# Possible number of Trees having N vertex

Given an array arr[] of N positive integers. The task is to find the number of trees possible having N vertices such that the distance between vertex 1 and vertex i is arr[i]. The total number of such trees can be very large so return the answer with modulo 109 + 7.

Examples:

Input: arr[] = {0, 1, 1, 2}
Output: 2
Explanation:
Below is the tree: Input: arr[] = { 0, 3, 2, 1, 2, 2, 1 }
Output: 24

Naive Approach: Below are the steps:

1. If A1 !=0 and if element from A2 . . . AN is 0 then the tree will not be able to be made so in this case, the count of trees will be 0.
2. If we take the first node as root node then all nodes on level 1 must be a child of the root node and all nodes on level 2 must be a child of 1 level nodes, so there is only one possibility to place them in level-wise.
3. Now let x be the number of nodes on ith level and y is the number of nodes on (i + 1)th level so the number of possibilities of arranging nodes on level i and (i + 1) is xy.
4. Hence, the count of trees will be  Xi X(i + 1) where Xi Is the number of nodes in ith level.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` `const` `int` `mod = 1e9 + 7; ` ` `  `// Function to count the total number ` `// of trees possible ` `int` `NumberOfTrees(``int` `arr[], ``int` `N) ` `{ ` `    ``// Find the max element in the ` `    ``// given array ` `    ``int` `maxElement = *max_element(arr, arr + N); ` ` `  `    ``// Level array store the number of ` `    ``// nodes on level i initially all ` `    ``// values are zero ` `    ``int` `level[maxElement + 1] = { 0 }; ` `    ``for` `(``int` `i = 0; i < N; i++) { ` ` `  `        ``level[arr[i]]++; ` `    ``} ` ` `  `    ``// In this case tree can not be created ` `    ``if` `(arr != 0 || level != 1) { ` ` `  `        ``return` `0; ` `    ``} ` ` `  `    ``// To store the count of trees ` `    ``int` `ans = 1; ` ` `  `    ``// Iterate until maxElement ` `    ``for` `(``int` `i = 0; i < maxElement; i++) { ` ` `  `        ``// Calculate level[i]^level[i+1] ` `        ``for` `(``int` `j = 0; j < level[i + 1]; j++) { ` ` `  `            ``// Update the count of tree ` `            ``ans = (ans * level[i]) % mod; ` `        ``} ` `    ``} ` ` `  `    ``// Return the final count of trees ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `N = 7; ` ` `  `    ``// Given array arr[] ` `    ``int` `arr[] = { 0, 3, 2, 1, 2, 2, 1 }; ` ` `  `    ``// Function Call ` `    ``cout << NumberOfTrees(arr, N); ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `static` `int` `mod = (``int``)(1e9 + ``7``); ` ` `  `// Function to count the total number ` `// of trees possible ` `static` `int` `NumberOfTrees(``int` `arr[], ``int` `N) ` `{ ` `     `  `    ``// Find the max element in the ` `    ``// given array ` `    ``int` `maxElement = Arrays.stream(arr).max().getAsInt(); ` ` `  `    ``// Level array store the number of ` `    ``// nodes on level i initially all ` `    ``// values are zero ` `    ``int` `level[] = ``new` `int``[maxElement + ``1``]; ` `    ``for``(``int` `i = ``0``; i < N; i++)  ` `    ``{ ` `        ``level[arr[i]]++; ` `    ``} ` ` `  `    ``// In this case tree can not be created ` `    ``if` `(arr[``0``] != ``0` `|| level[``0``] != ``1``) ` `    ``{ ` `        ``return` `0``; ` `    ``} ` ` `  `    ``// To store the count of trees ` `    ``int` `ans = ``1``; ` ` `  `    ``// Iterate until maxElement ` `    ``for``(``int` `i = ``0``; i < maxElement; i++) ` `    ``{ ` `         `  `        ``// Calculate level[i]^level[i+1] ` `        ``for` `(``int` `j = ``0``; j < level[i + ``1``]; j++) ` `        ``{ ` `             `  `            ``// Update the count of tree ` `            ``ans = (ans * level[i]) % mod; ` `        ``} ` `    ``} ` ` `  `    ``// Return the final count of trees ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``7``; ` ` `  `    ``// Given array arr[] ` `    ``int` `arr[] = { ``0``, ``3``, ``2``, ``1``, ``2``, ``2``, ``1` `}; ` ` `  `    ``// Function call ` `    ``System.out.print(NumberOfTrees(arr, N)); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program for the above approach ` `mod ``=` `int``(``1e9` `+` `7``) ` ` `  `# Function to count the total number ` `# of trees possible ` `def` `NumberOfTrees(arr, N): ` `     `  `    ``# Find the max element in the ` `    ``# given array ` `    ``maxElement ``=` `max``(arr); ` ` `  `    ``# Level array store the number of ` `    ``# Nodes on level i initially all ` `    ``# values are zero ` `    ``level ``=` `[``0``] ``*` `(maxElement ``+` `1``); ` `    ``for` `i ``in` `range``(N): ` `        ``level[arr[i]] ``+``=` `1``; ` ` `  `    ``# In this case tree can not be created ` `    ``if` `(arr[``0``] !``=` `0` `or` `level[``0``] !``=` `1``): ` `        ``return` `0``; ` ` `  `    ``# To store the count of trees ` `    ``ans ``=` `1``; ` ` `  `    ``# Iterate until maxElement ` `    ``for` `i ``in` `range``(maxElement): ` ` `  `        ``# Calculate level[i]^level[i+1] ` `        ``for` `j ``in` `range``(level[i ``+` `1``]): ` `             `  `            ``# Update the count of tree ` `            ``ans ``=` `(ans ``*` `level[i]) ``%` `mod; ` ` `  `    ``# Return the final count of trees ` `    ``return` `ans; ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``N ``=` `7``; ` ` `  `    ``# Given array arr ` `    ``arr ``=` `[ ``0``, ``3``, ``2``, ``1``, ``2``, ``2``, ``1` `]; ` ` `  `    ``# Function call ` `    ``print``(NumberOfTrees(arr, N)); ` ` `  `# This code is contributed by shikhasingrajput `

## C#

 `// C# program for the above approach ` `using` `System; ` `using` `System.Linq; ` `class` `GFG{ ` `     `  `static` `int` `mod = (``int``)(1e9 + 7); ` ` `  `// Function to count the total number ` `// of trees possible ` `static` `int` `NumberOfTrees(``int` `[]arr, ``int` `N) ` `{ ` `     `  `    ``// Find the max element in the ` `    ``// given array ` `    ``int` `maxElement = arr.Max(); ` ` `  `    ``// Level array store the number of ` `    ``// nodes on level i initially all ` `    ``// values are zero ` `    ``int` `[]level = ``new` `int``[maxElement + 1]; ` `    ``for``(``int` `i = 0; i < N; i++)  ` `    ``{ ` `        ``level[arr[i]]++; ` `    ``} ` ` `  `    ``// In this case tree can not be created ` `    ``if` `(arr != 0 || level != 1) ` `    ``{ ` `        ``return` `0; ` `    ``} ` ` `  `    ``// To store the count of trees ` `    ``int` `ans = 1; ` ` `  `    ``// Iterate until maxElement ` `    ``for``(``int` `i = 0; i < maxElement; i++) ` `    ``{ ` `         `  `        ``// Calculate level[i]^level[i+1] ` `        ``for` `(``int` `j = 0; j < level[i + 1]; j++) ` `        ``{ ` `             `  `            ``// Update the count of tree ` `            ``ans = (ans * level[i]) % mod; ` `        ``} ` `    ``} ` ` `  `    ``// Return the readonly count of trees ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `N = 7; ` ` `  `    ``// Given array []arr ` `    ``int` `[]arr = { 0, 3, 2, 1, 2, 2, 1 }; ` ` `  `    ``// Function call ` `    ``Console.Write(NumberOfTrees(arr, N)); ` `} ` `} ` ` `  `// This code is contributed by gauravrajput1`

Output:

```24
```

Time Complexity: O(N*K) where K is the number of vertices in a level.
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by optimizing the value of (ans*level[i]) using Modular Exponentiation. Below are the steps:

1. If the exponent is odd, make it even by subtracting 1 from it and multiply the base with the answer.
2. If the exponent is even, divide the exponent by 2 and square the base.
3. Return 1 when the exponent becomes 0.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` `const` `int` `mod = 1e9 + 7; ` ` `  `// Function that finds the value of x^y ` `// using Modular Exponentiation ` `int` `power(``int` `x, ``int` `y) ` `{ ` `    ``// Base Case ` `    ``if` `(y == 0) ` `        ``return` `1; ` ` `  `    ``int` `p = power(x, y / 2) % mod; ` ` `  `    ``p = (p * p) % mod; ` ` `  `    ``// If y is odd, multiply ` `    ``// x with result ` `    ``if` `(y & 1) ` `        ``p = (x * p) % mod; ` ` `  `    ``// Return the value ` `    ``return` `p; ` `} ` ` `  `// Function that counts the total ` `// number of trees possible ` `int` `NumberOfTrees(``int` `arr[], ``int` `N) ` `{ ` ` `  `    ``// Find the max element in array ` `    ``int` `maxElement ` `        ``= *max_element(arr, arr + N); ` ` `  `    ``// Level array store the number nodes ` `    ``// on level i initially all values are 0 ` `    ``int` `level[maxElement + 1] = { 0 }; ` ` `  `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``level[arr[i]]++; ` `    ``} ` ` `  `    ``// In this case tree can not be created ` `    ``if` `(arr != 0 || level != 1) { ` ` `  `        ``return` `0; ` `    ``} ` ` `  `    ``int` `ans = 1; ` ` `  `    ``for` `(``int` `i = 0; i < maxElement; i++) { ` ` `  `        ``// Calculate level[i]^level[i+1] ` `        ``ans = (ans * power(level[i], ` `                           ``level[i + 1])) ` `              ``% mod; ` `    ``} ` ` `  `    ``// Return the final count ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `N = 7; ` ` `  `    ``// Given array arr[] ` `    ``int` `arr[] = { 0, 3, 2, 1, 2, 2, 1 }; ` ` `  `    ``// Function Call ` `    ``cout << NumberOfTrees(arr, N); ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `static` `int` `mod = (``int``)1e9 + ``7``; ` ` `  `// Function that finds the value of x^y ` `// using Modular Exponentiation ` `static` `int` `power(``int` `x, ``int` `y) ` `{ ` `     `  `    ``// Base Case ` `    ``if` `(y == ``0``) ` `        ``return` `1``; ` ` `  `    ``int` `p = power(x, y / ``2``) % mod; ` ` `  `    ``p = (p * p) % mod; ` ` `  `    ``// If y is odd, multiply ` `    ``// x with result ` `    ``if` `((y & ``1``) != ``0``) ` `        ``p = (x * p) % mod; ` ` `  `    ``// Return the value ` `    ``return` `p; ` `} ` ` `  `// Function that counts the total ` `// number of trees possible ` `static` `int` `NumberOfTrees(``int` `arr[], ``int` `N) ` `{ ` ` `  `    ``// Find the max element in array ` `    ``int` `maxElement = Arrays.stream(arr).max().getAsInt(); ` ` `  `    ``// Level array store the number nodes ` `    ``// on level i initially all values are 0 ` `    ``int` `[]level = ``new` `int``[maxElement + ``1``]; ` ` `  `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `        ``level[arr[i]]++; ` `    ``} ` ` `  `    ``// In this case tree can not be created ` `    ``if` `(arr[``0``] != ``0` `|| level[``0``] != ``1``) ` `    ``{ ` `        ``return` `0``; ` `    ``} ` `     `  `    ``int` `ans = ``1``; ` ` `  `    ``for``(``int` `i = ``0``; i < maxElement; i++)  ` `    ``{ ` `        ``// Calculate level[i]^level[i+1] ` `        ``ans = (ans * power(level[i], ` `                           ``level[i + ``1``])) % mod; ` `    ``} ` ` `  `    ``// Return the final count ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``7``; ` ` `  `    ``// Given array arr[] ` `    ``int` `arr[] = { ``0``, ``3``, ``2``, ``1``, ``2``, ``2``, ``1` `}; ` ` `  `    ``// Function call ` `    ``System.out.print(NumberOfTrees(arr, N)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar  `

## Python3

 `# Python3 program for the above approach  ` `mod ``=` `int``(``1e9` `+` `7``) ` ` `  `# Function that finds the value of x^y ` `# using Modular Exponentiation ` `def` `power(x, y): ` ` `  `    ``# Base Case ` `    ``if``(y ``=``=` `0``): ` `        ``return` `1` ` `  `    ``p ``=` `power(x, y ``/``/` `2``) ``%` `mod ` ` `  `    ``p ``=` `(p ``*` `p) ``%` `mod ` ` `  `    ``# If y is odd, multiply ` `    ``# x with result ` `    ``if``(y & ``1``): ` `        ``p ``=` `(x ``*` `p) ``%` `mod ` ` `  `    ``# Return the value ` `    ``return` `p ` ` `  `# Function that counts the total ` `# number of trees possible ` `def` `NumberOfTrees(arr, N): ` ` `  `    ``# Find the max element in array ` `    ``maxElement ``=` `max``(arr) ` ` `  `    ``# Level array store the number nodes ` `    ``# on level i initially all values are 0 ` `    ``level ``=` `[``0``] ``*` `(maxElement ``+` `1``) ` ` `  `    ``for` `i ``in` `range``(N): ` `        ``level[arr[i]] ``+``=` `1` ` `  `    ``# In this case tree can not be created ` `    ``if``(arr[``0``] !``=` `0` `or` `level[``0``] !``=` `1``): ` `        ``return` `0` ` `  `    ``ans ``=` `1` ` `  `    ``for` `i ``in` `range``(maxElement): ` ` `  `        ``# Calculate level[i]^level[i+1] ` `        ``ans ``=` `(ans ``*` `power(level[i], ` `                           ``level[i ``+` `1``])) ``%` `mod ` ` `  `    ``# Return the final count ` `    ``return` `ans ` ` `  `# Driver Code ` `N ``=` `7` ` `  `# Given Queries ` `arr ``=` `[ ``0``, ``3``, ``2``, ``1``, ``2``, ``2``, ``1` `] ` ` `  `# Function call ` `print``(NumberOfTrees(arr, N)) ` ` `  `# This code is contributed by Shivam Singh `

## C#

 `// C# program for the above approach ` `using` `System; ` `using` `System.Linq; ` ` `  `class` `GFG{ ` `     `  `static` `int` `mod = (``int``)1e9 + 7; ` ` `  `// Function that finds the value of x^y ` `// using Modular Exponentiation ` `static` `int` `power(``int` `x, ``int` `y) ` `{ ` `     `  `    ``// Base Case ` `    ``if` `(y == 0) ` `        ``return` `1; ` ` `  `    ``int` `p = power(x, y / 2) % mod; ` ` `  `    ``p = (p * p) % mod; ` ` `  `    ``// If y is odd, multiply ` `    ``// x with result ` `    ``if` `((y & 1) != 0) ` `        ``p = (x * p) % mod; ` ` `  `    ``// Return the value ` `    ``return` `p; ` `} ` ` `  `// Function that counts the total ` `// number of trees possible ` `static` `int` `NumberOfTrees(``int` `[]arr, ``int` `N) ` `{ ` ` `  `    ``// Find the max element in array ` `    ``int` `maxElement = arr.Max(); ` ` `  `    ``// Level array store the number nodes ` `    ``// on level i initially all values are 0 ` `    ``int` `[]level = ``new` `int``[maxElement + 1]; ` ` `  `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{ ` `        ``level[arr[i]]++; ` `    ``} ` ` `  `    ``// In this case tree can not be created ` `    ``if` `(arr != 0 || level != 1) ` `    ``{ ` `        ``return` `0; ` `    ``} ` `     `  `    ``int` `ans = 1; ` ` `  `    ``for``(``int` `i = 0; i < maxElement; i++)  ` `    ``{ ` `         `  `        ``// Calculate level[i]^level[i+1] ` `        ``ans = (ans * power(level[i], ` `                           ``level[i + 1])) % mod; ` `    ``} ` ` `  `    ``// Return the readonly count ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `N = 7; ` ` `  `    ``// Given array []arr ` `    ``int` `[]arr = { 0, 3, 2, 1, 2, 2, 1 }; ` ` `  `    ``// Function call ` `    ``Console.Write(NumberOfTrees(arr, N)); ` `} ` `}  ` ` `  `// This code is contributed by 29AjayKumar`

Output:

```24
```

Time Complexity: O(N*log K) where K is the number of vertices in a level.
Auxiliary Space: O(N)

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