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// 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<int> &nums) {
if (nums.size() < 3)
return 1;
vector<int> 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<int> &nums) { return (int)dfs(nums) - 1; }
};
int main() {
Solution obj;
vector<int> 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
}
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