lopespm

// Using a a arduino keypad module from scratch, without using extra libraries. 
// In this case, one columns is targeted.
// This gist is a complement to https://youtu.be/1iPnFEWHnqo, where this is explained

int pinSource = 3;
int pinReceiver1 = 9;
int pinReceiver2 = 8;
int pinReceiver3 = 7;
int pinReceiver4 = 6;

lopespm

// Using a a arduino keypad module from scratch, without using extra libraries. 
// In this case, two columns are targeted
// This gist is a complement to https://youtu.be/1iPnFEWHnqo, where this is explained

int pinSource1 = 3;
int pinSource2 = 4;
int pinReceiver1 = 9;
int pinReceiver2 = 8;
int pinReceiver3 = 7;
int pinReceiver4 = 6;

lopespm

#!/bin/zsh

# See this blog article on how to create Obsidian automatic encrypted backups, on Obsidian application quit, or any other event:
# https://lopespm.com/notes/2024/09/11/obsidian-backup.html
#
obsidian_notes_folder="<your_obsidian_folder>" ; # For example, /Users/yourusername/Library/Application Support/obsidian
obsidian_notes_tar_archive="${obsidian_notes_folder}/obsidian_backup.tar.gz" ;
backup_folder="<folder_where_the_final_encrypted_backup_will_be_placed>"; # For example, /Users/yourusername/Library/CloudStorage/GoogleDrive/MyDrive/backup_folder

echo "Starting to compress obsidian notes..." ;

lopespm

# In this example, we get the maximum content length for the intfloat/multilingual-e5-large model (https://huggingface.co/intfloat/multilingual-e5-large)

from transformers import AutoConfig
checkpoint = "intfloat/multilingual-e5-large"
config = AutoConfig.from_pretrained(checkpoint)
print(f"Maximum context length for this model: {config.max_position_embeddings}")
### Output "Maximum context length for this model: 514"

lopespm

# https://stackoverflow.com/a/34435207/1765893
git rm -r --cached .
git add .
git commit -m "Remove all ignored files in .gitignore"

lopespm

# This solution to compute a random subset avoids the use of extra auxiliary space, with O(k + n) time complexity and O(1) space complexity, in Python (5.15 Compute a random subset, on EPI (Elements of Programming Interviews)) (September 2018 edition)).
# The idea here is to pick the r-th combination, and find its constituents by incrementing them in a Odometer like fashion, and taking into account that the next digit in the combination will be greater than the previous one.
# For example, the combination sequence for n=5 and k=2 is:
# 0 - [0,1]
# 1 - [0,2]
# 2 - [0,3]
# 3 - [0,4]
# 4 - [1,2]
# 5 - [1,3]
# 6 - [1,4]

lopespm

# Alternative python solution for 9.12 Reconstruct a binary tree from a preorder traversal with markers on EPI (Elements of Programming Interviews)) (September 2018 edition)
# Time complexity: O(n)
# Space complexity: O(n + h) - the size of the hash table plus the maximum depth of the function call stack

class BstNode:
    def __init__(self, data, left=None, right=None):
        self.data = data
        self.left = left
        self.right = right
    def __repr__(self):

lopespm

# Alternative python solution for 5.17 The Sudoku Check Problem on EPI (Elements of Programming Interviews)) (September 2018 edition)
# For an nxn Sudoku grid:
# Time complexity: O(n^2)
# Space complexity: O(n)

from typing import List
import math

from typing import List, Set
import math

lopespm

# (Variant #4 for exercise 16.2 on EPI (Elements of Programming Interviews)) (September 2018 edition)
# The core idea is calculate the levenshtein distance, while taking into account the special cases of the regex expression
# *, +, ? and . were taken into account for the regex expression. Expression blocks are not supported
# This algorithm uses dynamic programming, yielding a O(mn) time complexity, O(m) auxiliary space for cache
# (m and n are the lengths of regex and target strings respectively)
#
# Version using recursion with memoization: https://gist.github.com/lopespm/53a215d0b2b0518b52b6bb6687bdaff6

def regex_dist(regex: str, target: str):

lopespm

# (Variant #4 for exercise 16.2 on EPI (Elements of Programming Interviews)) (September 2018 edition)
# The core idea is calculate the levenshtein distance, while taking into account the special cases of the regex expression
# *, +, ? and . were taken into account for the regex expression. Expression blocks are not supported
# This algorithm uses recursion with memoization (could be transposed to a DP solution), yielding a O(mn) time complexity, O(mn) auxiliary space for cache and O(max(m,n)) function call stack
# (m and n are the lengths of regex and target strings respectively)
#
# Version using dynamic programming: https://gist.github.com/lopespm/2362a77e7bd230a4622a43709c195826

def regex_dist(regex: str, target: str):
    def regex_dist_aux(r_i, t_i):