Number of Ways to Reorder Array to Get Same BST

class Solution {
public:
    int countWays(vector<int>& nums) {
        int n = nums.size();
        // Generate Pascal's Triangle for nCk calculation
        pascalTriangle.resize(n + 1);
        for(int i = 0; i <= n; ++i) {
            pascalTriangle[i] = vector<long long>(i + 1, 1);
            for(int j = 1; j < i; ++j) {
                pascalTriangle[i][j] = (pascalTriangle[i - 1][j - 1] + pascalTriangle[i - 1][j]) % MODULO;
            }
        }
            
        return (computeWays(nums) - 1) % MODULO;
    }
        
private:
    vector<vector<long long>> pascalTriangle;
    const long long MODULO = 1e9 + 7;
        
    long long computeWays(vector<int> &nums) {
        int n = nums.size();
        if(n <= 2) {
            return 1;
        }
    
        vector<int> left, right;
        for(int i = 1; i < n; ++i) {
            if (nums[i] < nums[0]) {
                left.push_back(nums[i]);
            } else {
                right.push_back(nums[i]);
            }
        }
    		
        long long waysLeft = computeWays(left) % MODULO;
        long long waysRight = computeWays(right) % MODULO;
    		
        return (((waysLeft * waysRight) % MODULO) * pascalTriangle[n - 1][left.size()]) % MODULO;
    }
};

The solution in C++ revolves around finding the number of ways to reorder an array to obtain the same Binary Search Tree (BST) structure. Here is the breakdown of the implemented approach:

1. Calculate combinations using Pascal's Triangle: Initialization of a matrix (pascalTriangle) to store binomial coefficients, which will help in computing combinations efficiently. This matrix is built dynamically using the properties of Pascal's Triangle for nCk calculation.

2. Define a primary function countWays to count all possible reorders. This function reduces the result modulo 1e9 + 7 (MODULO) and subtracts one to account for the initial configuration.

3. Implement a recursive helper function computeWays:

   • It determines the base case when the size of the subarray (nums) is less than or equal to two, in which there is only one arrangement possible.
   • Splits the tree's root elements into left and right subtrees.
   • Recursively calculates the number of ways for left and right subtrees.
   • Uses the modular arithmetic rules to ensure numbers do not overflow and remain within bounds.

4. The multiplication of the number of ways for left and right subtrees is combined using the respective binomial coefficient from the precomputed pascalTriangle. This multiplication gives the total ways to merge the left and right subtrees while maintaining the original BST structure.

5. Return the result from the parent function after adjusting it with the MODULO.

This approach efficiently leverages dynamic programming and combinatorial mathematics to solve the problem, ensuring optimal performance on larger datasets.

class Solution {
    private long modulus = (long)1e9 + 7;
    private long[][] pascalsTriangle;
    
    public int numberOfWays(int[] elements) {
        int n = elements.length;
            
        // Initializing Pascal's triangle.
        pascalsTriangle = new long[n][n];
        for (int i = 0; i < n; ++i) {
            pascalsTriangle[i][0] = pascalsTriangle[i][i] = 1;
        }
        for (int i = 2; i < n; i++) {
            for (int j = 1; j < i; j++) {
                pascalsTriangle[i][j] = (pascalsTriangle[i - 1][j - 1] + pascalsTriangle[i - 1][j]) % modulus;
            }
        }
        List<Integer> elementsList = Arrays.stream(elements).boxed().collect(Collectors.toList());
        return (int)((findWays(elementsList) - 1) % modulus);
    }
    
    private long findWays(List<Integer> elementsList) {
        int n = elementsList.size();
        if (n < 3) {
            return 1;
        }
    
        List<Integer> smaller = new ArrayList<>();
        List<Integer> larger = new ArrayList<>();
        for (int i = 1; i < n; ++i) {
            if (elementsList.get(i) < elementsList.get(0)) {
                smaller.add(elementsList.get(i));
            } else {
                larger.add(elementsList.get(i));
            }
        }
        long waysSmaller = findWays(smaller) % modulus;
        long waysLarger = findWays(larger) % modulus;
    
        return (((waysSmaller * waysLarger) % modulus) * pascalsTriangle[n - 1][smaller.size()]) % modulus;
    }
}

This Java solution tackles the problem of counting the number of ways to reorder an array so that it forms the same Binary Search Tree (BST) by organizing the task into a series of combinatorial steps. The solution relies heavily on combinatorics and the use of Pascal’s triangle to find the combinatorial coefficients efficiently.

The provided code proceeds as follows:

• Initialize a 2D array named pascalsTriangle to store binomial coefficients which will be used later to compute the number of combinations possible at each step while splitting the tree.
• In the main function numberOfWays, it first represents the input array as a list of integers, then calls findWays to recursively calculate the number of valid reorderings.
• In the recursive method findWays, the code segments:
  • Identifies the root of the current subtree (first element of the list).
  • Distributes the remaining elements into smaller and larger lists depending on whether elements are less than or greater than the root, respectively.
  • Computes the number of ways to rearrange both the smaller and larger list recursively.
  • Combines the number of ways to reorder elements in both subtrees using the combinatorial values from pascalsTriangle. This combination considers how many ways the smaller and larger subsets can be interleaved around the root while maintaining the BST property.
• The computation considers modulo 1e9 + 7 to avoid overflow issues with large numbers.

Overall, this solution breaks down the array based on BST properties and uses dynamic programming with a precomputed Pascal's triangle for efficient combinatorial calculations. This approach allows the solution to handle potentially large arrays by addressing factorial growth using modular arithmetic.

class Solution:
    def calculateWays(self, numbers: List[int]) -> int:
        modulus = 1000000007
            
        def explore(nums):
            n = len(nums)
            if n <= 2:
                return 1
            left_subtree = [x for x in nums if x < nums[0]]
            right_subtree = [x for x in nums if x > nums[0]]
            return (explore(left_subtree) * explore(right_subtree) * comb(n - 1, len(left_subtree))) % modulus
            
        return (explore(numbers) - 1) % modulus

This summary explains the solution to the problem of calculating the number of ways to reorder an array such that the resulting Binary Search Tree (BST) remains the same as the BST formed by the original array.

The solution is implemented in Python. The main functionality resides within the calculateWays method of the Solution class. The method takes a list of integers numbers as input, representing the array.

• Inside calculateWays, a nested function explore is defined. It recursively determines the number of ways the subtree arrays can be reordered.
• The approach involves dividing the input array into two subtrees - the left subtree containing elements less than the root (first element of the array) and the right subtree containing elements greater than the root.
• Base case for recursion is defined when the length of the list is 2 or less, in which case there is only one way to arrange the elements (either one element or the elements are the same as the BST).
• The Binomial coefficient, represented by comb(n - 1, len(left_subtree)), calculates the number of ways to choose positions for the left subtree elements in the reordered array.
• The result from explore function is adjusted with modulo 1000000007 to prevent overflow and ensure the result fits within standard data ranges.

Finally, the total number of ways calculated by explore is reduced by one (explore(numbers) - 1), as it includes the original order of the array itself, which should be excluded according to the problem requirements. The final result is also taken modulo 1000000007 to keep it within the range.

This recursive algorithm efficiently calculates the number of valid reorderings using memoization provided by recursion and combinatorial calculations with the binomial coefficient, ensuring an optimized and effective solution to the problem.