Problem bei der Paarung von Freunden
Bei n Freunden kann jeder Single bleiben oder mit einem anderen Freund zusammengebracht werden. Jeder Freund kann nur einmal gekoppelt werden. Finden Sie heraus, wie viele Möglichkeiten es gibt, wie Freunde Single bleiben oder ein Paar bilden können.
Beispiele:
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(111) 2) a pair and individual (21) In case of n = 4 we have 3 ways to form a group: 1) all elements are individual (1111) 2) 2 individuals and one pair (211) 3) 2 separate pairs (22) tinytextbf{Mathematische Formel:} newline{frac{n!}{((g_1!)^xtimes (g_2!)^ytimes ... (g_n!)^z)times(x!times y!times...z!)}}spacespacetinytext{ ----- (1)} newline{tinytext{Wenn dieselbe Gruppengröße xy...z mal wiederholt wird, müssen wir unsere Antwort zusätzlich durch x!y!...z!}} teilen newline{tinytext{betrachten wir nun unseren Fall n=3 und die maximale Gruppengröße von Größe 2 und die minimale Größe 1:}} newline{frac{3!}{(1!)^3times(3!)} space+space frac{3!}{(2!)^1times(1!)^1times(1!)^2}space = 4} newline{text{betrachten wir nun unseren Fall n=4:}} newline{frac{4!}{(1!)^4times(4!)} space+ frac{4!}{(2!)^1times(1!)^2times(2!)}space + space frac{4!}{(2!)^2times(2!)}space = 10}
Empfohlene Praxis Problem bei der Paarung von Freunden Probieren Sie es aus!Für die n-te Person gibt es zwei Möglichkeiten: 1) Die n-te Person bleibt Single, wir wiederholen uns für f(n - 1) 2) Die n-te Person paart sich mit einer der verbleibenden n - 1 Personen. Wir erhalten (n – 1) * f(n – 2). Daher können wir f(n) rekursiv schreiben als: f(n) = f(n – 1) + (n – 1) * f(n – 2)
Da die obige rekursive Formel gilt überlappende Teilprobleme Wir können es mit dynamischer Programmierung lösen.
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 code int main () { int n = 4 ; cout < < countFriendsPairings ( n ) < < endl ; return 0 ; }
Java // Java program for solution of // friends pairing problem import 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 code n = 4 print ( countFriendsPairings ( n )) # This code is contributed # by Soumen Ghosh.
C# // C# program solution for // friends pairing problem using 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 // PHP program solution for // friends pairing problem // Returns count of ways n people // can remain single or paired up. function countFriendsPairings ( $n ) { $dp [ $n + 1 ] = 0 ; // Filling dp[] in bottom-up // manner using recursive formula // explained above. for ( $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 $n = 4 ; echo countFriendsPairings ( $n ) ; // This code is contributed // by nitin mittal. ?>
JavaScript < script > // Javascript program for solution of // friends pairing problem // Returns count of ways n people // can remain single or paired up. function countFriendsPairings ( n ) { let dp = []; // Filling dp[] in bottom-up manner using // recursive formula explained above. for ( let 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 let n = 4 ; document . write ( countFriendsPairings ( n )); < /script>
Ausgabe
10
Zeitkomplexität: An)
Hilfsraum: An)
Ein anderer Ansatz: (Verwendung von Rekursion)
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 code int 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 Recursion import 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 Code n = 4 print ( countFriendsPairings ( n )) # This code contributed by PrinciRaj1992
C# // C# program for solution of friends // pairing problem Using Recursion using 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 // PHP program for solution of friends // pairing problem Using Recursion // Returns count of ways n people // can remain single or paired up. function countFriendsPairings ( $n ) { $dp = array_fill ( 0 1000 - 1 ); 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 Code $n = 4 ; echo countFriendsPairings ( $n ) // This code is contributed by Ryuga ?>
JavaScript < script > // Javascript program for solution of friends // pairing problem Using Recursion let dp = new Array ( 1000 ); // Returns count of ways n people // can remain single or paired up. function countFriendsPairings ( 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 for ( let i = 0 ; i < 1000 ; i ++ ) dp [ i ] = - 1 ; let n = 4 ; document . write ( countFriendsPairings ( n )); // This code is contributed by rag2127 < /script>
Ausgabe
10
Zeitkomplexität: An)
Hilfsraum: An)
Da die obige Formel ähnlich ist Fibonacci-Zahl Wir können den Raum mit einer iterativen Lösung optimieren.
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 code int 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 code n = 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 // Returns count of ways n people // can remain single or paired up. function countFriendsPairings ( $n ) { $a = 1 ; $b = 2 ; $c = 0 ; if ( $n <= 2 ) { return $n ; } for ( $i = 3 ; $i <= $n ; $i ++ ) { $c = $b + ( $i - 1 ) * $a ; $a = $b ; $b = $c ; } return $c ; } // Driver code $n = 4 ; print ( countFriendsPairings ( $n )); // This code is contributed by mits ?>
JavaScript < script > // Returns count of ways n people // can remain single or paired up. function countFriendsPairings ( n ) { let a = 1 b = 2 c = 0 ; if ( n <= 2 ) { return n ; } for ( let i = 3 ; i <= n ; i ++ ) { c = b + ( i - 1 ) * a ; a = b ; b = c ; } return c ; } // Driver code let n = 4 ; document . write ( countFriendsPairings ( n )); // This code is contributed by avanitrachhadiya2155 < /script>
Ausgabe
10
Zeitkomplexität: An)
Hilfsraum: O(1)
Ein anderer Ansatz: Da wir das obige Problem mithilfe von Mathematik lösen können, erfolgt die Lösung unten ohne Verwendung dynamischer Programmierung.
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 code int 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 approach import java.util.* ; class GFG { static Vector < Integer > 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 code public 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 dynamically fact = [ 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 factorial preComputeFact ( 1000 ) n = 4 print ( countFriendsPairings ( n )) # solution contributed by adarsh_007
C# // C# program to implement the approach using 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 < script > // Javascript soln using mathematical approach // factorial array is stored dynamically let fact = [ 1 ]; function preComputeFact ( n ) { for ( let i = 1 ; i < n + 1 ; i ++ ) { fact . push (( fact [ i - 1 ] * i )); } } // Returns count of ways n people // can remain single or paired up. function countFriendsPairings ( n ) { let ones = n let twos = 1 ; let ans = 0 ; while ( ones >= 0 ) { // pow of 1 will always be one ans = ans + Math . floor ( fact [ n ] / ( twos * fact [ ones ] * fact [( n - ones ) / 2 ])) ones = ones - 2 twos = twos * 2 } return ans ; } // Driver Code // pre-compute factorial preComputeFact ( 1000 ) n = 4 document . write ( countFriendsPairings ( n )) // This code is contributed by ab2127 < /script>
Ausgabe
10
Zeitkomplexität: An)
Hilfsraum: An)
