# Friends Pairing Problem

Last Updated : 12 Dec, 2022

Given n friends, each one can remain single or can be paired up with some other friend. Each friend can be paired only once. Find out the total number of ways in which friends can remain single or can be paired up.

Examples:

Input  : n = 3
Output : 4
Explanation:
{1}, {2}, {3} : all single
{1}, {2, 3} : 2 and 3 paired but 1 is single.
{1, 2}, {3} : 1 and 2 are paired but 3 is single.
{1, 3}, {2} : 1 and 3 are paired but 2 is single.
Note that {1, 2} and {2, 1} are considered same.

Mathematical Explanation:
The problem is simplified version of how many ways we can divide n elements into multiple groups.
(here group size will be max of 2 elements).
In case of n = 3, we have only 2 ways to make a group:
1) all elements are individual(1,1,1)
2) a pair and individual (2,1)
In case of n = 4, we have 3 ways to form a group:
1) all elements are individual (1,1,1,1)
2) 2 individuals and one pair (2,1,1)
3) 2 separate pairs (2,2)

Recommended Practice

For n-th person there are two choices:1) n-th person remains single, we recur for f(n – 1)2) n-th person pairs up with any of the remaining n – 1 persons. We get (n – 1) * f(n – 2)Therefore we can recursively write f(n) as:f(n) = f(n – 1) + (n – 1) * f(n – 2)

Since the above recursive formula has overlapping subproblems, we can solve it using Dynamic Programming.

## C++

 // C++ program for solution of// friends pairing problem#include using namespace std; // Returns count of ways n people// can remain single or paired up.int countFriendsPairings(int n){    int dp[n + 1];     // Filling dp[] in bottom-up manner using    // recursive formula explained above.    for (int i = 0; i <= n; i++) {        if (i <= 2)            dp[i] = i;        else            dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];    }     return dp[n];} // Driver codeint main(){    int n = 4;    cout << countFriendsPairings(n) << endl;    return 0;}

## Java

 // Java program for solution of// friends pairing problemimport java.io.*; class GFG {     // Returns count of ways n people    // can remain single or paired up.    static int countFriendsPairings(int n)    {        int dp[] = new int[n + 1];         // Filling dp[] in bottom-up manner using        // recursive formula explained above.        for (int i = 0; i <= n; i++) {            if (i <= 2)                dp[i] = i;            else                dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];        }         return dp[n];    }     // Driver code    public static void main(String[] args)    {        int n = 4;        System.out.println(countFriendsPairings(n));    }} // This code is contributed by vt_m

## Python3

 # Python program solution of# friends pairing problem # Returns count of ways# n people can remain# single or paired up.def countFriendsPairings(n):     dp = [0 for i in range(n + 1)]     # Filling dp[] in bottom-up manner using    # recursive formula explained above.    for i in range(n + 1):         if(i <= 2):            dp[i] = i        else:            dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]     return dp[n] # Driver coden = 4print(countFriendsPairings(n)) # This code is contributed# by Soumen Ghosh.

## C#

 // C# program solution for// friends pairing problemusing System; class GFG {     // Returns count of ways n people    // can remain single or paired up.    static int countFriendsPairings(int n)    {        int[] dp = new int[n + 1];         // Filling dp[] in bottom-up manner using        // recursive formula explained above.        for (int i = 0; i <= n; i++) {            if (i <= 2)                dp[i] = i;            else                dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];        }         return dp[n];    }     // Driver code    public static void Main()    {        int n = 4;        Console.Write(countFriendsPairings(n));    }} // This code is contributed by nitin mittal.

## PHP

 

## Javascript

 

Output
10


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

Another approach: (Using recursion)

## C++

 // C++ program for solution of friends// pairing problem Using Recursion#include using namespace std; int dp[1000]; // Returns count of ways n people// can remain single or paired up.int countFriendsPairings(int n){    if (dp[n] != -1)        return dp[n];     if (n > 2)        return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);    else        return dp[n] = n;} // Driver codeint main(){    memset(dp, -1, sizeof(dp));    int n = 4;    cout << countFriendsPairings(n) << endl;    // this code is contributed by Kushdeep Mittal}

## Java

 // Java program for solution of friends// pairing problem Using Recursionimport java.io.*;class GFG {    static int[] dp = new int[1000];     // Returns count of ways n people    // can remain single or paired up.    static int countFriendsPairings(int n)    {        if (dp[n] != -1)            return dp[n];         if (n > 2)            return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);        else            return dp[n] = n;    }     // Driver code    public static void main(String[] args)    {        for (int i = 0; i < 1000; i++)            dp[i] = -1;        int n = 4;        System.out.println(countFriendsPairings(n));    }} // This code is contributed by Ita_c.

## Python3

 # Python3 program for solution of friends # pairing problem Using Recursion dp = [-1] * 1000# Returns count of ways n people # can remain single or paired up. def countFriendsPairings(n):     global dp         if(dp[n] != -1):         return dp[n]      if(n > 2):          dp[n] = (countFriendsPairings(n - 1) +                (n - 1) * countFriendsPairings(n - 2))         return dp[n]     else:        dp[n] = n         return dp[n]     # Driver Coden = 4print(countFriendsPairings(n)) # This code contributed by PrinciRaj1992

## C#

 // C# program for solution of friends// pairing problem Using Recursionusing System; class GFG {    static int[] dp = new int[1000];     // Returns count of ways n people    // can remain single or paired up.    static int countFriendsPairings(int n)    {        if (dp[n] != -1)            return dp[n];         if (n > 2)            return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);        else            return dp[n] = n;    }     // Driver code    static void Main()    {        for (int i = 0; i < 1000; i++)            dp[i] = -1;        int n = 4;        Console.Write(countFriendsPairings(n));    }} // This code is contributed by DrRoot_

## PHP

  2)     {        $dp[$n] = countFriendsPairings($n - 1) + ($n - 1) *                   countFriendsPairings($n - 2);  return $dp[$n]; } else { $dp[$n] = $n;         return $dp[$n];    }}      // Driver Code$n = 4; echo countFriendsPairings($n)  // This code is contributed by Ryuga?>

## Javascript

 

Output
10


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

Since the above formula is similar to fibonacci number, we can optimize the space with an iterative solution.

## C++

 #include using namespace std; // Returns count of ways n people// can remain single or paired up.int countFriendsPairings(int n){    int a = 1, b = 2, c = 0;    if (n <= 2) {        return n;    }    for (int i = 3; i <= n; i++) {        c = b + (i - 1) * a;        a = b;        b = c;    }    return c;} // Driver codeint main(){    int n = 4;    cout << countFriendsPairings(n);    return 0;} // This code is contributed by mits

## Java

 import java.io.*;class GFG {    // Returns count of ways n people    // can remain single or paired up.    static int countFriendsPairings(int n)    {        int a = 1, b = 2, c = 0;        if (n <= 2) {            return n;        }        for (int i = 3; i <= n; i++) {            c = b + (i - 1) * a;            a = b;            b = c;        }        return c;    }     // Driver code    public static void main(String[] args)    {        int n = 4;        System.out.println(countFriendsPairings(n));    }} // This code is contributed by Ravi Kasha.

## Python3

 # Returns count of ways n people# can remain single or paired up.def countFriendsPairings(n):    a, b, c = 1, 2, 0;    if (n <= 2):        return n;    for i in range(3, n + 1):        c = b + (i - 1) * a;        a = b;        b = c;    return c; # Driver coden = 4;print(countFriendsPairings(n)); # This code contributed by Rajput-Ji

## C#

 using System; class GFG {    // Returns count of ways n people    // can remain single or paired up.    static int countFriendsPairings(int n)    {        int a = 1, b = 2, c = 0;        if (n <= 2) {            return n;        }        for (int i = 3; i <= n; i++) {            c = b + (i - 1) * a;            a = b;            b = c;        }        return c;    }     // Driver code    public static void Main(String[] args)    {        int n = 4;        Console.WriteLine(countFriendsPairings(n));    }} // This code has been contributed by 29AjayKumar

## PHP

 

## Javascript

 

Output
10

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

Another Approach: Since we can solve the above problem using maths, the solution below is done without using dynamic programming.

## C++

 // C++ soln using mathematical approach#include using namespace std; void preComputeFact(vector<long long int>& fact, int n){    for(int i = 1; i <= n; i++)        fact.push_back(fact[i - 1] * i);} // Returns count of ways n people// can remain single or paired up.int countFriendsPairings(vector<long long int> fact,                          int n){    int ones = n, twos = 1, ans = 0;         while (ones >= 0)     {                 // pow of 1 will always be one        ans += fact[n] / (twos * fact[ones] *                fact[(n - ones) / 2]);        ones -= 2;        twos *= 2;    }    return ans;} // Driver codeint main(){    vector<long long int> fact;    fact.push_back(1);     preComputeFact(fact, 100);    int n = 4;     cout << countFriendsPairings(fact, n) << endl;    return 0;} // This code is contributed by rajsanghavi9.

## Java

 // Java soln using mathematical approachimport java.util.*; class GFG{static   Vector fact;static void preComputeFact( int n){    for(int i = 1; i <= n; i++)        fact.add(fact.elementAt(i - 1) * i);} // Returns count of ways n people// can remain single or paired up.static int countFriendsPairings(int n){    int ones = n, twos = 1, ans = 0;         while (ones >= 0)     {                 // pow of 1 will always be one        ans += fact.elementAt(n) / (twos * fact.elementAt(ones) *                fact.elementAt((n - ones) / 2));        ones -= 2;        twos *= 2;    }    return ans;} // Driver codepublic static void main(String[] args){  fact = new Vector<>();    fact.add(1);     preComputeFact(100);    int n = 4;     System.out.print(countFriendsPairings(n) +"\n");}} // This code is contributed by umadevi9616

## Python3

 # Python3 soln using mathematical approach# factorial array is stored dynamicallyfact = [1]def preComputeFact(n):    for i in range(1, n+1):        fact.append((fact[i-1]*i)) # Returns count of ways n people# can remain single or paired up.def countFriendsPairings(n):    ones = n    twos = 1    ans = 0    while(ones >= 0):        # pow of 1 will always be one        ans = ans + (fact[n]//(twos*fact[ones]*fact[(n-ones)//2]))        ones = ones - 2        twos = twos * 2    return(ans)  # Driver Code# pre-compute factorialpreComputeFact(1000)n = 4print(countFriendsPairings(n)) # solution contributed by adarsh_007

## C#

 // C# program to implement the approachusing System;using System.Collections.Generic;public class GFG {   // initializing the fact list  static List<int> fact = new List<int>();   // computing the next n values of fact  static void preComputeFact(int n)  {    for (int i = 1; i <= n; i++) {      fact.Add(fact[i - 1] * i);    }  }   // Returns count of ways n people  // can remain single or paired up.  static int countFriendsPairings(int n)  {    int ones = n;    int twos = 1;    int ans = 0;     while (ones >= 0) {       // pow of 1 will always be one      ans += fact[n]        / (twos * fact[ones]           * fact[(n - ones) / 2]);      ones -= 2;      twos *= 2;    }    return ans;  }   // driver code  static public void Main()  {     // initializing the first element of fact    fact.Add(1);    preComputeFact(100);    int n = 4;    Console.Write(countFriendsPairings(n));  }} // this code is contributed by phasing17

## Javascript

 

Output
10


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

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