**Permutations to Generate Binary Search Tree - Google Top Interview Questions**

### Problem Statement :

You are given a list of unique integers nums. We can create a binary search tree by taking each number in order and inserting it to an initially null binary search tree. Return the number of permutations of nums from which we can generate the same binary search tree. Mod the result by 10 ** 9 + 7. You can assume that the binary search tree does no rebalancing. Constraints 0 ≤ n ≤ 1,000 where n is the length of nums Example 1 Input nums = [2, 1, 3] Output 1 Explanation The generated tree looks like 2 / \ 1 3 And we can also generate this with [2, 3, 1].

### Solution :

` ````
Solution in C++ :
using ll = long long int;
ll mod = 1e9 + 7;
vector<long long> fact;
int N = 1005;
long long power(long long a, long long b) {
long long result = 1;
while (b > 0) {
if (b % 2) result = (result * a) % mod;
a = (a * a) % mod;
b /= 2;
}
return result;
}
long long nCr(int n, int r) {
ll num = fact[n];
ll den = (fact[r] * fact[n - r]) % mod;
ll ans = (num * power(den, mod - 2)) % mod;
return ans;
}
int dfs(vector<int> curr, int l, int r) {
int n = curr.size();
if (n <= 1) return 1;
ll start = curr[0];
int p = start - l + 1;
vector<int> left;
vector<int> right;
for (int i = 1; i < curr.size(); i++) {
if (curr[i] < start)
left.push_back(curr[i]);
else
right.push_back(curr[i]);
}
ll ans = nCr(left.size() + right.size(), right.size());
(ans *= dfs(left, l, start - 1)) %= mod;
(ans *= dfs(right, start + 1, r)) %= mod;
return ans;
}
int solve(vector<int>& nums) {
int n = nums.size();
fact.resize(N + 1, 1);
for (int i = 1; i <= N; i++) {
fact[i] = (fact[i - 1] * i) % mod;
}
long long ans = dfs(nums, 1, n);
int res = (ans - 1 + mod) % mod;
return (int)res;
return 0;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
private class Node {
int val;
int size;
Node left;
Node right;
public Node(int val) {
this.val = val;
this.size = 1;
this.left = null;
this.right = null;
}
}
private static final long mod = 1000000007;
public int solve(int[] nums) {
if (nums.length == 0) {
return 0;
}
Node root = null;
for (int num : nums) {
root = insertIntoBST(root, num);
}
int size = nums.length;
long[][] DP = new long[size + 1][size + 1];
for (int j = 0; j <= size; j++) {
DP[0][j] = 1;
}
for (int i = 0; i <= size; i++) {
DP[i][0] = 1;
}
for (int j = 1; j <= size; j++) {
for (int i = 1; i <= size; i++) {
DP[i][j] = ((DP[i - 1][j] + DP[i][j - 1]) % mod);
}
}
return (((int) getValidPermutationsCount(root, DP)) - 1);
}
private long getValidPermutationsCount(Node root, long[][] DP) {
if (root == null) {
return 1L;
}
int L = 0, R = 0;
if (root.left != null) {
L = root.left.size;
}
if (root.right != null) {
R = root.right.size;
}
long TL = getValidPermutationsCount(root.left, DP);
long TR = getValidPermutationsCount(root.right, DP);
return ((((TL * DP[L][R]) % mod) * TR) % mod);
}
private Node insertIntoBST(Node root, int val) {
if (root == null) {
return new Node(val);
}
root.size++;
if (val < root.val) {
root.left = insertIntoBST(root.left, val);
} else {
root.right = insertIntoBST(root.right, val);
}
return root;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
Mod = 1000000007
n = len(nums)
p = [1] * (n + 1)
for i in range(n):
p[i + 1] = p[i] * (i + 1)
def combi(x, y):
return p[x] * pow(p[y] * p[x - y] % Mod, Mod - 2, Mod) % Mod
def go(r):
if len(r) < 2:
return 1
x = r[0]
a, b = [], []
for i in range(1, len(r)):
if r[i] < x:
a += (r[i],)
else:
b += (r[i],)
return go(a) * go(b) % Mod * combi(len(r) - 1, len(a)) % Mod
return (go(nums) - 1 + Mod) % Mod
```

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