DEV Community: Andrew

18.08.2025.

Andrew — Mon, 18 Aug 2025 10:29:16 +0000

A few weeks passed. I did a lot of work, and also rested well. I will write not every week, but once in a while with some new happenings.
For now, I am in a constant process, so a lot of progress will soon be made. I can understand up to 30% of the Mandarin Chinese after 4 months of learning it, while other dialects are still challenging for me at that stage. But, the more grammar I will comprehend, the more it will explain everything for me. I also listen to the audio with simple day to day phrases a lot, so at the end of the planned practice year, I am going to understand around 90 different daily topics (while on average there are ~100 of them being in our lives throughout the waking hours basis).

Next Week - 28.07.2025

Andrew — Mon, 28 Jul 2025 07:11:55 +0000

I make a little change in my overall structure. From that moment I'm going to have one week for preparation - collecting words in my notes, links to different topics for learning etc.
I will return to previous amounts of materials for learning (e.g. 84 Chinese words), just with that difference of one resting week. Right now, I've completed all my routines and I look forward to the next week to relax a little bit. I will still do my daily tasks btw.

That week — Update.

Andrew — Tue, 22 Jul 2025 15:08:45 +0000

I had a highly important trip (Wǒ yǒu yī gè fēicháng zhòngyào de lǚxíng — haha, that’s what I am capable of saying right now in Chinese). Therefore, I could not learn this time. I’ve already started the next week from Monday, and also…I could not maintain my mood with that much of activity, so people around were disoriented with me. That could not work, I wish there was some place where I could improve myself that crazy, without the need to foster relationships, e.g. for an year or two, but at the same time I am grateful for everything. I had to take my plan, and divide it by two in order to stay fresh even after my undertakings. So, now I learn 42 words, instead of 84… That’s a shame, I will know only ~2300 words after an year, which is a very basic level, compared to 4600 I wanted before. But I might still understand 70% of every needed conversation for functioning in a foreign environment.

I’ve changed everything.

Andrew — Mon, 14 Jul 2025 14:14:16 +0000

Right now, I am not using AI to translate my Russian writings of this blog. I am writing it right away, since I don’t have time for that bullsh. I have too much work and other responsibilities, that I am managing to complete, when most of others would probably give up. I am on one path for 4 weeks right now.

I did 4 algorithms as always. Last one was Fibonacci Number, and before that Climbing Stairs. I still used AI to give me a full and amazing explanation of every result’s code part. I read it, fully understand it, then write my own pseudocode with thoughts and explanations to myself.

I can leave here these two last algos, how I’ve approached them when I forgot some details after reading with understanding the explanation.

class Solution {
public:
    int fib(int n) {
        /* 
            *В задачах на Fibonacci используется как показано в условии формула (n-1)+(n-2)
            Но начинается данная последовательность с 0 и 1.
            = Мне нужно пройтись по всему массиву чисел до n. Есть n допустим равное 3.
              В таком случае (F(3-1) + F(3-2)) + (F(2-1)+F(2-2)), но в начале должны быть 0 и 1.
            = Нужно создать переменную - вектор чисел для хранения резу...Либо, можно пойти с конца
              и тогда вместо сохранения элементов в массиве, можно будет сначала запихнуть в
              в две переменные (prev и current) 0 и 1 соответственно.
            = В начатом цикле...ага или же можно идти с конца и присвоить prev = n-1, а 
              current = n - 1; В цикле, если n меньше 2, то сразу вывести n, в другом случае
              result = prev + current; А потом prev--, и current--;
            = Для n равного 4, будет 1.)result = 3+2; 2.)2+1; 3.)1
            = Нет, легче будет сделать: prev = 0; current = 1; В цикле: result = prev + current;
              prev = current; current = result. А в конце вывести result
            = Для n равного 4, будет 0.)result = 0+1=1; 1.)1+1=2; 2.)1+2=3; 3.)2+3=5;
            = Тоже нужно вывести в конце current (тогда вместо result можно назвать next)
            = Можно использовать while (--n) {} цикл для такого случая, тк индекс не используется
            Итак:
            - int prev = 0; int current = 1; int next;
            - while (--n) {}, или скорее while (n--), чтобы также дойти до 0-вого индекса
            - Внутри цикла next = prev + current; А также prev = current; current = next;
            - В конце вывод return previous (да, ведь в конце current становится next...);
            - А так как next постоянно меняется внутри цикла и ему только нужны новые значения,
              то можно для читаемости и контроля кода сделать int next = ... в каждой итерации
        */

        int prev = 0;
        int current = 1;
        while (n--) {
            int next = prev + current;
            prev = current;
            current = next;
        }
        return prev;
    }
};

class Solution {
public:
    int climbStairs(int n) {
        /* 
            * Есть n шагов до верха лестницы (количество шагов)
            * Каждый раз можно сделать либо 1, либо 2 шага
            * n от 1 до 45
            * каждый шаг имеет два способа поднятия на него (шаг на 1, или шаг на 2), кроме 
              первого шага, который имеет только один способ оказаться на нём.
            = Мне нужно пройтись по циклу от 0(шага) до n(последнего шага)
            = На каждой итерации мне нужно будет засчитать возможность попадания на эту ступень
              из поз-апрошлой ступени, и также из прошлой. Формулой было бы -> step = (n-2) + (n-1)
            = В конце мне нужно будет вывести текущее значение всех посчитанных шагов вместе после
              полного прохождения цикла.
            = В самом начале нужно сделать так, чтобы посчитался только один шаг - первый шаг (до 
              него ступеней ещё не было)
            = Для этого, нужно иметь переменную, хранящую прошлый шаг и текущий.
            = Имея прошлый шаг сначала равный 0, и текущий, тоже допустим! равный 0, в таком случае
              в первой итерации steps = currentStep + previousStep. Но именно поэтому currentStep
              может быть в начале равен 1. Либо, можно попробовать сделать steps = i + previousStep 
              (где i допустим начнётся не с нуля, а с 1).
            = Во втором способе, steps будет при массиве [0, 3] = 1.)1; 2.)...
              Ага, и тут нужно добавлять также и в previousStep, чтобы он не оставался равным 
              всегда 0, а также двигался по лестице - соответственно, previousStep = i в конце.
            = Теперь, steps будет при массиве [0, 3] = 1.)1; 2.)i+(i-1)=1; 3.)...
            = Всё таки нужно что-то изменить. Добавить переменную previousStep = 0, и 
              currentStep = 1; в самом начале
            = Теперь steps будет при массиве [0, 3] = 1.)0+1; 2.)...;
            = Нужно переопределить previousStep и currentStep после текущей итерации, что nextStep
             тоже учитывался, и в итоге производился переход на следующую ступень.
            = previousStep = currentStep; currentStep = step;
            = Теперь steps будет при массиве [0, 3] = 0.)0+1=1; 1.)1+1=2; 2.)1+2=3 3.)2+3=5;
            = В конце вывести currentStep? (ведь 5 это уже следующий шаг, своего рода nextStep)
            = Да, в таком случае можно назвать вместо step - nextStep.
            Итак:
            - Две переменные previousStep = 0; currentStep = 1;
            - Цикл for (int i = 0; i < n; ++i) {}
            - Внутри цикла nextStep = previousStep + currentStep
            - После этого также внутри цикла previousStep = currentStep; currentStep = nextStep;
            - После цикла вывод return currentStep.

            P.S - Если бы я лучше знал шаблоны, то распознал бы задачу на Fibonnaci sequence, 
            а тогда я мог бы дойти до решения в 2-4 раза быстрее.
        */

        int prevStep = 0;
        int currentStep = 1;
        int nextStep;
        for (int i = 0; i < n; ++i) {
            nextStep = prevStep + currentStep;
            prevStep = currentStep;
            currentStep = nextStep;
        }
        return currentStep;
    }
};

Yea it’s Russian, but I don’t give a d. Nobody would read it anyway. I will probably just re-read this some day and recollect some more.

I was also reading about 5 CS topics, including: Von Neumann and Harvard’s architectures; Registers (flags, general purpose); Pipelining, Branch prediction and an Arithmetic logic unit ALU.

With Chinese words it’s a little bit trickier, since after I’ve added more to my work schedule, I could not manage to find time for everything, so I still know 500–550 words, but I need to repeat and learn already familiar words, so I could get to 600 and even 700 this time. I’ll do that, since I have slowly found my pase and flow in everything.

I also do my own startup, which is just in a well…starting point. But I am already proud of doing next steps and staying on one course no matter what. I have my own Notes that include the year+ plan, and all the motivation with other important stuff keeping me on that road to first success. At the end of it all, you need to succeed only once, just like Drew Houston, and Mark Cuban were saying.

That’s it, I am going to complete the tone of my job from the previous week, and also start that new one from Monday. I was actually afraid of becoming a bad programmer, but that fear made me think more about: security, maintenance, cleanliness (readability), correctness (best practices) and complete understanding of what I am doing, why, how, where etc.

You can follow my path to a huge one of a Tech Entrepreneur with good programming skills person. Independent and fully organized in terms of everything in life. Well, maybe except of my zero interest on any romantic relationships (probably I am asexual, but never mind…).

Yea, let’s get it then.

Weekly Dev – Algorithms, Language Learning, and Life

Andrew — Tue, 08 Jul 2025 10:29:36 +0000

Crazy week. I didn't even have time to write this log. It's already Tuesday night, and I finally have a moment to breathe.

🧠 Algorithms This Week
As usual — 4 algorithms. This time:

Move Zeroes

Contains Duplicate

Intersection of Two Arrays

Majority Element

I'm still working through them, focusing on recognizing patterns and reinforcing muscle memory for problem-solving.

🎧 Language Learning Strategy
I’ve added a small twist to my daily English learning: listening to one 7-minute topic over the course of three days before switching to the next.
Why? I realized that I often miss even familiar words in fast, natural speech.

By doing this consistently for a year, I should be able to understand about 120 topics, covering most day-to-day situations — and roughly 4,600 words.
Right now I know about 500. Each week, I can deeply memorize 60–70 of the 84 I review. The rest usually settle in after a week or two.

🏓 Life Events
This period included:

My birthday 🎉

Meetings with some amazing people

A couple of tennis sessions

This meant I wasn’t home a lot and couldn’t stick to my usual routine. I managed to finish the weekly plan in just 3–4 days out of 7.

One practice that helps a lot: daily detailed memory recall of the previous 24 hours. Soon, I’ll try doing that for two days instead of one — memory training is part of the process too.

⚡ Electronics & Computer Fundamentals
I'm gradually working through basic electronics — essential for understanding how computers and other hardware actually work.

I’m not diving too deep; just getting the fundamentals. This will help me later when I move on to more practical, higher-level tech topics. I’m about to finish the current chapter and get to the fun stuff soon.

For the next log, I might share more details about my learning framework. For now, though, I’m slowing down the pace a bit — taking a couple of low-pressure days to recharge.

2nd Week - Algorithms

Andrew — Mon, 30 Jun 2025 11:54:27 +0000

5. Merge Two Sorted Lists (Easy)

🔗 Problem Link

🧱 Context:

Written in C++.
Merges two sorted singly linked lists into a new sorted list.
A singly linked list is a structure where each node holds a value (val) and a pointer to the next node (next).

🧩 ListNode Structure

struct ListNode {
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

val: value of the node
next: pointer to the next node

Three constructors:

Default: val = 0, next = nullptr
With value only
With value and next node pointer

🧠 Solution Class

class Solution {
public:

Class for our solution
public: means the method is accessible externally

🔧 `mergeTwoLists` Method

ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {

Takes two pointers to sorted lists
Returns pointer to a new merged sorted list

🧱 Step 1: Create dummy starting node

ListNode* dummyHead = new ListNode();

Used as a fake starting point to simplify appending new nodes.

📍 Step 2: Create traversal pointer

ListNode* current = dummyHead;

current moves along the new list.

🔄 Step 3: While both lists aren't finished

while (list1 && list2) {

Continue while both lists have nodes.

🔁 Step 4: Compare values

if (list1->val <= list2->val) {

Choose the smaller value.

➡️ Step 5: Append from list1

current->next = list1;
list1 = list1->next;

🔁 Otherwise, take from list2

} else {
    current->next = list2;
    list2 = list2->next;
}

🧭 Step 6: Move forward in the new list

current = current->next;

🔚 Step 7: Attach remaining nodes

current->next = list1 ? list1 : list2;

🏁 Step 8: Return the merged list

return dummyHead->next;

Return the first real node (skip dummy).

✅ Summary:

Create a dummy node
Traverse both lists
Choose the smaller node each time
Attach the rest when one list ends
Return merged result

6. Reverse Linked List (Easy)

🔗 Problem Link

🔹 ListNode Structure

struct ListNode {
    int val;
    ListNode *next;

    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

🌀 `reverseList` Function

ListNode* reverseList(ListNode* head) {

Input: pointer to start of the list
Output: pointer to start of reversed list

Steps:

Create dummy head

ListNode* dummy = new ListNode();

Pointer for traversal

ListNode* current = head;

Loop through list

while (current != nullptr) {

Save the next node

ListNode* nextNode = current->next;

Reverse pointer

current->next = dummy->next;

Insert at start of new list

dummy->next = current;

Move to next

current = nextNode;

Return reversed list

return dummy->next;

🧩 Visual Walkthrough

Original:

head → 1 → 2 → 3 → 4 → nullptr

After iterations:

dummy → 1
dummy → 2 → 1
dummy → 3 → 2 → 1
dummy → 4 → 3 → 2 → 1

🟡 Recap:

Create empty list
Take nodes one by one from original
Insert at beginning of new list
Return result

7. Linked List Cycle (Easy)

🔗 Problem Link

📦 Node Structure

struct ListNode {
    int value;
    ListNode *next;
    ListNode(int x) : value(x), next(nullptr) {}
};

🧠 Cycle Detection (`hasCycle`)

bool hasCycle(ListNode *head) {

🐢🐇 Use two pointers:

ListNode *slowPointer = head;
ListNode *fastPointer = head;

slowPointer moves 1 step
fastPointer moves 2 steps

🔁 Traverse the list

while (fastPointer && fastPointer->next) {
    slowPointer = slowPointer->next;
    fastPointer = fastPointer->next->next;

If they meet:

if (slowPointer == fastPointer)
    return true;

No cycle?

return false;

📊 Visual

Without cycle:

1 → 2 → 3 → 4 → nullptr

With cycle:

1 → 2 → 3 → 4
          ↑
          ↓
          2 (loop)

🧩 Summary:

Two pointers moving at different speeds
If cycle exists → they'll meet
If not → fast pointer hits end

8. Maximum Subarray (Medium)

🔗 Problem Link

📦 The Problem:

Given an array, find the maximum sum of any continuous subarray.

Example:
Input: [-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output: 6 → from [4, -1, 2, 1]

🧠 Kadane’s Algorithm

int maxSubArray(vector<int>& nums) {

🔹 Step 1: Initialize max values

int currentMax = nums[0];
int globalMax = nums[0];

🔁 Step 2: Iterate

for (int i = 1; i < nums.size(); ++i) {
    currentMax = std::max(currentMax, 0) + nums[i];
    globalMax = std::max(globalMax, currentMax);
}

✅ Step 3: Return result

return globalMax;

🧩 Example Breakdown

i	nums[i]	currentMax	globalMax
0	-2	-2	-2
1	1	1	1
2	-3	-2	1
3	4	4	4
4	-1	3	4
5	2	5	5
6	1	6 ✅	6
7	-5	1	6
8	4	5	6

🟡 Summary:

currentMax: sum of current subarray
Reset to 0 if it turns negative
globalMax: max seen so far
One pass: O(n) complexity

✅ Follow-up

If you liked this breakdown, let me know — I plan to continue this series with more Leetcode problems, especially in C++ with deep explanations for beginners.

2 Week Of : “Spark” Project

Andrew — Sun, 29 Jun 2025 17:37:26 +0000

Friday. Time – 12:53 AM (in China).

Yes, in practice I didn’t manage to complete my plan within the week, but I had a lot of business stuff going on — one night I didn’t even sleep at all, nearly 24 hours straight. And on the other days I slept 5, 6, 7, or at least once even 8 hours. In reality though, I still have tonight — and no one canceled that. Time is a myth, just like thoughts, the future, the past, or even the present. If you look deeper into time, you’ll realize we don’t have it — and at the same time, we do. I believe there’s no point wasting something we kind of don’t even have, on things that are completely meaningless. Although, everyone does as they please, but I’ve chosen my own path. Below is the text I wrote, I think, on Thursday.

This is how I learn Chinese. (The mnemonic method is weird, but it’s from the book Unlimited Memory. I’ve already memorized over 400 words in 3 months and one week. I also write down abbreviations for these words with later pronunciation of them, in addition to the more obvious reasons to remember these words for real-life situations → a method I got from YouTube. For once that platform was actually useful, and not just triggering FOMO.) I’ve been living in China for 3 months now, so I occasionally have conversations with native speakers, plus minimal practice with them. 400 words is very little… even a child knows way more. 1,000 words is the first minimum to understand most basic topics, and I still have a lot to learn. Also, sentence construction is often not obvious, which means I’ll need to memorize lots of phrases, particles, forms, structures, etc. Oh, almost forgot. I’m from Russia, so it’s much easier for me to come up with homonyms and associations in Russian… I’m baring my soul with this post, because many of the things below came to me by chance and initially in the form of a weird “mini-story.” I should also mention that even tones are included in my mnemonics, because if you pronounce even a single vowel at the wrong pitch, you probably won’t be understood…

The words themselves (absurd, but it works):

bǎi - hundred, numerous (Можно сказать бай огромному количеству знакомых, наверное сотне) *Например, это стоило 100 юаней
bào zhǐ (бАо джи) - newspaper, newsprint (Да он избалованный джином словно, вот так пишет эти газеты)
guò - to pass (time), to celebrate (Гула то сколько было, но время прошло, вы же отмечали?)

Next, I have 2 tests awaiting me from both sides (Chinese – English), which I might pass immediately, but usually I need to revise certain unlearned words first and only then pass everything. That’s how I test myself.

And I still have 3 projects, and possibly tennis coaching (the last one is not certain yet). One of them is a joint project, the other 2 are my personal ones. I don’t know how I manage everything, but I’m extremely motivated, because nothing else in this life or the entire world interests me.

And here are my 5 behind-the-scenes topics on computer theory (all links are in Russian, for an obvious reason… Previously, I tried learning completely new topics without knowing the basics and in a foreign language, especially English — it was pointless and difficult):

1. Basics of Electronics

Questions:

What is electric current?
What’s the difference between alternating current (AC) and direct current (DC)?
What are a resistor, capacitor, and inductor?
How do a diode and a transistor work?
What is Ohm’s law? Write down the formula.
What are logic levels in digital electronics (0 and 1)?

Answers:

Electric current is the directional flow of charged particles (electrons or ions).
AC changes direction (e.g., in a wall outlet), DC flows in one direction (e.g., a battery).
* Resistor – resistance, limits current.
- Capacitor – stores charge.
- Inductor – creates a magnetic field when current flows through it.
* Diode – allows current to pass in only one direction.
- Transistor – amplifies or switches current (e.g., NPN/PNP types).
Ohm’s Law: $V = I \times R$ (Voltage = Current × Resistance).
Logic 0 – low voltage (~0V), logic 1 – high voltage (~3.3V or 5V).

2. Boolean Algebra (Logic Algebra)

Questions:

What is Boolean algebra and where is it used?
What are the basic logical operations?
What is a truth table?
How are logical expressions written?

Answers:

Boolean algebra is a branch of mathematics that studies logical operations on statements (true/false). It is used in digital electronics, programming.
Basic operations: AND, OR, NOT, XOR (exclusive OR).
Truth table – a table showing the result of an operation for all possible input combinations.
Logical expressions are written using variables (A, B) and operators (∧, ∨, ¬).

3. Logical Addition (OR)

Questions:

How does the OR operation work?
What is the truth table for OR?
What does the logical expression for OR look like?
How is OR implemented using transistors or relays?

Answers:

OR outputs 1 if at least one input is 1.
OR truth table:

| A | B | A∨B |
| - | - | --- |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |

Expression: $A + B$ or $A \lor B$.
In electronics, OR can be implemented with diodes or transistors (parallel connection).

4. Logical Multiplication (AND)

Questions:

How does the AND operation work?
What is the truth table for AND?
What does the logical expression for AND look like?
How is AND implemented with transistors?

Answers:

AND outputs 1 only if both inputs are 1.
AND truth table:

| A | B | A∧B |
| - | - | --- |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |

Expression: $A \cdot B$ or $A \land B$.
In electronics, AND is implemented using transistors connected in series.

5. Logical Negation (NOT)

Questions:

How does the NOT operation work?
What is the truth table for NOT?
What does the logical expression for NOT look like?
How is NOT implemented with transistors?

Answers:

NOT inverts the signal (0 → 1, 1 → 0).
NOT truth table:

| A | ¬A |
| - | -- |
| 0 | 1 |
| 1 | 0 |

Expression: $\overline{A}$ or $\neg A$.
In electronics, NOT is implemented with a transistor (inverter).

And 1 C++ topic.

📘 Original tutorial source (in Russian)

1. Basics of Loops in C++

Questions:

What is a loop in programming?
What types of loops exist in C++?
What’s the difference between while, do-while, and for?
How do you avoid an infinite loop?

Answers:

A loop is a construct that repeats a block of code as long as a condition is true.
C++ offers:

while – loop with a pre-condition
do-while – loop with a post-condition
for – counter-controlled loop
1. Differences:
while checks the condition before execution
do-while executes the body at least once, then checks the condition
for is a compact loop with init, condition, and update
1. To avoid infinite loops:
Ensure the exit condition is reachable
In for, verify the counter is changing correctly
In while/do-while, avoid conditions that are always true

2. The `while` Loop

Questions:

How does while work?
What is the syntax of while?
When is while better than for?
Can while run zero times?

Answers:

while executes code as long as the condition is true
Syntax:

while (condition) {
    // loop body
}

Use while when the number of iterations is unknown (e.g., reading a file)
Yes — if the condition is initially false, it won’t run at all

3. The `do-while` Loop

Questions:

How does do-while work?
How is it different from while?
When should you use do-while?
Is ; required after while()?

Answers:

do-while executes the body first, then checks the condition
Difference:

do-while guarantees at least one execution
1. Use it when you need one execution minimum (e.g., menu input)
2. Yes, semicolon is required after while():

do {
    // body
} while (condition);  // ← don't forget ;

4. The `for` Loop

Questions:

How does for work?
What are the parts of a for loop declaration?
Can parts of for be skipped? (e.g., initialization)
How do you count backward (10 to 1)?

Answers:

for executes code while the condition is true, updating a counter
Structure:

for (init; condition; update) {
    // body
}

Yes, parts can be skipped, but semicolons are required:

int i = 0;
for (; i < 5; i++) {
    ...
}

Reverse countdown:

for (int i = 10; i >= 1; i--) {
    cout << i << " ";
}

5. Loop Control (`break`, `continue`)

Questions:

What does break do?
What does continue do?
Can break and continue be used in all loops?
How do you exit a nested loop?

Answers:

break immediately exits the loop
continue skips the current iteration, moving to the next
Yes, both work in while, do-while, and for
To exit nested loops:

Use break in the inner loop
Or use goto (not recommended) or flags:

bool exitFlag = false;

for (int i = 0; i < 10 && !exitFlag; i++) {
    for (int j = 0; j < 10; j++) {
        if (condition) {
            exitFlag = true;
            break;
        }
    }
}

6. Practical Tasks (To check your understanding)

1. `for` loop: Print even numbers from 2 to 20

for (int i = 2; i <= 20; i += 2) {
    cout << i << " ";
}

2. `while` loop: Sum numbers until user enters 0

int num, sum = 0;
while (true) {
    cin >> num;
    if (num == 0) break;
    sum += num;
}

3. `do-while`: Ask for password until "12345" is entered

string password;
do {
    cout << "Enter password: ";
    cin >> password;
} while (password != "12345");

🔍 Optional Topics to Explore Further

Nested loops
Loop optimization (++i vs i++)
Range-based for loop (C++11 and newer)

🌙 Now, Back to Monday Night...
I still have 4 algorithms left to write and need to pass a test on Chinese vocabulary.
Also, I’ll review all CS and C++ topics one more time.

"Spark" - 1 Week (22.06.2025)

Andrew — Wed, 25 Jun 2025 14:03:38 +0000

Week 1 – Learning Log (22.06.2025)

🗓 Thursday

How's the first week going? Honestly, not bad at all.

I came up with a lot of new project ideas (not the focus this week though).
Created mnemonics for nearly all 84 vocabulary phrases, and I’m close to finishing.
I’ll soon start translation tests (Chinese <-> English) to measure real retention.
Will write down reasons why remembering them matters.

This week brought another intense wave of introspection and existential thoughts—but I made it through. Oddly enough, The Social Network movie helps me regain focus and motivation.

Algorithms I'm Starting With

I’ve selected 4 beginner-level problems to master:

✅ Two Sum
✅ Valid Parentheses
✅ Best Time to Buy and Sell Stock
✅ Valid Palindrome

I'll:

Study the most efficient solutions
Write pseudocode
Implement them in C++

My approach prioritizes speed, clarity, and adaptability. Critics might say I’m copying solutions. Fair. But my aim is to absorb as many techniques as possible before building things completely solo.

I've also started learning 5 computer theory topics—right down to electrical components. Why? Because I want to understand how things work beneath the code. I’ve also added a C++ topic for brushing up on fundamentals.

And despite it being 2 AM... I feel energized. Let's go. Even if AI takes over coding, I’ll keep living my narrative.

🗓 Sunday

It's been a tough but meaningful week.
⏰ 9:35 PM – Still finishing up:

Chinese vocabulary
LeetCode solutions
A bit of C++ and CS theory

I can't say I’ve mastered the technical theory yet, but now, when I revisit these ideas in the future, I’ll recognize and understand them much faster.

Meanwhile, I’m juggling some business projects too.
I registered my new domain: unpackly.com — development starts next week.

Still need to:

Identify the target audience
Find communities where they hang out
Begin engaging authentically

🛠️ Weekly Resources + LeetCode Review

One of my ongoing project ideas is building a tool that summarizes all your study content in seconds—so here's a "manual version" of that for now.

My LeetCode ID: feelsgood121
I used the following materials to study and solve these 4 problems:

🧠 4 Algorithms

1. Two Sum (Easy)

🔗 Link

One-Pass Hash Table stands out as the most time-efficient option, especially useful for larger datasets.

class Solution {
👉 Here we create a class named Solution. In C++, classes are like containers that hold functions and variables. On LeetCode, code is usually written inside such a class.

public:
👉 This is an access specifier. It means everything that follows inside the class is publicly accessible — other parts of the program can use these functions.

vector<int> twoSum(vector<int>& nums, int target) {
👉 Here we declare the twoSum function, which:

Returns a vector<int> (a list of numbers like [1, 2])
Takes two arguments:
vector<int>& nums — a list of numbers (e.g., [2, 7, 11, 15])
int target — the number we want as the sum of two elements from nums

unordered_map<int, int> numMap;
👉 We create a hash map (a fast data structure) that:

Stores pairs: number → its index
unordered_map<int, int> means: key and value are both int
Example: if nums = [2, 7], numMap might become {2: 0, 7: 1}

for (int i = 0; i < nums.size(); i++) {
👉 This starts a loop over all numbers in nums:
i is the index (0, 1, 2…)
nums.size() gives the total count

int complement = target - nums[i];
👉 We calculate the “complement” — the number needed to reach the target with the current number.
E.g., target = 9, nums[i] = 2 → complement = 7

if (numMap.count(complement)) {
👉 We check if this complement already exists in numMap.
numMap.count(x) returns 1 if x is in the map, 0 if not

return {numMap[complement], i};
👉 If the complement is found, we return the pair of indices:
numMap[complement] — index of the number we’ve already seen
i — current number’s index

🧠 This means we’ve found the two numbers that sum to target.
}

numMap[nums[i]] = i;
👉 If no match found yet, add the current number to the map:
Key = nums[i], Value = i
}

return {};
}
};
👉 Close the loop, function, and class. If no pair was found, we return an empty list.

💡 Example

nums = [2, 7, 11, 15];
target = 9;
i = 0 → nums[i] = 2 → complement = 7 → not in map → add 2:0  
i = 1 → nums[i] = 7 → complement = 2 → found → return {0, 1}

2. Valid Parentheses (Easy)

🔗 Link

This code checks whether brackets in a string are correctly closed.

Examples:

"()" → valid
"(]" → invalid
"({[]})" → valid

🔹 Main Code

class Solution {
👉 Creates the class Solution.

public:
👉 Makes everything below publicly accessible.

bool isValid(string s) {
👉 This is the function isValid:
Takes a string s (e.g., "(())")
Returns true if the parentheses are valid, else false

string stack;
👉 We use this string as a stack:
A stack lets you:
Push to the end
Pop from the end
Access the last added element

for (char c : s) {
👉 Loop through each character in s

if (c == '(' || c == '{' || c == '[') {
    stack.push_back(c);
}
👉 If it’s an opening bracket, we add it to the stack

else if (stack.empty() || !match(stack.back(), c)) {
    return false;
}
👉 If it’s a closing bracket:
Check if the stack is empty (then it’s invalid)
Or check if the bracket matches the last open one (match function)

else {
    stack.pop_back();
}
👉 If brackets match — remove the last open one
}

return stack.empty();
👉 If the stack is empty at the end — valid; otherwise — invalid
}

bool match(char left, char right) {
    return (left == '(' && right == ')') ||
           (left == '[' && right == ']') ||
           (left == '{' && right == '}');
}
👉 Checks if left and right form a valid pair.
};

🔹 Example

string s = "({[]})";
( → push  
{ → push  
[ → push  
] → matches [ → pop  
} → matches { → pop  
) → matches ( → pop ✅  
→ Valid → returns true

3. Best Time to Buy and Sell Stock (Easy)

🔗 Link

🔹 Purpose

This code solves the problem:

📈 Find the maximum profit from buying and selling a stock once.
(You can buy once and sell once — sell must be after buy)

Example:
If prices = [7, 1, 5, 3, 6, 4] → max profit = 6 - 1 = 5

🔹 Line-by-Line

#include <vector>
#include <algorithm>
👉 Includes libraries for vectors and functions like min/max

class Solution {
public:
    int maxProfit(vector<int>& prices) {
👉 The function receives a list of stock prices and returns max profit.

int max_profit = 0;
int min_price = prices[0];
👉 Start with 0 profit, and first price as the initial minimum

for (int price : prices) {
    max_profit = max(max_profit, price - min_price);
    min_price = min(min_price, price);
}
👉 For each price:
Calculate potential profit → update max_profit if it's better
Update min_price if today’s price is lower

return max_profit;
👉 Return the best profit found
}
};

🔹 Example

prices = [7, 1, 5, 3, 6, 4]
→ max profit = 5 ✅

4. Valid Palindrome (Easy)

🔗 Link

Checks if a string is a palindrome, ignoring spaces, punctuation, and case.

🔹 What’s a palindrome?

A string that reads the same forward and backward.
✅ "A man, a plan, a canal: Panama"
❌ "race a car"

🔹 Line-by-Line

class Solution {
public:
    bool isPalindrome(string s) {
👉 Takes a string s and returns true if it’s a valid palindrome

int left = 0, right = s.size() - 1;
👉 Two pointers: left from start, right from end

while (left < right) {
👉 Loop until the pointers meet

if (!isalnum(s[left])) { ++left; }
else if (!isalnum(s[right])) { --right; }
👉 Skip non-alphanumeric characters

else if (tolower(s[left]) != tolower(s[right])) {
    return false;
}
👉 Compare characters (case-insensitive). If not equal → return false

else {
    ++left;
    --right;
}
👉 If they match → move inward
}

return true;
👉 If all characters matched → it’s a palindrome
}
};

🔹 Example

"A man, a plan, a canal: Panama"
→ Cleaned: "amanaplanacanalpanama"
→ All characters match → returns true ✅

🧠 Computer Science Topics I'm Studying

Focusing on low-level basics to truly understand how computers work:

Capacitor
Diode
Transistor
LED
Phototransistor

Maybe I’m going too deep. But who knows—maybe I’ll be an engineer one day. In any case, I’m enjoying it.

💻 C++ Focus This Week

I’m revisiting data types — still not 100% confident when solving real problems. But it’s getting better.

🈴 Weekly Chinese Vocabulary – 84 Words/Phrases
Here are most of the 84 words/phrases I’ve memorized this week:

Nǐ yǒu jǐ gè háizi?

Nǐ yǒu shénme àihào?

Nǐ qùguo Měiguó ma?

Nǐ de shǒujī hàomǎ shì duōshao?

Nǐ de yóuxiāng dìzhǐ shì shènme?

Xiànzài yǒu kòng ma?

Zuìjìn gōnzuò máng ma?

Nǐ měitiān jǐ diǎn shàngbāng?

Nǐ měitiān jǐ diǎn xiàbān?

Nǐ yǐqián zuòguo shénme gōngzuò?

Duì, bàituō le

Méi wèntí

Zhīdào le

Kěyǐ

Dāngrán kěyǐ

Yuánlái rúcǐ

Wǒ yě zhème rènwéi

Zhēn búcuò

Búyòng le

Wǒ bù míngbai

Wǒ bú zhème rènwéi

Bùxíng

Kěyǐ ma?

Nǐ néng bāng wǒ gè máng ma?

Bǎ mén guānshàng hǎo ma?

lái

Fēijī

shǎo

diěn

jiào

xiào

zài

cháng

ràng

xiūxi

zhāng

dào

yǐjīng

jué de

chàng gē

děng

hēi

bǐ

zhe

wài

zhǔnbèi

mǎshàng

zuì

tí

chū

yīn

hóng

shēng bìng

kuài lè

yuán

zhēn

duì

pǎobù

yánsè

lù

jièshào

yùndòng

zhèngzài

lǚyóu

kěnéng

qīng

shíjiān

měi

jiàn

qiāng

yǎnjing

zìxíngchē

kè

zhǎo

xīwàng

tiàowǔ

hái

cuò

shǒubiǎo

yīqǐ

gěi

gōngjīn

lí

🎯 Final Thoughts

This week wasn't perfect, but it was real.
I didn’t try to do everything solo. I focused on learning deeply, making steady progress, and setting up systems that will pay off long-term.

Next up: Refining my learning tools, developing unpackly.com, and pushing further into CS theory and algorithms.

Even if the world changes, I'll keep building.

Personal Plan - Project "Spark".

Andrew — Wed, 25 Jun 2025 14:02:37 +0000

The first week of my project, which I decided to name “Spark”, is about to begin. This blog serves only one purpose — to track my progress over the course of one year. I’ll be starting almost from scratch, as I don’t have deep knowledge of programming or the IT field as a whole. Right now, I’m only capable of building a simple web application with basic front-end, back-end, and simple algorithms.

There won’t be much text here, since my work will be focused, stripped of doubts and worries. In order not to burn out over the year, I’ve simplified the process as much as possible, using my analytical type of thinking only in the most critical and important moments. After all, one cannot completely shut off their mind — I don’t plan to become dull in the digital era by watching short videos and endlessly scrolling feeds.

Alongside my tech studies, I’m also learning Chinese — 336 words per month, to be exact — and two test weeks have already been successfully completed. My plan is ambitious, but I believe it’s achievable. Let’s go. The project details will be revealed along the way.

In a year or a year and a half, I might become qualified for FAANG companies, though I don’t have a strong desire to work there — or I might end up as no one. But I have nothing to lose; otherwise, I’ll simply continue as a tennis coach with over 10 years of professional career experience.

DEV Community: Andrew

18.08.2025.

Next Week - 28.07.2025

That week — Update.

I’ve changed everything.

Weekly Dev – Algorithms, Language Learning, and Life

2nd Week - Algorithms

5. Merge Two Sorted Lists (Easy)

🧱 Context:

🧩 ListNode Structure

🧠 Solution Class

🔧 mergeTwoLists Method

🧱 Step 1: Create dummy starting node

📍 Step 2: Create traversal pointer

🔄 Step 3: While both lists aren't finished

🔁 Step 4: Compare values

➡️ Step 5: Append from list1

🔁 Otherwise, take from list2

🧭 Step 6: Move forward in the new list

🔚 Step 7: Attach remaining nodes

🏁 Step 8: Return the merged list

✅ Summary:

6. Reverse Linked List (Easy)

🔹 ListNode Structure

🌀 reverseList Function

Steps:

🧩 Visual Walkthrough

🟡 Recap:

7. Linked List Cycle (Easy)

📦 Node Structure

🧠 Cycle Detection (hasCycle)

🐢🐇 Use two pointers:

🔁 Traverse the list

📊 Visual

🧩 Summary:

8. Maximum Subarray (Medium)

📦 The Problem:

🧠 Kadane’s Algorithm

🔹 Step 1: Initialize max values

🔁 Step 2: Iterate

✅ Step 3: Return result

🧩 Example Breakdown

🟡 Summary:

✅ Follow-up

2 Week Of : “Spark” Project

Basics of Electronics

Boolean Algebra (Logic Algebra)

Logical Addition (Disjunction)

Logical Multiplication (Conjunction)

Logical Negation (Inversion)

1. Basics of Electronics

2. Boolean Algebra (Logic Algebra)

3. Logical Addition (OR)

4. Logical Multiplication (AND)

5. Logical Negation (NOT)

1. Basics of Loops in C++

Questions:

Answers:

2. The while Loop

Questions:

Answers:

3. The do-while Loop

Questions:

Answers:

4. The for Loop

Questions:

Answers:

5. Loop Control (break, continue)

Questions:

Answers:

6. Practical Tasks (To check your understanding)

1. for loop: Print even numbers from 2 to 20

2. while loop: Sum numbers until user enters 0

3. do-while: Ask for password until "12345" is entered

🔍 Optional Topics to Explore Further

"Spark" - 1 Week (22.06.2025)

Week 1 – Learning Log (22.06.2025)

🗓 Thursday

Algorithms I'm Starting With

🗓 Sunday

🔧 `mergeTwoLists` Method

🌀 `reverseList` Function

🧠 Cycle Detection (`hasCycle`)

2. The `while` Loop

3. The `do-while` Loop

4. The `for` Loop

5. Loop Control (`break`, `continue`)

1. `for` loop: Print even numbers from 2 to 20

2. `while` loop: Sum numbers until user enters 0

3. `do-while`: Ask for password until "12345" is entered