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# Posts published in December 2018

Given a binary tree, we install cameras on the nodes of the tree.

Each camera at a node can monitor its parent, itself, and its immediate children.

Calculate the minimum number of cameras needed to monitor all nodes of the tree.

Example 1:

Input: [0,0,null,0,0]
Output: 1
Explanation: One camera is enough to monitor all nodes if placed as shown.


Example 2:

Input: [0,0,null,0,null,0,null,null,0]
Output: 2
Explanation: At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.


Note:

1. The number of nodes in the given tree will be in the range [1, 1000].
2. Every node has value 0.

## Solution: Greedy + Recursion

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

## Problem

Return all non-negative integers of length N such that the absolute difference between every two consecutive digits is K.

Note that every number in the answer must not have leading zeros except for the number 0 itself. For example, 01 has one leading zero and is invalid, but 0 is valid.

You may return the answer in any order.

Example 1:

Input: N = 3, K = 7
Output: [181,292,707,818,929]
Explanation: Note that 070 is not a valid number, because it has leading zeroes.


Example 2:

Input: N = 2, K = 1
Output: [10,12,21,23,32,34,43,45,54,56,65,67,76,78,87,89,98]

Note:

1. 1 <= N <= 9
2. 0 <= K <= 9

## Solution: Search

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

## Problem

Given a wordlist, we want to implement a spellchecker that converts a query word into a correct word.

For a given query word, the spell checker handles two categories of spelling mistakes:

• Capitalization: If the query matches a word in the wordlist (case-insensitive), then the query word is returned with the same case as the case in the wordlist.
• Example: wordlist = ["yellow"]query = "YellOw"correct = "yellow"
• Example: wordlist = ["Yellow"]query = "yellow"correct = "Yellow"
• Example: wordlist = ["yellow"]query = "yellow"correct = "yellow"
• Vowel Errors: If after replacing the vowels (‘a’, ‘e’, ‘i’, ‘o’, ‘u’) of the query word with any vowel individually, it matches a word in the wordlist (case-insensitive), then the query word is returned with the same case as the match in the wordlist.
• Example: wordlist = ["YellOw"]query = "yollow"correct = "YellOw"
• Example: wordlist = ["YellOw"]query = "yeellow"correct = "" (no match)
• Example: wordlist = ["YellOw"]query = "yllw"correct = "" (no match)

In addition, the spell checker operates under the following precedence rules:

• When the query exactly matches a word in the wordlist (case-sensitive), you should return the same word back.
• When the query matches a word up to capitlization, you should return the first such match in the wordlist.
• When the query matches a word up to vowel errors, you should return the first such match in the wordlist.
• If the query has no matches in the wordlist, you should return the empty string.

Given some queries, return a list of words answer, where answer[i] is the correct word for query = queries[i].

Example 1:

Input: wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"]
Output: ["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"]

Note:

• 1 <= wordlist.length <= 5000
• 1 <= queries.length <= 5000
• 1 <= wordlist[i].length <= 7
• 1 <= queries[i].length <= 7
• All strings in wordlist and queries consist only of english letters.

## Solution: HashTable

Using 3 hashtables: original words, lower cases, lower cases with vowels replaced to “*”

Time complexity: O(|W|+|Q|)
Space complexity: O(|W|)

## Problem

A binary tree is univalued if every node in the tree has the same value.

Return true if and only if the given tree is univalued.

Example 1:

Input: [1,1,1,1,1,null,1]
Output: true


Example 2:

Input: [2,2,2,5,2]
Output: false


Note:

1. The number of nodes in the given tree will be in the range [1, 100].
2. Each node’s value will be an integer in the range [0, 99].

## Solution: Recursion

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

## Related Problems

Given a single positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x ... where each operator op1op2, etc. is either addition, subtraction, multiplication, or division (+-*, or /).  For example, with x = 3, we might write 3 * 3 / 3 + 3 - 3 which is a value of 3.

When writing such an expression, we adhere to the following conventions:

1. The division operator (/) returns rational numbers.
2. There are no parentheses placed anywhere.
3. We use the usual order of operations: multiplication and division happens before addition and subtraction.
4. It’s not allowed to use the unary negation operator (-).  For example, “x - x” is a valid expression as it only uses subtraction, but “-x + x” is not because it uses negation.

We would like to write an expression with the least number of operators such that the expression equals the given target.  Return the least number of operators used.

Example 1:

Input: x = 3, target = 19 Output: 5 Explanation: 3 * 3 + 3 * 3 + 3 / 3.  The expression contains 5 operations.

Example 2:

Input: x = 5, target = 501 Output: 8 Explanation: 5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5.  The expression contains 8 operations.

Example 3:

Input: x = 100, target = 100000000 Output: 3 Explanation: 100 * 100 * 100 * 100.  The expression contains 3 operations.

Note:

• 2 <= x <= 100
• 1 <= target <= 2 * 10^8

# Solution: Dijkstra

Find the shortest path from target to 0 using ops.

Time complexity: O(nlogn)
Space complexity: O(nlogn)
n = x * log(t) / log(x)

# Solution 2: DFS + Memoization

Time complexity: O(x * log(t)/log(x))
Space complexity: O(x * log(t)/log(x))

## C++

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