Consider a coding competition on neveropen practice. Now there are n distinct participants taking part in the competition. A single participant can make pair with at most one other participant. We need count the number of ways in which n participants participating in the coding competition.
Examples :
Input : n = 2
Output : 2
2 shows that either both participant
can pair themselves in one way or both
of them can remain single.
Input : n = 3
Output : 4
One way : Three participants remain single
Three More Ways : [(1, 2)(3)], [(1), (2,3)]
and [(1,3)(2)]
Method 1:
1) Every participant can either pair with another participant or can remain single.
2) Let us consider X-th participant, he can either remain single or he can pair up with someone from [1, x-1].
Below is the implementation of the above idea:
C++
// Number of ways in which participant can take part.#include<iostream>using namespace std;int numberOfWays(int x){ // Base condition if (x==0 || x==1) return 1; // A participant can choose to consider // (1) Remains single. Number of people // reduce to (x-1) // (2) Pairs with one of the (x-1) others. // For every pairing, number of people // reduce to (x-2). else return numberOfWays(x-1) + (x-1)*numberOfWays(x-2);}// Driver codeint main(){ int x = 3; cout << numberOfWays(x) << endl; return 0;} |
Java
// Number of ways in which// participant can take part.import java.io.*;class GFG {static int numberOfWays(int x){ // Base condition if (x==0 || x==1) return 1; // A participant can choose to consider // (1) Remains single. Number of people // reduce to (x-1) // (2) Pairs with one of the (x-1) others. // For every pairing, number of people // reduce to (x-2). else return numberOfWays(x-1) + (x-1)*numberOfWays(x-2);}// Driver codepublic static void main (String[] args) {int x = 3;System.out.println( numberOfWays(x)); }}// This code is contributed by vt_m. |
Python3
# Python program to find Number of ways # in which participant can take part.# Function to calculate number of ways.def numberOfWays (x): # Base condition if x == 0 or x == 1: return 1 # A participant can choose to consider # (1) Remains single. Number of people # reduce to (x-1) # (2) Pairs with one of the (x-1) others. # For every pairing, number of people # reduce to (x-2). else: return (numberOfWays(x-1) + (x-1) * numberOfWays(x-2))# Driver codex = 3print (numberOfWays(x))# This code is contributed by "Sharad_Bhardwaj" |
C#
// Number of ways in which// participant can take part.using System;class GFG { static int numberOfWays(int x) { // Base condition if (x == 0 || x == 1) return 1; // A participant can choose to // consider (1) Remains single. // Number of people reduce to // (x-1) (2) Pairs with one of // the (x-1) others. For every // pairing, number of people // reduce to (x-2). else return numberOfWays(x - 1) + (x - 1) * numberOfWays(x - 2); } // Driver code public static void Main () { int x = 3; Console.WriteLine(numberOfWays(x)); }}// This code is contributed by vt_m. |
Javascript
<script>// Number of ways in which// participant can take part. function numberOfWays(x) { // Base condition if (x == 0 || x == 1) return 1; // A participant can choose to consider // (1) Remains single. Number of people // reduce to (x-1) // (2) Pairs with one of the (x-1) others. // For every pairing, number of people // reduce to (x-2). else return numberOfWays(x - 1) + (x - 1) * numberOfWays(x - 2); } // Driver code var x = 3; document.write(numberOfWays(x));// This code is contributed by gauravrajput1 </script> |
PHP
<?php// Number of ways in which // participant can take part.function numberOfWays($x){ // Base condition if ($x == 0 || $x == 1) return 1; // A participant can choose // to consider (1) Remains single. // Number of people reduce to (x-1) // (2) Pairs with one of the (x-1) // others. For every pairing, number // of peopl reduce to (x-2). else return numberOfWays($x - 1) + ($x - 1) * numberOfWays($x - 2);}// Driver code$x = 3;echo numberOfWays($x);// This code is contributed by Sam007?> |
Output
4
Time Complexity: O(2x)
Auxiliary Space: O(log(x)), due to recursive call stack
Since there are overlapping subproblems, we can optimize it using dynamic programming.
C++
// Number of ways in which participant can take part.#include<iostream>using namespace std;int numberOfWays(int x){ int dp[x+1]; dp[0] = dp[1] = 1; for (int i=2; i<=x; i++) dp[i] = dp[i-1] + (i-1)*dp[i-2]; return dp[x];}// Driver codeint main(){ int x = 3; cout << numberOfWays(x) << endl; return 0;} |
Java
// Number of ways in which// participant can take part.import java.io.*;class GFG {static int numberOfWays(int x){ int dp[] = new int[x+1]; dp[0] = dp[1] = 1; for (int i=2; i<=x; i++) dp[i] = dp[i-1] + (i-1)*dp[i-2]; return dp[x];}// Driver codepublic static void main (String[] args) {int x = 3;System.out.println(numberOfWays(x)); }}// This code is contributed by vipinyadav15799 |
Python3
# Python program to find Number of ways # in which participant can take part.# Function to calculate number of ways.def numberOfWays (x): dp=[] dp.append(1) dp.append(1) for i in range(2,x+1): dp.append(dp[i-1]+(i-1)*dp[i-2]) return(dp[x]) # Driver codex = 3print (numberOfWays(x))# This code is contributed by "Sharad_Bhardwaj" |
C#
// Number of ways in which// participant can take part.using System;class GFG { static int numberOfWays(int x) { int []dp = new int[x+1]; dp[0] = dp[1] = 1; for (int i = 2; i <= x; i++) dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]; return dp[x]; } // Driver code public static void Main () { int x = 3; Console.WriteLine(numberOfWays(x)); }}// This code is contributed by vt_m. |
Javascript
<script>// Number of ways in which// participant can take part. function numberOfWays( x) { let dp = Array(x + 1).fill(0); dp[0] = dp[1] = 1; for ( i = 2; i <= x; i++) dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]; return dp[x]; } // Driver code let x = 3; document.write(numberOfWays(x));// This code is contributed by gauravrajput1</script> |
PHP
<?php// PHP program for Number of ways // in which participant can take part.function numberOfWays($x){ $dp[0] = 1; $dp[1] = 1; for ($i = 2; $i <= $x; $i++) $dp[$i] = $dp[$i - 1] + ($i - 1) * $dp[$i - 2]; return $dp[$x];} // Driver code $x = 3; echo numberOfWays($x) ; // This code is contributed by Sam007?> |
Output
4
Time Complexity : O( x )
Auxiliary Space: O( x )
Efficient approach: Space optimization O(1)
In previous approach we the current value dp[i] is only depend upon the previous 2 values i.e. dp[i-1] and dp[i-2]. So to optimize the space we can keep track of previous and current values by the help of three variables prev1, prev2 and curr which will reduce the space complexity from O(x) to O(1).
Implementation Steps:
- Create 2 variables prev1 and prev2 to keep track o previous values of DP.
- Initialize base case prev1 = prev2 = 1.
- Create a variable curr to store current value.
- Iterate over subproblem using loop and update curr.
- After every iteration update prev1 and prev2 for further iterations.
- At last return curr.
Implementation:
C++
// C++ code for above approach#include <iostream>using namespace std;// function to find the Number of// ways in which participant can take part.int numberOfWays(int x){ // current value int curr; // to keep track of previous values int prev1 = 1, prev2 = 1; // iterate to get current values // from previousyl stored values. for (int i = 2; i <= x; i++) { curr = prev2 + (i - 1) * prev1; // assigning values to iterate further prev1 = prev2; prev2 = curr; } // return final answer return curr;}// Driver codeint main(){ int x = 3; cout << numberOfWays(x) << endl; return 0;} |
Java
import java.util.*;public class Main { public static int numberOfWays(int x) { // current value int curr = 0; // to keep track of previous values int prev1 = 1, prev2 = 1; // iterate to get current values // from previously stored values. for (int i = 2; i <= x; i++) { curr = prev2 + (i - 1) * prev1; // assigning values to iterate further prev1 = prev2; prev2 = curr; } // return final answer return curr; } public static void main(String[] args) { int x = 3; System.out.println(numberOfWays(x)); }} |
Python3
# function to find the Number of# ways in which participant can take part.def numberOfWays(x): # current value curr = 0 # to keep track of previous values prev1, prev2 = 1, 1 # iterate to get current values # from previously stored values. for i in range(2, x+1): curr = prev2 + (i-1)*prev1 # assigning values to iterate further prev1, prev2 = prev2, curr # return final answer return curr# Driver codeif __name__ == '__main__': x = 3 print(numberOfWays(x)) |
C#
using System;namespace NumberOfWays{ class Program { // Function to find the Number of ways in which a participant can take part. static int NumberOfWays(int x) { // Current value int curr = 0; // To keep track of previous values int prev1 = 1, prev2 = 1; // Iterate to get current values from previously stored values. for (int i = 2; i <= x; i++) { curr = prev2 + (i - 1) * prev1; // Assigning values to iterate further prev1 = prev2; prev2 = curr; } // Return the final answer return curr; } // Main driver code static void Main(string[] args) { int x = 3; Console.WriteLine(NumberOfWays(x)); // Pause execution to see the result Console.ReadKey(); } }} |
Javascript
function numberOfWays(x) {// current value let curr = 0; // to keep track of previous values let prev1 = 1, prev2 = 1; // iterate to get current values // from previously stored values. for (let i = 2; i <= x; i++) { curr = prev2 + (i - 1) * prev1; prev1 = prev2; prev2 = curr; }// return final answer return curr;}let x = 3;console.log(numberOfWays(x)); |
Output
4
Time Complexity : O( x )
Auxiliary Space: O( 1 )
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