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# Count permutations of given array that generates the same Binary Search Tree (BST)

Given an array, arr[] of size N consisting of elements from the range [1, N], that represents the order, in which the elements are inserted into a Binary Search Tree, the task is to count the number of ways to rearrange the given array to get the same BST.

Examples:

Input: arr[ ] ={3, 4, 5, 1, 2}
Output: 6
Explanation :
The permutations of the array which represent the same BST are:{{3, 4, 5, 1, 2}, {3, 1, 2, 4, 5}, {3, 1, 4, 2, 5}, {3, 1, 4, 5, 2}, {3, 4, 1, 2, 5}, {3, 4, 1, 5, 2}}. Therefore, the output is 6.

Input: arr[ ] ={2, 1, 6, 5, 4, 3}
Output: 5

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach: The idea is to first fix the root node and then recursively count the number of ways to rearrange the elements of the left subtree and the elements of the right subtree in such a way that the relative order within the elements of the left subtree and right subtree must be same. Here is the recurrence relation:

countWays(arr) = countWays(left) * countWays(right) * combinations(N, X).
left: Contains all the elements in the left subtree(Elements which are lesser than the root)
right: Contains all the elements in the right subtree(Elements which are greater than the root)
N = Total number of elements in arr[]
X = Total number of elements in left subtree.

Follow the steps below to solve the problem:

1. Fix the root node of BST, and store the elements of the left subtree(Elements which are lesser than arr[0]), say ctLeft[], and store the elements of the right subtree(Elements which are greater than arr[0]), say ctRight[].
2. To generate identical BST, maintain the relative order within the elements of left subtree and the right subtree.
3. Calculate the number of ways to rearrange the array to generate BST using the above-mentioned recurrence relation.

## C++

 // C++ program to implement // the above approach   #include using namespace std;   // Function to precompute the // factorial of 1 to N void calculateFact(int fact[], int N) {     fact[0] = 1;     for (long long int i = 1; i < N; i++) {         fact[i] = fact[i - 1] * i;     } }   // Function to get the value of nCr int nCr(int fact[], int N, int R) {     if (R > N)         return 0;       // nCr= fact(n)/(fact(r)*fact(n-r))     int res = fact[N] / fact[R];     res /= fact[N - R];       return res; }   // Function to count the number of ways // to rearrange the array to obtain same BST int countWays(vector& arr, int fact[]) {     // Store the size of the array     int N = arr.size();       // Base case     if (N <= 2) {         return 1;     }       // Store the elements of the     // left subtree of BST     vector leftSubTree;       // Store the elements of the     // right subtree of BST     vector rightSubTree;       // Store the root node     int root = arr[0];       for (int i = 1; i < N; i++) {           // Push all the elements         // of the left subtree         if (arr[i] < root) {             leftSubTree.push_back(                 arr[i]);         }           // Push all the elements         // of the right subtree         else {             rightSubTree.push_back(                 arr[i]);         }     }       // Store the size of leftSubTree     int N1 = leftSubTree.size();       // Store the size of rightSubTree     int N2 = rightSubTree.size();       // Recurrence relation     int countLeft         = countWays(leftSubTree,                     fact);     int countRight         = countWays(rightSubTree,                     fact);       return nCr(fact, N - 1, N1)            * countLeft * countRight; }   // Driver Code int main() {       vector arr;     arr = { 3, 4, 5, 1, 2 };       // Store the size of arr     int N = arr.size();       // Store the factorial up to N     int fact[N];       // Precompute the factorial up to N     calculateFact(fact, N);       cout << countWays(arr, fact);       return 0; }

## Java

 // Java program to implement // the above approach import java.util.*;   class GFG{   // Function to precompute the // factorial of 1 to N static void calculateFact(int fact[], int N) {     fact[0] = 1;     for(int i = 1; i < N; i++)     {         fact[i] = fact[i - 1] * i;     } }   // Function to get the value of nCr static int nCr(int fact[], int N, int R) {     if (R > N)         return 0;       // nCr= fact(n)/(fact(r)*fact(n-r))     int res = fact[N] / fact[R];     res /= fact[N - R];       return res; }   // Function to count the number of ways // to rearrange the array to obtain same BST static int countWays(Vector arr,                      int fact[]) {           // Store the size of the array     int N = arr.size();       // Base case     if (N <= 2)     {         return 1;     }       // Store the elements of the     // left subtree of BST     Vector leftSubTree = new Vector();       // Store the elements of the     // right subtree of BST     Vector rightSubTree = new Vector();       // Store the root node     int root = arr.get(0);       for(int i = 1; i < N; i++)     {                   // Push all the elements         // of the left subtree         if (arr.get(i) < root)         {             leftSubTree.add(arr.get(i));         }           // Push all the elements         // of the right subtree         else         {             rightSubTree.add(arr.get(i));         }     }       // Store the size of leftSubTree     int N1 = leftSubTree.size();       // Store the size of rightSubTree     int N2 = rightSubTree.size();       // Recurrence relation     int countLeft = countWays(leftSubTree,                               fact);     int countRight = countWays(rightSubTree,                                fact);       return nCr(fact, N - 1, N1) *              countLeft * countRight; }   // Driver Code public static void main(String[] args) {     int []a = { 3, 4, 5, 1, 2 };           Vector arr = new Vector();     for(int i : a)         arr.add(i);               // Store the size of arr     int N = a.length;       // Store the factorial up to N     int []fact = new int[N];       // Precompute the factorial up to N     calculateFact(fact, N);       System.out.print(countWays(arr, fact)); } }   // This code is contributed by Amit Katiyar

## Python3

 # Python3 program to implement # the above approach   # Function to precompute the # factorial of 1 to N def calculateFact(fact: list, N: int) -> None:       fact[0] = 1     for i in range(1, N):         fact[i] = fact[i - 1] * i   # Function to get the value of nCr def nCr(fact: list, N: int, R: int) -> int:       if (R > N):         return 0       # nCr= fact(n)/(fact(r)*fact(n-r))     res = fact[N] // fact[R]     res //= fact[N - R]       return res   # Function to count the number of ways # to rearrange the array to obtain same BST def countWays(arr: list, fact: list) -> int:       # Store the size of the array     N = len(arr)       # Base case     if (N <= 2):         return 1       # Store the elements of the     # left subtree of BST     leftSubTree = []       # Store the elements of the     # right subtree of BST     rightSubTree = []       # Store the root node     root = arr[0]       for i in range(1, N):           # Push all the elements         # of the left subtree         if (arr[i] < root):             leftSubTree.append(arr[i])           # Push all the elements         # of the right subtree         else:             rightSubTree.append(arr[i])       # Store the size of leftSubTree     N1 = len(leftSubTree)       # Store the size of rightSubTree     N2 = len(rightSubTree)       # Recurrence relation     countLeft = countWays(leftSubTree, fact)     countRight = countWays(rightSubTree, fact)       return (nCr(fact, N - 1, N1) *             countLeft * countRight)   # Driver Code if __name__ == '__main__':       arr = [ 3, 4, 5, 1, 2 ]       # Store the size of arr     N = len(arr)       # Store the factorial up to N     fact = [0] * N       # Precompute the factorial up to N     calculateFact(fact, N)           print(countWays(arr, fact))   # This code is contributed by sanjeev2552

## C#

 // C# program to implement // the above approach using System; using System.Collections.Generic;   class GFG{   // Function to precompute the // factorial of 1 to N static void calculateFact(int []fact, int N) {     fact[0] = 1;     for(int i = 1; i < N; i++)     {         fact[i] = fact[i - 1] * i;     } }   // Function to get the value of nCr static int nCr(int []fact, int N, int R) {     if (R > N)         return 0;       // nCr= fact(n)/(fact(r)*fact(n-r))     int res = fact[N] / fact[R];     res /= fact[N - R];       return res; }   // Function to count the number of ways // to rearrange the array to obtain same BST static int countWays(List arr,                      int []fact) {           // Store the size of the array     int N = arr.Count;       // Base case     if (N <= 2)     {         return 1;     }       // Store the elements of the     // left subtree of BST     List leftSubTree = new List();       // Store the elements of the     // right subtree of BST     List rightSubTree = new List();       // Store the root node     int root = arr[0];       for(int i = 1; i < N; i++)     {                   // Push all the elements         // of the left subtree         if (arr[i] < root)         {             leftSubTree.Add(arr[i]);         }           // Push all the elements         // of the right subtree         else         {             rightSubTree.Add(arr[i]);         }     }       // Store the size of leftSubTree     int N1 = leftSubTree.Count;       // Store the size of rightSubTree     int N2 = rightSubTree.Count;       // Recurrence relation     int countLeft = countWays(leftSubTree,                               fact);     int countRight = countWays(rightSubTree,                                fact);       return nCr(fact, N - 1, N1) *              countLeft * countRight; }   // Driver Code public static void Main(String[] args) {     int []a = { 3, 4, 5, 1, 2 };           List arr = new List();     foreach(int i in a)         arr.Add(i);               // Store the size of arr     int N = a.Length;       // Store the factorial up to N     int []fact = new int[N];       // Precompute the factorial up to N     calculateFact(fact, N);       Console.Write(countWays(arr, fact)); } }   // This code is contributed by Amit Katiyar

## Javascript



Output:

6

Time Complexity: O(N2)
Auxiliary Space: O(N)

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