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
View More Similar Problems
Array-DS
An array is a type of data structure that stores elements of the same type in a contiguous block of memory. In an array, A, of size N, each memory location has some unique index, i (where 0<=i<N), that can be referenced as A[i] or Ai. Reverse an array of integers. Note: If you've already solved our C++ domain's Arrays Introduction challenge, you may want to skip this. Example: A=[1,2,3
View Solution →2D Array-DS
Given a 6*6 2D Array, arr: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print t
View Solution →Dynamic Array
Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.
View Solution →Left Rotation
A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d
View Solution →Sparse Arrays
There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun
View Solution →Array Manipulation
Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu
View Solution →