In mathematics, the telephone numbers involution numbers are a sequence of integers that count the number of connection patterns in a telephone system with n subscribers, where connections are made between pairs of subscribers. These numbers also describe the number of matchings of a complete graph of n vertices, the number of permutations on n elements that are involutions, the sum of absolute value of coefficients of the Hermite polynomials, the number of standard Young tableaux with n cells, and the sum of the degrees of the irreducible representations of the symmetric group.
The telephone numbers are also used to count the number of ways to place n rooks on an n x n chessboard in such a way that no two rooks attack each other and in such a way the configuration of the rooks is symmetric under a diagonal reflection of the board.
The telephone number can be evaluated by the following recurrence relation:
Given a positive integer n. The task is to find the nth telephone number.
Examples :
Input : n = 4 Output : 10 Input : n = 6 Output : 76
Below is naive implementation of finding the nth telephone number based on above recursive formula.
C++
// CPP Program to find the nth telephone number.#include <bits/stdc++.h>using namespace std;// return the nth telephone numberint telephonenumber(int n){ // base step if (n == 0 || n == 1) return 1; // recursive step return telephonenumber(n - 1) + (n - 1) * telephonenumber(n - 2);}// Driven Programint main(){ int n = 6; cout << telephonenumber(n) << endl; return 0;} |
Java
// JAVA Code to find the nth// telephone number.import java.util.*;class GFG { // return the nth telephone number static int telephonenumber(int n) { // base step if (n == 0 || n == 1) return 1; // recursive step return telephonenumber(n - 1) + (n - 1) * telephonenumber(n - 2); } /* Driver program to test above function */ public static void main(String[] args) { int n = 6; System.out.println(telephonenumber(n)); }}// This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 code to find the # nth telephone number.# return the nth telephone numberdef telephonenumber (n): # base step if n == 0 or n == 1: return 1 # recursive step return (telephonenumber(n - 1) + (n - 1) * telephonenumber(n - 2))# Driven Programn = 6print(telephonenumber(n))# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# Code to find the nth// telephone number.using System;class GFG { // return the nth telephone number static int telephonenumber(int n) { // base step if (n == 0 || n == 1) return 1; // recursive step return telephonenumber(n - 1) + (n - 1) * telephonenumber(n - 2); } /* Driver program to test above function */ public static void Main() { int n = 6; Console.Write(telephonenumber(n)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP Program to find // the nth telephone number// return the nth // telephone numberfunction telephonenumber( $n){ // base step if ($n == 0 or $n == 1) return 1; // recursive step return telephonenumber($n - 1) + ($n - 1) * telephonenumber($n - 2);}// Driven Code$n = 6;echo telephonenumber($n) ;// This code is contributed by anuj_67.?> |
Javascript
<script>// Javascript Program to find // the nth telephone number.// return the nth telephone numberfunction telephonenumber(n){ // base step if (n == 0 || n == 1) return 1; // recursive step return telephonenumber(n - 1) + (n - 1) * telephonenumber(n - 2);}// Driven Programvar n = 6;document.write( telephonenumber(n));</script> |
Output:
76
Time complexity: O(2n)
Auxiliary Space: O(2n)
Below is efficient implementation of finding the nth telephone number using Dynamic Programming:
C++
// CPP Program to find the nth telephone number.#include <bits/stdc++.h>using namespace std;// return the nth telephone numberint telephonenumber(int n){ int dp[n + 1]; memset(dp, 0, sizeof(dp)); // Base case dp[0] = dp[1] = 1; // finding ith telephone number, where 2 <= i <= n. for (int i = 2; i <= n; i++) dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]; return dp[n];}// Driver Programint main(){ int n = 6; cout << telephonenumber(n) << endl; return 0;} |
Java
// JAVA Code to find nth Telephone Numberimport java.util.*;class GFG { // return the nth telephone number static int telephonenumber(int n) { int dp[] = new int[n + 1]; // Base case dp[0] = dp[1] = 1; // finding ith telephone number, // where 2 <= i <= n. for (int i = 2; i <= n; i++) dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]; return dp[n]; } /* Driver program to test above function */ public static void main(String[] args) { int n = 6; System.out.println(telephonenumber(n)); }}// This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 code to find the# nth telephone number.# return the nth telephone numberdef telephonenumber (n): dp = [0] * (n + 1) # Base case dp[0] = dp[1] = 1 # finding ith telephone number, # where 2 <= i <= n. for i in range(2, n + 1): dp[i] = dp[i - 1] + (i - 1) * dp[i - 2] return dp[n] # Driver Coden = 6print(telephonenumber(n))# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# Code to find nth Telephone Numberusing System;class GFG { // return the nth telephone number static int telephonenumber(int n) { int[] dp = new int[n + 1]; // Base case dp[0] = dp[1] = 1; // finding ith telephone number, // where 2 <= i <= n. for (int i = 2; i <= n; i++) dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]; return dp[n]; } /* Driver program to test above function */ public static void Main() { int n = 6; Console.Write(telephonenumber(n)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP Program to find // the nth telephone number.// return the nth telephone numberfunction telephonenumber($n){ $dp = array(); // Base case $dp[0] = $dp[1] = 1; // finding ith telephone number, // where 2 <= i <= n. for ( $i = 2; $i <= $n; $i++) $dp[$i] = $dp[$i - 1] + ($i - 1) * $dp[$i - 2]; return $dp[$n];}// Driver Code$n = 6;echo telephonenumber($n);// This code is contributed by anuj_67.?> |
Javascript
<script>// JavaScript Program to find nth Telephone Number // return the nth telephone number function telephonenumber(n) { let dp = []; // Base case dp[0] = dp[1] = 1; // finding ith telephone number, // where 2 <= i <= n. for (let i = 2; i <= n; i++) dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]; return dp[n]; }// Driver code let n = 6; document.write(telephonenumber(n)); // This code is contributed by sanjoy_62.</script> |
Output:
76
Time complexity: O(n)
Auxiliary space: O(n)
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