Counting pairs when a person can form pair with at most one

Consider a coding competition on geeksforgeeks practice. Now there are n distinct participants taking part in the competition. A single participant can make pair with at most one other participant. We need count the number of ways in which n participants participating in the coding competition.

Examples :

Input : n = 2
Output : 2
2 shows that either both participant 
can pair themselves in one way or both 
of them can remain single.

Input : n = 3 
Output : 4
One way : Three participants remain single
Three More Ways : [(1, 2)(3)], [(1), (2,3)]
and [(1,3)(2)]

Method 1:

1) Every participant can either pair with another participant or can remain single. 
2) Let us consider X-th participant, he can either remain single or he can pair up with someone from [1, x-1].

Below is the implementation of the above idea:

C++

// Number of ways in which participant can take part.
#include<iostream>
using namespace std;

int numberOfWays(int x)
{
    // Base condition
    if (x==0 || x==1)
        return 1;

    // A participant can choose to consider
    // (1) Remains single. Number of people
    //     reduce to (x-1)
    // (2) Pairs with one of the (x-1) others.
    //     For every pairing, number of people
    //     reduce to (x-2).
    else
        return numberOfWays(x-1) +
               (x-1)*numberOfWays(x-2);
}

// Driver code
int main()
{
    int x = 3;
    cout << numberOfWays(x) << endl;
    return 0;
}

Java

// Number of ways in which
// participant can take part.
import java.io.*;

class GFG {

static int numberOfWays(int x)
{
    // Base condition
    if (x==0 || x==1)
        return 1;

    // A participant can choose to consider
    // (1) Remains single. Number of people
    //     reduce to (x-1)
    // (2) Pairs with one of the (x-1) others.
    //     For every pairing, number of people
    //     reduce to (x-2).
    else
        return numberOfWays(x-1) +
            (x-1)*numberOfWays(x-2);
}

// Driver code
public static void main (String[] args) {
int x = 3;
System.out.println( numberOfWays(x));

    }
}

// This code is contributed by vt_m.

Python3

# Python program to find Number of ways
# in which participant can take part.

# Function to calculate number of ways.
def numberOfWays (x):

    # Base condition
    if x == 0 or x == 1:
        return 1

    # A participant can choose to consider
    # (1) Remains single. Number of people
    # reduce to (x-1)
    # (2) Pairs with one of the (x-1) others.
    # For every pairing, number of people
    # reduce to (x-2).
    else:
        return (numberOfWays(x-1) +
              (x-1) * numberOfWays(x-2))

# Driver code
x = 3
print (numberOfWays(x))

# This code is contributed by "Sharad_Bhardwaj"

C#

// Number of ways in which
// participant can take part.
using System;

class GFG {

    static int numberOfWays(int x)
    {

        // Base condition
        if (x == 0 || x == 1)
            return 1;

        // A participant can choose to
        // consider (1) Remains single.
        // Number of people reduce to
        // (x-1) (2) Pairs with one of
        // the (x-1) others. For every
        // pairing, number of people
        // reduce to (x-2).
        else
            return numberOfWays(x - 1) +
            (x - 1) * numberOfWays(x - 2);
    }

    // Driver code
    public static void Main ()
    {
        int x = 3;

        Console.WriteLine(numberOfWays(x));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// Number of ways in which
// participant can take part.

function numberOfWays($x)
{
    // Base condition
    if ($x == 0 || $x == 1)
        return 1;

    // A participant can choose
    // to consider (1) Remains single.
    // Number of people reduce to (x-1)
    // (2) Pairs with one of the (x-1)
    // others. For every pairing, number
    // of peopl reduce to (x-2).
    else
        return numberOfWays($x - 1) +
            ($x - 1) * numberOfWays($x - 2);
}

// Driver code
$x = 3;
echo numberOfWays($x);

// This code is contributed by Sam007
?>

Javascript

<script>
// Number of ways in which
// participant can take part.

    function numberOfWays(x)
    {

        // Base condition
        if (x == 0 || x == 1)
            return 1;

        // A participant can choose to consider
        // (1) Remains single. Number of people
        // reduce to (x-1)
        // (2) Pairs with one of the (x-1) others.
        // For every pairing, number of people
        // reduce to (x-2).
        else
            return numberOfWays(x - 1) + (x - 1) * numberOfWays(x - 2);
    }

    // Driver code
    var x = 3;
    document.write(numberOfWays(x));

// This code is contributed by gauravrajput1
</script>

Output

4

Time Complexity: O(2^x)
Auxiliary Space: O(log(x)), due to recursive call stack

Since there are overlapping subproblems, we can optimize it using dynamic programming. 

C++

// Number of ways in which participant can take part.
#include<iostream>
using namespace std;

int numberOfWays(int x)
{
    int dp[x+1];
    dp[0] = dp[1] = 1;

    for (int i=2; i<=x; i++)
       dp[i] = dp[i-1] + (i-1)*dp[i-2];

    return dp[x];
}

// Driver code
int main()
{
    int x = 3;
    cout << numberOfWays(x) << endl;
    return 0;
}

Java

// Number of ways in which
// participant can take part.
import java.io.*;
class GFG {

static int numberOfWays(int x)
{
    int dp[] = new int[x+1];
    dp[0] = dp[1] = 1;

    for (int i=2; i<=x; i++)
    dp[i] = dp[i-1] + (i-1)*dp[i-2];

    return dp[x];
}

// Driver code
public static void main (String[] args) {
int x = 3;
System.out.println(numberOfWays(x));

    }
}
// This code is contributed by vipinyadav15799

Python3

# Python program to find Number of ways
# in which participant can take part.

# Function to calculate number of ways.
def numberOfWays (x):

    dp=[]
    dp.append(1)
    dp.append(1)
    for i in range(2,x+1):
        dp.append(dp[i-1]+(i-1)*dp[i-2])
    return(dp[x])


# Driver code
x = 3
print (numberOfWays(x))

# This code is contributed by "Sharad_Bhardwaj"

C#

// Number of ways in which
// participant can take part.
using System;

class GFG {

    static int numberOfWays(int x)
    {
        int []dp = new int[x+1];
        dp[0] = dp[1] = 1;

        for (int i = 2; i <= x; i++)
            dp[i] = dp[i - 1] +
                 (i - 1) * dp[i - 2];

        return dp[x];
    }

    // Driver code
    public static void Main ()
    {
        int x = 3;

        Console.WriteLine(numberOfWays(x));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program for Number of ways
// in which participant can take part.

function numberOfWays($x)
{

    $dp[0] = 1;
    $dp[1] = 1;
    for ($i = 2; $i <= $x; $i++)
    $dp[$i] = $dp[$i - 1] + ($i - 1) *
                         $dp[$i - 2];

    return $dp[$x];
}

    // Driver code
    $x = 3;
    echo numberOfWays($x) ;

// This code is contributed by Sam007
?>

Javascript

<script>
// Number of ways in which
// participant can take part.

    function numberOfWays( x)
    {
        let dp = Array(x + 1).fill(0);
        dp[0] = dp[1] = 1;

        for ( i = 2; i <= x; i++)
            dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];

        return dp[x];
    }

    // Driver code
    let x = 3;
    document.write(numberOfWays(x));

// This code is contributed by gauravrajput1
</script>

Output

4

Time Complexity : O( x )
Auxiliary Space: O( x )

Efficient approach: Space optimization O(1)

In previous approach we the current value dp[i] is only depend upon the previous 2 values i.e. dp[i-1] and dp[i-2]. So to optimize the space we can keep track of previous and current values by the help of three variables prev1, prev2 and curr which will reduce the space complexity from O(x) to O(1).  

Implementation Steps:

• Create 2 variables prev1 and prev2 to keep track o previous values of DP.
• Initialize base case prev1 = prev2 = 1.
• Create a variable curr to store current value.
• Iterate over subproblem using loop and update curr.
• After every iteration update prev1 and prev2 for further iterations.
• At last return curr.

Implementation:

C++

// C++ code for above approach

#include <iostream>
using namespace std;

// function to find the Number of
// ways in which participant can take part.
int numberOfWays(int x)
{

    // current value
    int curr;

    // to keep track of previous values
    int prev1 = 1, prev2 = 1;

    // iterate to get current values
    // from previousyl stored values.
    for (int i = 2; i <= x; i++) {
        curr = prev2 + (i - 1) * prev1;

        // assigning values to iterate further
        prev1 = prev2;
        prev2 = curr;
    }

    // return final answer
    return curr;
}

// Driver code
int main()
{
    int x = 3;
    cout << numberOfWays(x) << endl;
    return 0;
}

Java

import java.util.*;

public class Main {
  public static int numberOfWays(int x)
  {
    // current value
    int curr = 0;

    // to keep track of previous values
    int prev1 = 1, prev2 = 1;

    // iterate to get current values
    // from previously stored values.
    for (int i = 2; i <= x; i++) {
      curr = prev2 + (i - 1) * prev1;

      // assigning values to iterate further
      prev1 = prev2;
      prev2 = curr;
    }

    // return final answer
    return curr;
  }

  public static void main(String[] args)
  {
    int x = 3;
    System.out.println(numberOfWays(x));
  }
}

Python3

# function to find the Number of
# ways in which participant can take part.


def numberOfWays(x):
    # current value
    curr = 0

    # to keep track of previous values
    prev1, prev2 = 1, 1

    # iterate to get current values
    # from previously stored values.
    for i in range(2, x+1):
        curr = prev2 + (i-1)*prev1

        # assigning values to iterate further
        prev1, prev2 = prev2, curr

    # return final answer
    return curr


# Driver code
if __name__ == '__main__':
    x = 3
    print(numberOfWays(x))

Javascript

function numberOfWays(x) {
// current value
    let curr = 0;

    // to keep track of previous values
    let prev1 = 1,
        prev2 = 1;


     // iterate to get current values
    // from previously stored values.
    for (let i = 2; i <= x; i++) {
        curr = prev2 + (i - 1) * prev1;
        prev1 = prev2;
        prev2 = curr;
    }
// return final answer
    return curr;
}

let x = 3;
console.log(numberOfWays(x));

Output

4

Time Complexity : O( x )
Auxiliary Space: O( 1 )

This article is contributed by nikunj_agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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