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

06 Jul, 2022
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)]
 
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]. 
C++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    int x = 3;    cout << numberOfWays(x) << endl;    return 0;}
// 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 codepublic static void main (String[] args) {int x = 3;System.out.println( numberOfWays(x));         }} // This code is contributed by vt_m.
# 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 codex = 3print (numberOfWays(x)) # This code is contributed by "Sharad_Bhardwaj"
// 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// 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?>
<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
Since there are overlapping subproblems, we can optimize it using dynamic programming.  
C++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    int x = 3;    cout << numberOfWays(x) << endl;    return 0;}
// 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 codepublic static void main (String[] args) {int x = 3;System.out.println(numberOfWays(x));         }}// This code is contributed by vipinyadav15799
# 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 codex = 3print (numberOfWays(x)) # This code is contributed by "Sharad_Bhardwaj"
// 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 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?>
<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
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vt_m
Sam007
vipinyadav15799
GauravRajput1
mitalibhola94
Combinatorial
Dynamic Programming
Dynamic Programming
Combinatorial
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