Posts tagged as “recursion”

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 3 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 very large, return it modulo 10^9 + 7.

Example 1:

Input: nums = [2,1,3]
Output: 1
Explanation: We can reorder nums to be [2,3,1] which will yield the same BST. There are no other ways to reorder nums which will yield the same BST.


Example 2:

Input: nums = [3,4,5,1,2]
Output: 5
Explanation: The following 5 arrays will yield the same BST:
[3,1,2,4,5]
[3,1,4,2,5]
[3,1,4,5,2]
[3,4,1,2,5]
[3,4,1,5,2]


Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: There are no other orderings of nums that will yield the same BST.


Example 4:

Input: nums = [3,1,2,5,4,6]
Output: 19


Example 5:

Input: nums = [9,4,2,1,3,6,5,7,8,14,11,10,12,13,16,15,17,18]
Output: 216212978
Explanation: The number of ways to reorder nums to get the same BST is 3216212999. Taking this number modulo 10^9 + 7 gives 216212978.


Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= nums.length
• All integers in nums are distinct.

Solution: Recursion + Combinatorics

For a given root (first element of the array), we can split the array into left children (nums[i] < nums[0]) and right children (nums[i] > nums[0]). Assuming there are l nodes for the left and r nodes for the right. We have C(l + r, l) different ways to insert l elements into a (l + r) sized array. Within node l / r nodes, we have ways(left) / ways(right) different ways to re-arrange those nodes. So the total # of ways is:
C(l + r, l) * ways(l) * ways(r)
Don’t forget to minus one for the final answer.

Time complexity: O(n^2)
Space complexity: O(n^2)

python3

Given a tree (i.e. a connected, undirected graph that has no cycles) consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges. The root of the tree is the node 0, and each node of the tree has a label which is a lower-case character given in the string labels (i.e. The node with the number i has the label labels[i]).

The edges array is given on the form edges[i] = [ai, bi], which means there is an edge between nodes ai and bi in the tree.

Return an array of size n where ans[i] is the number of nodes in the subtree of the ith node which have the same label as node i.

A subtree of a tree T is the tree consisting of a node in T and all of its descendant nodes.

Example 1:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], labels = "abaedcd"
Output: [2,1,1,1,1,1,1]
Explanation: Node 0 has label 'a' and its sub-tree has node 2 with label 'a' as well, thus the answer is 2. Notice that any node is part of its sub-tree.
Node 1 has a label 'b'. The sub-tree of node 1 contains nodes 1,4 and 5, as nodes 4 and 5 have different labels than node 1, the answer is just 1 (the node itself).


Example 2:

Input: n = 4, edges = [[0,1],[1,2],[0,3]], labels = "bbbb"
Output: [4,2,1,1]
Explanation: The sub-tree of node 2 contains only node 2, so the answer is 1.
The sub-tree of node 3 contains only node 3, so the answer is 1.
The sub-tree of node 1 contains nodes 1 and 2, both have label 'b', thus the answer is 2.
The sub-tree of node 0 contains nodes 0, 1, 2 and 3, all with label 'b', thus the answer is 4.


Example 3:

Input: n = 5, edges = [[0,1],[0,2],[1,3],[0,4]], labels = "aabab"
Output: [3,2,1,1,1]


Example 4:

Example 5:

Input: n = 7, edges = [[0,1],[1,2],[2,3],[3,4],[4,5],[5,6]], labels = "aaabaaa"
Output: [6,5,4,1,3,2,1]


Constraints:

• 1 <= n <= 10^5
• edges.length == n - 1
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• labels.length == n
• labels is consisting of only of lower-case English letters.

Solution: Post order traversal + hashtable

For each label, record the count. When visiting a node, we first record the current count of its label as before, and traverse its children, when done, increment the current count, ans[i] = current – before.

Time complexity: O(n)
Space complexity: O(n)

Python3

Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.

Return the number of good nodes in the binary tree.

Example 1:

Input: root = [3,1,4,3,null,1,5]
Output: 4
Explanation: Nodes in blue are good.
Root Node (3) is always a good node.
Node 4 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path
Node 3 -> (3,1,3) is the maximum value in the path.

Example 2:

Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.

Example 3:

Input: root = [1]
Output: 1
Explanation: Root is considered as good.

Constraints:

• The number of nodes in the binary tree is in the range [1, 10^5].
• Each node’s value is between [-10^4, 10^4].

Solution: Recursion

Time complexity: O(n)
Space complexity: O(n)

C++

Given the number kreturn the minimum number of Fibonacci numbers whose sum is equal to k, whether a Fibonacci number could be used multiple times.

The Fibonacci numbers are defined as:

• F1 = 1
• F2 = 1
• Fn = Fn-1 + Fn-2 , for n > 2.

It is guaranteed that for the given constraints we can always find such fibonacci numbers that sum k.

Example 1:

Input: k = 7
Output: 2
Explanation: The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ...
For k = 7 we can use 2 + 5 = 7.

Example 2:

Input: k = 10
Output: 2
Explanation: For k = 10 we can use 2 + 8 = 10.


Example 3:

Input: k = 19
Output: 3
Explanation: For k = 19 we can use 1 + 5 + 13 = 19.


Constraints:

• 1 <= k <= 10^9

Solution: Greedy

Find the largest fibonacci numbers x that x <= k, ans = 1 + find(k – x)

Time complexity: O(logk^2) -> O(logk)
Space complexity: O(logk) -> O(1)

Recursive

Iterative

C++

Given two binary trees original and cloned and given a reference to a node target in the original tree.

The cloned tree is a copy of the original tree.

Return a reference to the same node in the cloned tree.

Note that you are not allowed to change any of the two trees or the target node and the answer must be a reference to a node in the cloned tree.

Follow up: Solve the problem if repeated values on the tree are allowed.

Example 1:

Input: tree = [7,4,3,null,null,6,19], target = 3
Output: 3
Explanation: In all examples the original and cloned trees are shown. The target node is a green node from the original tree. The answer is the yellow node from the cloned tree.


Example 2:

Input: tree = [7], target =  7
Output: 7


Example 3:

Input: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4
Output: 4


Example 4:

Input: tree = [1,2,3,4,5,6,7,8,9,10], target = 5
Output: 5


Example 5:

Input: tree = [1,2,null,3], target = 2
Output: 2


Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• The values of the nodes of the tree are unique.
• target node is a node from the original tree and is not null.

Solution: Recursion

Traverse both trees in the same order, if original == target, return cloned.

Time complexity: O(n)
Space complexity: O(h)

Python3

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