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// 1569. Number of Ways to Reorder Array to Get Same BST
#include <stdio.h>
#include <stdlib.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
*/
long long factorial[1024], factorial_inv[1024], inv[1024];
int mod = 1e9 + 7;
typedef struct {
int val;
int left_cnt, right_cnt;
struct node *left, *right;
} node;
void insert(node *root, int val) {
if (root->val > val) {
if (root->left == NULL) {
node *new = (node *)malloc(sizeof(node));
new->val = val;
new->left = NULL;
new->right = NULL;
new->left_cnt = 0;
new->right_cnt = 0;
root->left = new;
root->left_cnt++;
return;
} else {
root->left_cnt++;
insert(root->left, val);
}
} else if (root->val < val) {
if (root->right == NULL) {
node *new = (node *)malloc(sizeof(node));
new->val = val;
new->left = NULL;
new->right = NULL;
new->left_cnt = 0;
new->right_cnt = 0;
root->right = new;
root->right_cnt++;
return;
} else {
root->right_cnt++;
insert(root->right, val);
}
}
}
void free_node(node *root) {
if (root->left)
free_node(root->left);
if (root->right)
free_node(root->right);
free(root);
}
long long ways(node *root) {
if (!root)
return 1;
if (!root->left && !root->right)
return 1;
if (!root->left)
return ways(root->right) % mod;
if (!root->right)
return ways(root->left) % mod;
long long tmp = factorial[root->left_cnt + root->right_cnt];
tmp = (tmp * factorial_inv[root->left_cnt]) % mod;
tmp = (tmp * factorial_inv[root->right_cnt]) % mod;
return ((tmp * ways(root->left) % mod) * ways(root->right)) % mod;
}
int numOfWays(int *nums, int nums_size) {
factorial[0] = factorial[1] = 1;
factorial_inv[0] = factorial_inv[1] = 1;
inv[0] = inv[1] = 1;
for (int i = 2; i < 1024; i++) {
factorial[i] = (factorial[i - 1] * i) % mod;
inv[i] = mod - (mod / i * inv[mod % i]) % mod;
factorial_inv[i] = (factorial_inv[i - 1] * inv[i]) % mod;
}
node *root = (node *)malloc(sizeof(node));
root->val = nums[0];
root->left = NULL;
root->right = NULL;
root->left_cnt = 0;
root->right_cnt = 0;
for (int i = 1; i < nums_size; i++)
insert(root, nums[i]);
int ans = (int)ways(root) - 1;
free_node(root);
return ans;
}
int main() {
int n1[] = {2, 1, 3};
int n2[] = {3, 4, 5, 1, 2};
int n3[] = {1, 2, 3};
printf("%d\n", numOfWays(n1, 3)); // expect: 1
printf("%d\n", numOfWays(n2, 5)); // expect: 5
printf("%d\n", numOfWays(n3, 3)); // expect: 0
}
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