Added solution to Count Permutations of BST

DynamicProgramming/CountPermutationsOfBST.cpp

@@ -0,0 +1,97 @@
+// https://www.interviewbit.com/problems/count-permutations-of-bst/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long
+
+const ll MOD = 1e9+7;
+const ll INF = 1e18;
+const ll N = 2e2+5;
+
+#define S second
+#define F first
+#define vt vector
+#define rep(i, st, en) for(ll i = (st); i < (en); ++i)
+#define repr(i, en, st) for(ll i = (en); i >= (st); --i)
+
+int Solution::cntPermBST(int a, int b) {
+
+	if (b >= a) {
+		return 0;
+	}
+
+
+	vt<ll> fac(N), ifac(N);
+	fac[0] = 1;
+	rep(i, 1, N) {
+		fac[i] = (fac[i-1]*i)%MOD;
+	}
+
+	auto bpow = [](ll x, ll p, ll m) -> ll {
+		ll r = 1;
+		while (p) {
+			if (p&1) {
+				r = (r*x)%m;
+			}
+			p >>= 1;
+			x = x*x%m;
+		}
+		return r;
+	};
+
+	ifac[N-1] = bpow(fac[N-1], MOD-2, MOD);
+	repr(i, N-2, 0) {
+		ifac[i] = (ifac[i+1]*(i+1))%MOD;
+	}
+
+	auto ncr = [&](ll n, ll r, ll m) -> ll {
+		if (r > n) {
+			return 0;
+		}
+		ll x = fac[n]*ifac[r]%m*ifac[n-r]%m;
+		return x;
+	};
+
+
+	vt<vt<ll>> dp(a+1, vt<ll>(a+1, 0)); // length of perm, height of tree
+	vt<vt<ll>> ps(a+1, vt<ll>(a+1, 0)); 
+	// stores number of valid such permutations for insertion
+
+	dp[1][0] = 1;
+	dp[0][0] = 1;
+
+	rep(i, 0, a+1) {
+		ps[1][i] = ps[0][i] = 1; // presum
+	}
+
+	rep(i, 2, a+1) {
+		rep(j, 1, i) {// height can be from 1 to i-1
+			// split can be made on a point 1 to i
+
+			// add 0 length tree case first
+			// 0 on left i-1 on right
+			dp[i][j] = (dp[i][j] + 2*dp[i-1][j-1])%MOD;
+
+			rep(k, 2, i) {
+				ll add = ps[k-1][j-1]*ps[i-k][j-1]%MOD;
+				ll sub = 0;
+				if (j-2 >= 0) {
+					sub = ps[k-1][j-2]*ps[i-k][j-2]%MOD;
+				}
+				ll c = (add - sub + MOD) % MOD;
+
+				ll comb = ncr(i-1, k-1, MOD);
+				dp[i][j] = (dp[i][j] + comb*c%MOD)%MOD;
+
+			}
+
+
+		}
+		rep(j, 1, a+1) {
+			ps[i][j] = (ps[i][j-1]+dp[i][j])%MOD;
+		}
+	}
+
+	return dp[a][b];
+}