// 1569. Number of Ways to Reorder Array to Get Same BST #include "leetcode.h" /* * given an array 'nums' that represents a permutation of integers from '1 to * n'. we are going to construct a binary search tree (BST) by inserting the * elements of 'nums' in order into an initially empty BST. find the number of * different ways to reorder 'nums' so that the constructed BST is identical to * that formed from the original array 'nums'. for example given 'nums = * [2,1,3]', we will have 2 as the root, 1 as a left child, and three as a right * child. the array [2,3,1] also yields the same BST, but '[3,2,1]' yields a * different BST. return the number of ways to reorder 'nums' such that the BST * formed is identical to the original BST formed from 'nums'. since the answer * may be large, return it modulo 10^9+7 */ class Solution { int mod = 1e+9 + 7; long l[10001][10001] = {0}; bool init = false; long combination(int n, int r) { if (init) return l[n][r]; l[0][0] = 1; for (int i = 1; i < 1001; i++) { l[i][0] = 1; for (int j = 1; j < i + 1; j++) l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]) % mod; } init = true; return l[n][r]; } long dfs(vector &nums) { if (nums.size() < 3) return 1; vector l, r; for (auto n : nums) { if (n > nums[0]) l.push_back(n); if (n < nums[0]) r.push_back(n); } long c = combination(l.size() + r.size(), l.size()); return c * dfs(l) % mod * dfs(r) % mod; } public: int numOfWays(vector &nums) { return dfs(nums) - 1; } }; int main() { Solution obj; vector n1 = {2, 1, 3}, n2 = {3, 4, 5, 1, 2}, n3 = {1, 2, 3}; printf("%d\n", obj.numOfWays(n1)); // expect: 1 printf("%d\n", obj.numOfWays(n2)); // expect: 5 printf("%d\n", obj.numOfWays(n3)); // expect: 0 }