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.

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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