Given an integer **L** which is the sum of degrees of all the vertices of some tree. The task is to find the count of all such distinct trees (labeled trees). Two trees are distinct if they have at least a single different edge.**Examples:**

Input:L = 2Output:1Input:L = 6Output:16

**Simple Solution**: A simple solution is to find the number of nodes of the tree which has sum of degrees of all vertices as **L**. Number of nodes in such a tree is **n = (L / 2 + 1)** as described in this article.

Now the solution is to form all the labeled trees which can be formed using n nodes. This approach is quite complex and for larger values of n it is not possible to find out the number of trees using this process.**Efficient Solution**: An efficient solution is to find the number of nodes using Cayley’s formula which states that there are **n ^{(n – 2)}** trees with n labeled vertices. So the time complexity of the code now reduces to

**O(n)**which can be further reduced to

**O(logn)**using modular exponentiation.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <iostream>` `using` `namespace` `std;` `#define ll long long int` `// Iterative Function to calculate (x^y) in O(log y)` `ll power(` `int` `x, ll y)` `{` ` ` `// Initialize result` ` ` `ll res = 1;` ` ` `while` `(y > 0) {` ` ` `// If y is odd, multiply x with result` ` ` `if` `(y & 1)` ` ` `res = (res * x);` ` ` `// y must be even now` ` ` `// y = y / 2` ` ` `y = y >> 1;` ` ` `x = (x * x);` ` ` `}` ` ` `return` `res;` `}` `// Function to return the count` `// of required trees` `ll solve(` `int` `L)` `{` ` ` `// number of nodes` ` ` `int` `n = L / 2 + 1;` ` ` `ll ans = power(n, n - 2);` ` ` `// Return the result` ` ` `return` `ans;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `L = 6;` ` ` `cout << solve(L);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.io.*;` `class` `GFG` `{` ` ` `// Iterative Function to calculate (x^y) in O(log y)` `static` `long` `power(` `int` `x, ` `long` `y)` `{` ` ` `// Initialize result` ` ` `long` `res = ` `1` `;` ` ` `while` `(y > ` `0` `)` ` ` `{` ` ` `// If y is odd, multiply x with result` ` ` `if` `(y==` `1` `)` ` ` `res = (res * x);` ` ` `// y must be even now` ` ` `// y = y / 2` ` ` `y = y >> ` `1` `;` ` ` `x = (x * x);` ` ` `}` ` ` `return` `res;` `}` `// Function to return the count` `// of required trees` `static` `long` `solve(` `int` `L)` `{` ` ` `// number of nodes` ` ` `int` `n = L / ` `2` `+ ` `1` `;` ` ` `long` `ans = power(n, n - ` `2` `);` ` ` `// Return the result` ` ` `return` `ans;` `}` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` `int` `L = ` `6` `;` ` ` `System.out.println (solve(L));` `}` `}` `// This code is contributed by ajit.` |

## Python3

` ` `# Python implementation of the approach` `# Iterative Function to calculate (x^y) in O(log y)` `def` `power(x, y):` ` ` `# Initialize result` ` ` `res ` `=` `1` `;` ` ` `while` `(y > ` `0` `):` ` ` `# If y is odd, multiply x with result` ` ` `if` `(y ` `%` `2` `=` `=` `1` `):` ` ` `res ` `=` `(res ` `*` `x);` ` ` `# y must be even now` ` ` `#y = y / 2` ` ` `y ` `=` `int` `(y) >> ` `1` `;` ` ` `x ` `=` `(x ` `*` `x);` ` ` `return` `res;` `# Function to return the count` `# of required trees` `def` `solve(L):` ` ` ` ` `# number of nodes` ` ` `n ` `=` `L ` `/` `2` `+` `1` `;` ` ` `ans ` `=` `power(n, n ` `-` `2` `);` ` ` `# Return the result` ` ` `return` `int` `(ans);` `L ` `=` `6` `;` `print` `(solve(L));` `# This code has been contributed by 29AjayKumar` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` `// Iterative Function to calculate (x^y) in O(log y)` `static` `long` `power(` `int` `x, ` `long` `y)` `{` ` ` `// Initialize result` ` ` `long` `res = 1;` ` ` `while` `(y > 0)` ` ` `{` ` ` `// If y is odd, multiply x with result` ` ` `if` `(y == 1)` ` ` `res = (res * x);` ` ` `// y must be even now` ` ` `// y = y / 2` ` ` `y = y >> 1;` ` ` `x = (x * x);` ` ` `}` ` ` `return` `res;` `}` `// Function to return the count` `// of required trees` `static` `long` `solve(` `int` `L)` `{` ` ` `// number of nodes` ` ` `int` `n = L / 2 + 1;` ` ` `long` `ans = power(n, n - 2);` ` ` `// Return the result` ` ` `return` `ans;` `}` `// Driver code` `static` `public` `void` `Main ()` `{` ` ` `int` `L = 6;` ` ` `Console.WriteLine(solve(L));` `}` `}` `// This code is contributed by Tushil.` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Iterative Function to calculate (x^y) in O(log y)` `function` `power(x, y)` `{` ` ` `// Initialize result` ` ` `var` `res = 1;` ` ` `while` `(y > 0) {` ` ` `// If y is odd, multiply x with result` ` ` `if` `(y & 1)` ` ` `res = (res * x);` ` ` `// y must be even now` ` ` `// y = y / 2` ` ` `y = y >> 1;` ` ` `x = (x * x);` ` ` `}` ` ` `return` `res;` `}` `// Function to return the count` `// of required trees` `function` `solve(L)` `{` ` ` `// number of nodes` ` ` `var` `n = L / 2 + 1;` ` ` `var` `ans = power(n, n - 2);` ` ` `// Return the result` ` ` `return` `ans;` `}` `// Driver code` `var` `L = 6;` `document.write( solve(L));` `// This code is contributed by rutvik_56.` `</script>` |

**Output:**

16

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**