code to find number of
// ways to multiply n elements
// with an associative operation
import java.io.*;
 
class GFG
{
// Function to find the
// required factorial
static int fact(int n)
{
    if (n == 0 || n == 1)
        return 1 ;
 
    int ans = 1;
    for (int i = 1 ; i <= n; i++)
        ans = ans * i ;
 
    return ans ;
}
 
// Function to find nCr
static int nCr(int n, int r)
{
    int Nr = n , Dr = 1 , ans = 1;
    for (int i = 1 ; i <= r ; i++ )
    {
        ans = ( ans * Nr ) / ( Dr ) ;
        Nr-- ;
        Dr++ ;
    }
    return ans ;
}
 
// function to find
// the number of ways
static int solve ( int n )
{
    int N = 2 * n - 2 ;
    int R = n - 1 ;
    return nCr (N, R) * fact(n - 1) ;
}
 
// Driver Code
public static void main (String[] args)
{
int n = 6 ;
System.out.println( solve (n)) ;
}
}
 
// This code is contributed by anuj_67.


Python3




# Python3 code to find number
# of ways to multiply n
# elements with an
# associative operation
 
# Function to find the
# required factorial
def fact(n):
    if (n == 0 or n == 1):
        return 1;
 
    ans = 1;
    for i in range(1, n + 1):
        ans = ans * i;
 
    return ans;
 
# Function to find nCr
def nCr(n, r):
    Nr = n ; Dr = 1 ; ans = 1;
    for i in range(1, r + 1):
        ans = int((ans * Nr) / (Dr));
        Nr = Nr - 1;
        Dr = Dr + 1;
    return ans;
 
# function to find
# the number of ways
def solve ( n ):
    N = 2* n - 2;
    R = n - 1 ;
    return (nCr (N, R) *
            fact(n - 1));
 
# Driver code
n = 6 ;
print(solve (n) );
 
# This code is contributed
# by mits


C#




// C# code to find number of
// ways to multiply n elements
// with an associative operation
using System;
 
class GFG {
     
    // Function to find the
    // required factorial
    static int fact(int n)
    {
        if (n == 0 || n == 1)
            return 1 ;
     
        int ans = 1;
        for (int i = 1 ; i <= n; i++)
            ans = ans * i ;
     
        return ans ;
    }
     
    // Function to find nCr
    static int nCr(int n, int r)
    {
        int Nr = n , Dr = 1 , ans = 1;
        for (int i = 1 ; i <= r ; i++ )
        {
            ans = ( ans * Nr ) / ( Dr ) ;
            Nr-- ;
            Dr++ ;
        }
        return ans ;
    }
     
    // function to find
    // the number of ways
    static int solve ( int n )
    {
        int N = 2 * n - 2 ;
        int R = n - 1 ;
        return nCr (N, R) * fact(n - 1) ;
    }
     
    // Driver Code
    public static void Main ()
    {
        int n = 6 ;
        Console.WriteLine( solve (n)) ;
    }
}
 
// This code is contributed by anuj_67.


PHP




<?php
// PHP code to find number
// of ways to multiply n
// elements with an
// associative operation
 
// Function to find the
// required factorial
function fact($n)
{
    if ($n == 0 || $n == 1)
        return 1;
 
    $ans = 1;
    for ($i = 1 ; $i <= $n; $i++)
        $ans = $ans * $i;
 
    return $ans;
}
 
// Function to find nCr
function nCr($n, $r)
{
    $Nr = $n ; $Dr = 1 ; $ans = 1;
    for ($i = 1 ; $i <= $r ; $i++ )
    {
        $ans = ($ans * $Nr) /
                      ($Dr);
        $Nr--;
        $Dr++;
    }
    return $ans ;
}
 
// function to find
// the number of ways
function solve ( $n )
{
    $N = 2* $n - 2 ;
    $R = $n - 1 ;
    return nCr ($N, $R) *
           fact($n - 1) ;
}
 
// Driver code
$n = 6 ;
echo solve ($n) ;
 
// This code is contributed
// by ajit
?>


Javascript




<script>
 
// Javascript code to find number of
// ways to multiply n elements
// with an associative operation
 
// Function to find the
// required factorial
function fact(n)
{
    if (n == 0 || n == 1)
        return 1;
   
    let ans = 1;
    for(let i = 1 ; i <= n; i++)
        ans = ans * i;
   
    return ans;
}
   
// Function to find nCr
function nCr(n, r)
{
    let Nr = n , Dr = 1 , ans = 1;
    for(let i = 1 ; i <= r ; i++)
    {
        ans = parseInt((ans * Nr) / (Dr), 10);
        Nr--;
        Dr++;
    }
    return ans;
}
   
// Function to find
// the number of ways
function solve(n)
{
    let N = 2 * n - 2;
    let R = n - 1;
    return nCr (N, R) * fact(n - 1);
}
 
// Driver code
let n = 6;
document.write(solve(n));
 
// This code is contributed by decode2207
 
</script>


Output : 

30240

 

Time Complexity: O(n).
Auxiliary Space: O(1).


Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated : 03 Jan, 2023
Like Article
Save Article
Previous
Next
Similar Reads
Complete Tutorials