Givet n venner kan hver enkelt forblive single eller kan parres med en anden ven. Hver ven kan kun parres én gang. Find ud af det samlede antal måder, hvorpå venner kan forblive single eller kan parres.
Eksempler:
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{Matematisk formel:} newline{frac{n!}{((g_1!)^xtimes (g_2!)^ytimes ... (g_n!)^z)times(x!times y!times...z!)}}spacespacetinytext{ ----- (1)} newline{tinytext{hvis samme gruppestørrelse skal gentages og vores gruppestørrelse skal deles x!y!...z!}} nylinje{tinytext{nu lad os overveje vores case n=3 og gruppestørrelse maks. størrelse 2 og min størrelse 1:}} newline{frac{3!}{(1!)^3times(3!)} space+space frac{3!}{(2!)^1times(1!)^1times(1!)^1times(1!)nows" n=4:}} newline{frac{4!}{(1!)^4times(4!)} space+frac{4!}{(2!)^1times(1!)^2times(2!)}space + space frakt{4!}{(2!)^2time(2!)}space = 10}
Anbefalet praksis Venneparringsproblem Prøv det!
For n'te person er der to valgmuligheder: 1) n'te person forbliver single vi gentager sig for f(n - 1)2) n'te person parrer sig med en af de resterende n - 1 personer. Vi får (n - 1) * f(n - 2) Derfor kan vi rekursivt skrive f(n) som:f(n) = f(n - 1) + (n - 1) * f(n - 2)
Da ovenstående rekursive formel har overlappende delproblemer vi kan løse det ved hjælp af dynamisk programmering.
C++// C++ program for solution of // friends pairing problem #include
// 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
# 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# 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 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. ?> <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>
Produktion
10
Tidskompleksitet: På)
Hjælpeplads: På)
En anden tilgang: (Ved brug af rekursion)
C++// C++ program for solution of friends // pairing problem Using Recursion #include
// 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 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# 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 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 ?> <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>
Produktion
10
Tidskompleksitet: På)
Hjælpeplads: På)
Da ovenstående formel ligner fibonacci nummer vi kan optimere rummet med en iterativ løsning.
C++#include
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.
# 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
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
// 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 ?> <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>
Produktion
10
Tidskompleksitet: På)
Hjælpeplads: O(1)
En anden tilgang: Da vi kan løse ovenstående problem ved hjælp af matematik, udføres løsningen nedenfor uden brug af dynamisk programmering.
C++// C++ soln using mathematical approach #include
// 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 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# 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
<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>
Produktion
10
Tidskompleksitet: På)
Hjælpeplads: På)