# Friends Pairing Problem

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.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

```f(n) = ways n people can remain single
or pair up.

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

```// C++ program solution friends pairing problem
#include <bits/stdc++.h>
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 code
int main()
{
int n = 4;
cout << countFriendsPairings(n) << endl;
return 0;
}
```

Output:

```10
```

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

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