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

• Difficulty Level : Medium
• Last Updated : 18 Aug, 2021

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 <bits/stdc++.h>
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<int>& 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<int> leftSubTree;

// Store the elements of the
// right subtree of BST
vector<int> 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<int> 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<Integer> 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<Integer> leftSubTree = new Vector<Integer>();

// Store the elements of the
// right subtree of BST
Vector<Integer> rightSubTree = new Vector<Integer>();

// 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)
{
}

// Push all the elements
// of the right subtree
else
{
}
}

// 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<Integer> arr = new Vector<Integer>();
for(int i : a)

// 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<int> 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<int> leftSubTree = new List<int>();

// Store the elements of the
// right subtree of BST
List<int> rightSubTree = new List<int>();

// 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)
{
}

// Push all the elements
// of the right subtree
else
{
}
}

// 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<int> arr = new List<int>();
foreach(int i in a)

// 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

```<script>

// JavaScript program to implement
// the above approach

// Function to precompute the
// factorial of 1 to N
function calculateFact(fact, N)
{
fact[0] = 1;
for (var i = 1; i < N; i++) {
fact[i] = fact[i - 1] * i;
}
}

// Function to get the value of nCr
function nCr(fact, N, R)
{
if (R > N)
return 0;

// nCr= fact(n)/(fact(r)*fact(n-r))
var res = parseInt(fact[N] / fact[R]);
res = parseInt(res / fact[N - R]);

return res;
}

// Function to count the number of ways
// to rearrange the array to obtain same BST
function countWays(arr, fact)
{
// Store the size of the array
var N = arr.length;

// Base case
if (N <= 2) {
return 1;
}

// Store the elements of the
// left subtree of BST
var leftSubTree = [];

// Store the elements of the
// right subtree of BST
var rightSubTree = [];

// Store the root node
var root = arr[0];

for (var i = 1; i < N; i++) {

// Push all the elements
// of the left subtree
if (arr[i] < root) {
leftSubTree.push(
arr[i]);
}

// Push all the elements
// of the right subtree
else {
rightSubTree.push(
arr[i]);
}
}

// Store the size of leftSubTree
var N1 = leftSubTree.length;

// Store the size of rightSubTree
var N2 = rightSubTree.length;

// Recurrence relation
var countLeft
= countWays(leftSubTree,
fact);
var countRight
= countWays(rightSubTree,
fact);

return nCr(fact, N - 1, N1)
* countLeft * countRight;
}

// Driver Code

var arr = [];
arr = [3, 4, 5, 1, 2];

// Store the size of arr
var N = arr.length;

// Store the factorial up to N
var fact = Array(N);

// Precompute the factorial up to N
calculateFact(fact, N);
document.write( countWays(arr, fact));

</script>  ```
Output:
`6`

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

My Personal Notes arrow_drop_up