# HackerEarth Ways to a BST problem solution

In these HackerEarth Ways to a BST problem solution, You are given N distinct numbers in the form of an array Arri. You can form a Binary Search Tree(BST) by inserting them in the order they are given. Now find the number of different permutations of the array, so that the BST formed in each permutation is the same as that of the given array(including the given permutation). As the number can be large, output modulo (10 to power 9 + plus 7).

## HackerEarth Ways to a BST problem solution.

`#include <bits/stdc++.h>using namespace std;typedef long long ll;const ll maxN = 1000;const ll MOD = 1e9+7;struct node{  ll val, tot;  node *lc, *rc;};ll fact[maxN];node* init(ll val){  node *nd = (node*)malloc(sizeof(node));  nd->val = val;  nd->lc = nd->rc = NULL;  return nd;}ll mulMod(ll a, ll b){  return ((a%MOD) * (b%MOD))%MOD;}node* insert(node* r, ll val){  if(!r)  {    node *nd = init(val);//(node*)malloc(sizeof(node));    return nd;  }  if(val<=(r->val))    r->lc = insert(r->lc, val);  else    r->rc = insert(r->rc, val);  return r;}ll update(node *r){  if(!r)    return 0;  r->tot = 1 + update(r->lc) + update(r->rc);  return (r->tot);}ll power(ll a, ll b){  if(b==0)    return 1;  ll ret = power(mulMod(a, a), b/2)%MOD;  if(b%2)    ret = mulMod(a, ret);  return ret;}ll getInv(ll a){  return power(a, MOD-2);}ll getCnt(node *r){  if(!r)    return 1;  ll lCnt, rCnt, ans, x, y;  lCnt = getCnt(r->lc);  rCnt = getCnt(r->rc);  x = ((!(r->lc))?0:((r->lc)->tot));  y = ((!(r->rc))?0:((r->rc)->tot));  ans = mulMod(mulMod(getInv(fact[x]), getInv(fact[y])), mulMod(lCnt, rCnt));  ans = mulMod(ans, fact[x+y]);  return ans;}void preprocess(){  ll i;  fact[0] = 1;  for(i=1;i<=maxN;++i)  {    fact[i] = mulMod(i, fact[i-1]);  }}int main(){  ll t, n, m, i, j, ans, a;  preprocess();  scanf("%lld", &t);  while(t--)  {    scanf("%lld", &n);    ans = 0;    node *root = NULL;    for(i=0;i<n;++i)    {      scanf("%lld", &a);      root = insert(root, a);    }    update(root);    ans = getCnt(root);    printf("%lld\n", ans);  }  return 0;}`