Pragmatism

-

Pragmatism means laser focusing on what you're trying to accomplish, and get it done. This problem for example can be solved probably in O(N^3) or O(N^2), but if O(N^4) is easy to implement, and the limits allow it, why not? Cheers, ACC.

Count Special Quadruplets - LeetCode





Comments

Post a Comment

Popular posts from this blog

The power of pruning in backtracking V

-
Image
Backtracking is a great technique in algorithms, but it needs to be accompanied by aggressive pruning or else it may quickly become intractable. This was an interesting problem, and the small input numbers (10^5 and k<=5) tells me that the authors knew that this could quickly become intractable, even with pruning. The pruning here primarily happens in two occasions: first during the base case (k==0), second when the product becomes larger than n. Comparing my solution with Sonnet 4's and ChatGPT 5's (and I gave them all the hints too), there is a small gain in my solution compared to theirs. It was surprising to see that Anthropic's one didn't make use of pruning as well as mine's or ChatGPT 5's. Code is down below, cheers, ACC. Balanced K-Factor Decomposition - LeetCode Given two integers  n  and  k , split the number  n  into exactly  k  positive integers such that the  product  of these integers is equal to  n . Return  any ...
Read more

Quasi FSM (Finite State Machine) problem + Vibe

-
Image
Not really an FSM problem since the state isn't changing, it is just defined by the current input. Simply following the instructions should do it. Using VSCode IDE you can also engage the help of Cline or Copilot for a combo of coding and vibe coding, see below screenshot. Cheers, ACC. Process String with Special Operations I - LeetCode You are given a string  s  consisting of lowercase English letters and the special characters:  * ,  # , and  % . Build a new string  result  by processing  s  according to the following rules from left to right: If the letter is a  lowercase  English letter append it to  result . A  '*'   removes  the last character from  result , if it exists. A  '#'   duplicates  the current  result  and  appends  it to itself. A  '%'   reverses  the current  result . Return the final string  result  after processing all char...
Read more

The power and simplicity of IComparer V

-
Image
Easy problem but it shows the power of IComparer. There are multiple possible solutions here, hence you let the IComparer do the job by just comparing the absolute values in the (internal) QuickSort implementation. Code is down below, cheers, ACC. Sort Array By Absolute Value - LeetCode You are given an integer array  nums . Rearrange elements of  nums  in  non-decreasing  order of their absolute value. Return  any  rearranged array that satisfies this condition. Note : The absolute value of an integer x is defined as: x  if  x >= 0 -x  if  x < 0   Example 1: Input:   nums = [3,-1,-4,1,5] Output:   [-1,1,3,-4,5] Explanation: The absolute values of elements in  nums  are 3, 1, 4, 1, 5 respectively. Rearranging them in increasing order, we get 1, 1, 3, 4, 5. This corresponds to  [-1, 1, 3, -4, 5] . Another possible rearrangement is  [1, -1, 3, -4, 5]. Example 2: Input:   nums = [-100,100] ...
Read more