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

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)
        {
            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<Integer> arr = new Vector<Integer>();
    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<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)
        {
            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<int> arr = new List<int>();
    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

<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)