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

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: 6Explanation :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:
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[].To generate identical BST, maintain the relative order within the elements of left subtree and the right subtree.Calculate the number of ways to rearrange the array to generate BST using the above-mentioned recurrence relation.
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[].
To generate identical BST, maintain the relative order within the elements of left subtree and the right subtree.
Calculate the number of ways to rearrange the array to generate BST using the above-mentioned recurrence relation.
C++
Java
Python3
C#
Javascript
// C++ program to implement// the above approach #include <bits/stdc++.h>using namespace std; // Function to precompute the// factorial of 1 to Nvoid 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 nCrint 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 BSTint 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 Codeint 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 program to implement// the above approachimport java.util.*; class GFG{ // Function to precompute the// factorial of 1 to Nstatic 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 nCrstatic 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 BSTstatic 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 Codepublic 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 program to implement# the above approach # Function to precompute the# factorial of 1 to Ndef 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 nCrdef 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 BSTdef 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 Codeif __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# program to implement// the above approachusing System;using System.Collections.Generic; class GFG{ // Function to precompute the// factorial of 1 to Nstatic 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 nCrstatic 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 BSTstatic 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 Codepublic 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
<script> // JavaScript program to implement// the above approach // Function to precompute the// factorial of 1 to Nfunction 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 nCrfunction 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 BSTfunction 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 arrvar N = arr.length; // Store the factorial up to Nvar fact = Array(N); // Precompute the factorial up to NcalculateFact(fact, N);document.write( countWays(arr, fact)); </script>
6
Time Complexity: O(N2)Auxiliary Space: O(N)
amit143katiyar
sanjeev2552
rutvik_56
kushjmehta
binomial coefficient
factorial
Permutation and Combination
Arrays
Binary Search Tree
Combinatorial
Mathematical
Recursion
Tree
Arrays
Mathematical
Recursion
Combinatorial
Binary Search Tree
Tree
factorial
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