This problem looks hard at the beginning, however, if you read it carefully and figure our all possible things that can define the unique states, you will find the solution.

What we need are:

• A’s location, B’ location, A + B ‘s current value
• A and B’s moving directions
• Special case where a and…

2115. Find All Possible Recipes from Given Supplies

Naively find until can not find any new recipe

• If we find a new recipe, then add it to the supplies.
• Repeat this process until no new recipe can be found.

Time complexity:

The worst case each time we find one recipe. The total complexity will be

O(sizeof(recipes)*sizeof(recipes)*sizeof( ingredient)) = O(100³)…

131. Palindrome Partitioning

For this problem, we can put a separate after the position i or not put a separate. In total, we will have 2^(n-1) cases to put the separators.

We can solve it by using

• backtrack
• bit manipulation

The time complexity is:

O(n * 2^n ) where O(n)…

This is the 4th problem of this week’s LC weekly contest.

Explanation

We can change the problem to find the LIS for each subsequence with index of

• 0, k, 2k, …
• 1, k+1, k + 1 + k,…

A well know problem of longest increasing or non decreasing subsqeuence problem. By…

310. Minimum Height Trees

Explanation

For the problem listed here, it is quite interesting as we can:

• use the diameter of a tree template to solve it
• use the topological sort to solve this one.

Using diameter of a tree

Here is my observations:

1. The result at most have two nodes. 1) when the longest path’s…

3 steps

apt remove cmake
pip install cmake --upgrade
sudo ln /usr/local/bin/cmake /usr/bin/cmake

You can use the following command to check the cmake version

cmake --version

952. Largest Component Size by Common Factor

Union Find is a natural way to combine different sets with connection into one set. So the basic idea is:

• If two numbers have common factor, union them
• During the union process, maintain a set size as the final answer is asking for…

2081. Sum of k-Mirror Numbers

This week’s (weekly 268)last question can be solved by checking all the possible numbers in a brute force way. In order to construct all the possible candidates, we can use backtracking based on standard approaches. If you are using python, luckily, we can use itertools.porduct…