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++

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// C++ program for the above approach
#include <bits/stdc++.h>
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] != 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
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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] != 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 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

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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++

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// C++ program for the above approach
#include <bits/stdc++.h>
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] != 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
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;
}

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Java

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// 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 

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Python3

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# 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

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C#

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// 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] != 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 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

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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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