Learning CPP via Advent of Code 2024 [Days 1 to 5]

January 04, 2025   ·   AoC   C++   ·   Advent of Code

Learning CPP via Advent of Code 2024 [Days 1 to 5]

I’ve decided to use Advent of Code 2024 as a platform to improve my C++ skills. With a background in C, having worked on Embedded Systems for the past two years, and over five years of experience in Python where I consider myself between intermediate and advanced, I felt it was time to dive into C++. While I’m familiar with core programming concepts like loops, functions, and Object-Oriented Programming (OOP) from my experience in C and Python, I wanted a more practical and engaging way to learn C++ without retreading the basics.

Advent of Code, with its daily coding challenges, seemed like a good way to get hands-on with C++.

This is my attempt to document my learning journey through Advent of Code 2024. Instead of cluttering with each post for each day’s challenge, I’ll be grouping my progress into consolidated posts. This page is the first part of my progress log covering Days 1 to 5.

Full solutions are available on GitHub: ABD-01/AoC2024.

Updated on 21 February 2025

Added next part for Days 10 and 11.

Updated on 15 February 2025

Added next part for Day 9.

Updated on 18 January 2025

Added next part for Days 6 to 8.

Day 1: Historian Hysteria

Solution Overview

Part 1: Sorting and Pairing Numbers

Given two lists of numbers.

Pair up the smallest number in the left list with the smallest number in the right list, then the second-smallest left number with the second-smallest right number, and so on.

Sort both lists, take absolute difference of the values elementwise and sum it up.

    std::sort(vec1.begin(), vec1.end()); // sorts in ascending order
    std::sort(vec2.begin(), vec2.end());
    //std::ranges::sort(vec2);

    int result = 0;
    for (std::size_t i = 0; i < vec1.size(); i++)
        result += abs(vec1[i] - vec2[i]);

Part 2: Calculating Similarity Scores

Calculate a total similarity score by adding up each number in the left list after multiplying it by the number of times that number appears in the right list.

    for (std::size_t i = 0; i < vec1.size(); i++)
    {
        /**
        int count = 0;
        for (std::size_t j = 0; j < vec1.size(); j++)
            if (vec1[i] == vec2[j])
                count++;
        */
        int count = std::count(vec2.begin(), vec2.end(), vec1[i]);
        //int count = std::ranges::count(vec2, vec1[i]);
        result += (count * vec1[i]);
    }

Now here using std::count inside a loop results in a time complexity of \(O(n^2)\). This could be optimized further using a hash map (std::unordered_map) to count occurrences in \(O(n)\) time.

Concepts Learned

Reading from a file

Use file streams defined in header <fstream>

• ifstream: Input file stream for reading files.

• ofstream: Output file stream for writing files.

• fstream: Input-output file stream for both reading and writing.

String Streams (<sstream>): Used to read from or write to strings.

• istringstream: For input operations from a string.

• ostringstream: For output operations to a string.

• stringstream: For both input and output operations on a string.

standard input/output streams (<iostream>)

Vectors

// 1. Creates a vector 'a' with 'count' elements
vector<int> a(count); 

// 2. Creates a vector 'v' with 'count' elements, all initialized to 'value'
vector<int> v(count, value); 

// 3. Looping over a vector and printing its elements
for (auto i : vec) std::cout << i << " "; 

// 4. Iterators
std::vector<int>::iterator it = vec.begin();
auto it = vec.end();

// 5. Methods
a.size()
a.push_back // adds an element to the end

Example

#include <iostream>
#include <fstream>
#include <sstream>
#include <vector>

int main() {
    std::ifstream file("input.txt");
    if (!file.is_open()) {
        std::cerr << "Error opening file!" << std::endl;
        return 1;
    }

    std::vector<int> a, b;
    std::string line;
    while (std::getline(file, line)) {
        std::istringstream iss(line);
        int num1, num2;
        if (!(iss >> num1 >> num2)) {
            std::cerr << "Error parsing line!" << std::endl;
            continue;
        }
        a.push_back(num1);
        b.push_back(num2);
    }

    file.close();

    return 0;
}

Ranges

for ranges, use -std=c++20 during compilation

1. std::ranges::count

   1. count(v, target2);

   2. Same as std::count(v.begin(), v.end(), target1);

2. std::ranges::sort Same as std::sort(vec2.begin(), vec2.end());

Initializer List

# TODO

explicit specifier

When a constructor is marked as [explicit](explicit specifier - cppreference.com), the compiler will not use that constructor for implicit type conversions or copy-initialization, meaning it will not automatically convert or create an object from a single argument when the compiler thinks it’s needed (for example, in assignments or function calls).

Example

#include <iostream>

class MyClass {
public:
    explicit MyClass(int x) { std::cout << "Constructor called with " << x << std::endl; }
};

void display(MyClass obj) {
    std::cout << "In display function\n";
}

int main() {
    // display(10);  // Error: no implicit conversion from int to MyClass
    display(MyClass(10));  // Must explicitly create MyClass
    return 0;
}

Insertion Sort

Code

template<typename T>
void insertionSort(T& vec)
{
    typename T::iterator first = vec.begin();
    typename T::iterator last = vec.end();

    for(auto current = first + 1; current != last; current++)
    {
       auto key = *current;
       //auto position = current - 1;
       auto position = current;
       for (; position != first && *(position-1) > key; position--)
       {
            //*(position+1) = *position;
            *(position) = *(position-1);
       }
       //*(position+1) = key;
       *(position) = key;
    }
    return;
}

References and Resources

• File Handling in CPP: A Quick and Easy Guide

• Ranges: the STL to the Next Level - Fluent C++

• explicit specifier - cppreference.com

• Sorting (Bubble, Selection, Insertion, Merge, Quick, Counting, Radix) - VisuAlgo

• Classes Part 22 - Curly brace versus parenthesis and std::initializer_list| Modern cpp Series Ep. 59

---

Day 2: Red-Nosed Reports

Solution Overview

Part 1: Count Safe Reports

For a given list of numbers check if

• it is all increasing or all decreasing.

• Any two adjacent values differ by at least one and at most three.

bool isReportSafe(const std::vector<int>& v)
{
	if (v.size() < 2) return true; // A single-element or empty list is trivially safe.

    bool isSafe = true;
    // Must be all increasing or all decreasing.
    bool isSorted = (v[0] > v[1]) // If first element is greater than second, assume it should be decreasing
        ? std::is_sorted(v.begin(), v.end(), std::greater<int>{})
        : std::is_sorted(v.begin(), v.end()); 

    if(!isSorted) return false;

    for (std::size_t i = 1; i < v.size(); ++i)
    {
       int diff = abs(v[i]-v[i-1]); 
       // Any two adjacent values must differ by at least one and at most three.
       if (!(1 <= diff && diff<= 3)) 
       {
            isSafe = false;
            break;
       }
    }
    // Could also use std::adjacent_find 
    return isSafe;
}

Part 2: Tolerate a Single Unsafe Report

For a given list, remove one element and check is it safe or not

bool part2(const std::vector<int>& vec)
    for(std::size_t i = 0; i < vec.size(); ++i)
    {
        std::vector<int> temp = vec;
        temp.erase(temp.begin() + i);
        if (isReportSafe(temp)) 
            return true;
    }

Key Takeaways

Vector of Vectors

std::is_sorted and std::greater

std::adjacent_find

Source: gcc-mirror/gcc/libstdc++-v3/include/bits/stl_algo.h

Code

bool funct(const std::vector<int>& vec)
    
    std::vector<int> v{1, 2, 3, 4, 4, 5};
	auto it = std::adjacent_find(v.begin(), v.end());
	if (it != v.end())
		std::cout << "First pair of repeated elements: " << *it << '\n';
		
	// Solution using adjacent_find
    return std::adjacent_find(v.begin(), v.end(), 
        [](int a, int b) {
            int diff = std::abs(b - a);
            return diff < 1 || diff > 3;
        }) == v.end();

std::count_if

Example

bool funct(const std::vector<int>& vec)
#include <algorithm>

bool isSorted = (v[0] > v[1]) 
	? std::is_sorted(v.begin(), v.end(), std::greater<>{})
	: std::is_sorted(v.begin(), v.end());

return std::adjacent_find(v.begin(), v.end(), 
	[](int a, int b) {
		int diff = std::abs(b - a);
		return diff < MIN_DIFF || diff > MAX_DIFF;
	}) == v.end();

static int countSafeReports(const std::vector<std::vector<int>>& reports)
{
	return std::count_if(reports.begin(), reports.end(), isReportSafe);
}

References and Resources

• How to split a string in C++ - Fluent C++

• std::is_sorted - cppreference.com

• std::adjacent_find - cppreference.com

• std::count, std::count_if - cppreference.com

---

Day 3: Mull It Over

Solution Overview

Part 1: Multiplication in Trash

Parse the given corrupted string and look for pattern mul(X,Y), where X and Y are each 1-3 digit numbers. Add up all the results of the multiplications of X and Y.

    std::istreambuf_iterator<char> it(file), end;
    auto first = it;
    auto second = it;

	std::vector<std::pair<int, int>> mulVec;
    unsigned int result(0);

    for(first = it; it != end; ++it)
    {
        if (
            (char)*first == 'm' && 
            (char)*++first == 'u' &&
            (char)*++first == 'l' &&
            (char)*++first == '('
        )
        {
            bool secondNum = false;
            unsigned int num1(0), num2(0); 

            for(second=++first; second != end; ++second)    
            {
                if(std::isdigit(*second))
                {
                    if(secondNum)
                        num2 = num2*10 + (*second - '0');
                    else
                        num1 = num1*10 + (*second - '0');
                }
                else if ((char)*second == ',')
                    secondNum=true;
                else if ((char)*second == ')' && secondNum)
                {
                    mulVec.emplace_back(num1, num2);
                    result += (num1*num2);
                    break;
                }
                else
                    break;
            }
        }

    }

Part 2: Do/Don’t Multiply

There are two new instructions you’ll need to handle:

• The do() instruction enables future mul instructions.

• The don't() instruction disables future mul instructions.

    std::regex regex(R"((mul\(\d{1,3},\d{1,3}\))|(do\(\))|(don't\(\)))");
    
    std::sregex_iterator start{ std::sregex_iterator(contents.begin(), contents.end(), regex) };
    std::sregex_iterator end{ std::sregex_iterator() };

    unsigned int result(0);

    int mulEnabled(1);
    for (auto i = start; i != end; ++i)
    {
        //std::smatch match = *i;
        std::string match = (*i).str();

        if(match.compare("don't()") == 0)
            mulEnabled = 0;

        else if (match.compare("do()") == 0) 
            mulEnabled = 1;

        else
        {
            int num1(0), num2(0);
            auto j = 4;
            while(match[j] != ',')
                num1 = num1*10 + (match[j++] - '0'); 
            ++j;
            while(match[j] != ')')
                num2 = num2*10 + (match[j++] - '0');

            result += (mulEnabled * num1 * num2); 
        }
    }

Concepts Learned

Stream Iterators

1. std::istream_iterator

2. std::istreambuf_iterator

3. Read entire file into a string:

std::ifstream file(filename);
std::string contents{std::istreambuf_iterator<char>(file), std::istreambuf_iterator<char>()};
// or into a vector of strings

Tip

When reading characters, std::istream_iterator skips whitespace by default (unless disabled with std::noskipws or equivalent), while std::istreambuf_iterator does not. In addition, std::istreambuf_iterator is more efficient, since it avoids the overhead of constructing and destructing the sentry object once per character.

Regex

1. std::cmatch for std::match_results<const char*>: cannot be used with iterators.

2. std::smatch for std::match_results<std::string::const_iterator> to be used with iterators

3. std::regex_match for entire sequence

4. std::sregex_iterator for iterators, which dereferences to std::smatch

Examples

std::regex pattern(R"(mul\(\d{1,3},\d{1,3}\))");
std::string input = "mul(123,456)";
if (std::regex_match(input, pattern)) {
    std::cout << "Pattern matched!" << std::endl;
}

    std::regex regex(R"((mul\(\d{1,3},\d{1,3}\))|(do\(\))|(don't\(\)))");
    
    std::sregex_iterator start{ std::sregex_iterator(contents.begin(), contents.end(), regex) };
    std::sregex_iterator end{ std::sregex_iterator() };

    for (auto i = start; i != end; ++i)
    {
        std::smatch match = *i;
        cout << match.str() << endl;
    }

Strings

1. std::string::compare: returns 0 if equal

2. std::stoi - String to Int

References and Resources

• std::regex_iterator - cppreference.com

• C++ Weekly - Ep 62 - std::regex

---

Day4: Ceres Search

Solution Overview

Part 1: Find XMAS

Find the word XMAS in a given grid such that it can be horizontal, vertical, diagonal, written backwards, or even overlapping other words.

std::vector<std::string> contents = {std::istream_iterator<std::string>(file), std::istream_iterator<std::string>()};

int rows{(int)contents.size()}, cols{(int)contents[0].size()};
std::stringstream ss{};
for (auto j = 0; j < cols; ++j)
{
	for(auto i = 0; i < rows; ++i)
		ss << contents[i][j];
	ss << endl;
}
std::vector<std::string> rotated_contents = {std::istream_iterator<std::string>(ss), std::istream_iterator<std::string>()};

std::vector<std::string> diag_contents = get_diagonals(contents);
std::vector<std::string> antidiag_contents = get_diagonals(contents, true);

int result(0);
std::string xmas("XMAS");
std::string samx("SAMX");

// repeat this for rotated_contents, diag_contents, antidiag_contents
for(const auto& line : contents)
{
	for(auto i = 0; i <= line.size() - xmas.size(); ++i)
	{
		if(line.substr(i, xmas.size()) == xmas || line.substr(i, samx.size()) == samx)
			++result;
	}
}

Part 2: Find MAS in X

find two MAS in the shape of an X.

M.S
.A.
M.S

    std::vector<std::string> grid = {std::istream_iterator<std::string>(file), std::istream_iterator<std::string>()};

    int rows{(int)grid.size()}, cols{(int)grid[0].size()};
    int result = 0;

    for (int i = 1; i < rows - 1; ++i)
    {
        for(int j = 1; j < cols - 1; ++j)
        {
            if(grid[i][j] != 'A') continue;
            result += checkMAS(grid, i, j, 1) & checkMAS(grid, i, j, -1);
        }
    }

bool checkMAS(const std::vector<std::string>& grid, int x, int y, int dir)
{
    char c1 = grid[x-1][y-dir];
    char c2 = grid[x+1][y+dir];

    if(c1 == 'M' && c2 == 'S') return true;
    if(c2 == 'M' && c1 == 'S') return true;
    return false;
}

Key Takeaways

Get diagonals and AntiDiagonals of a 2d Grid

Number of diagonals in a Grid: \(R+C−1\)

Code

    int maxRow = 4;
    int maxCol = 3;

    std::cout << "Diagonals" << std::endl;
    int row_start = 0; // begin a 0
    int col_start = maxCol - 1; // begins at maxCol - 1 (the top-right corner).
    for (int i = 0; i < maxCol+maxRow-1; ++i) // Iterates over all diagonals
    { 
        std::cout << i << ": ";
        row_start += (col_start < 0); // clever mechanism to shift the starting row when column goes out of bounds using `(col_start < 0)`
        // row_start increments by 1 when col_start goes out of bounds (negative), ensuring the starting row shifts downward.
        for (
	        int j = std::max(0, col_start), k = row_start;  // `j` for columns and `k` for rows
	        j < maxCol && k < maxRow;
	        ++j, ++k
	    )
        {
                std::cout << k << "," << j << " ";
        }
        --col_start;  // shifts the starting column leftward
        std::cout << std::endl;
    }

    std::cout << "Anti Diagonals" << std::endl;
    row_start = 0;
    col_start = 0; //begins at 0 (the top-left corner)
    for (int i = 0; i < maxCol+maxRow-1; ++i) // Iterates over all diagonals
    { 
        std::cout << i << ": ";
        row_start += (col_start > maxCol-1); // shift row when col goes out of bounds
        for (
	        int j = std::min(maxCol-1, col_start), k = row_start;
	        j > -1 && k < maxRow;
	        --j, ++k
	    )
        {
                std::cout << k << "," << j << " ";
        }
        ++col_start; // shifts the starting column rightward
        std::cout << std::endl;
    }

/**
Output:
Diagonals
0: 0,2
1: 0,1 1,2     
2: 0,0 1,1 2,2 
3: 1,0 2,1 3,2 
4: 2,0 3,1     
5: 3,0
Anti Diagonals 
0: 0,0
1: 0,1 1,0     
2: 0,2 1,1 2,0 
3: 1,2 2,1 3,0
4: 2,2 3,1
5: 3,2

*/

Another method suggested by Claude 3.5 Instead of managing starting points (row_start, col_start) explicitly, we can use the property that:

• Diagonal Index: \(k=j−i\)

• Anti-Diagonal Index: \(k=i+j\)

• Diagonals will have \(k\in[−(C−1),(R−1)]\).

• Anti-diagonals will have \(k\in[0,R+C−2]\).

Code

void printDiagonals(int maxRow, int maxCol) {
    std::cout << "Diagonals\n";
    // For main diagonals, points on same diagonal have same (row-col)
    for (int diff = -(maxCol-1); diff < maxRow; ++diff) {
        std::cout << (diff + maxCol - 1) << ": ";
        int startRow = std::max(0, diff);
        int startCol = startRow - diff;
        for (int row = startRow; row < maxRow && (row-diff) < maxCol; ++row) {
            std::cout << row << "," << (row-diff) << " ";
        }
        std::cout << '\n';
    }
    
    std::cout << "\nAnti Diagonals\n";
    // For anti-diagonals, points on same diagonal have same (row+col)
    for (int sum = 0; sum < maxRow + maxCol - 1; ++sum) {
        std::cout << sum << ": ";
        int startRow = std::max(0, sum - (maxCol - 1));
        int endRow = std::min(maxRow - 1, sum);
        for (int row = startRow; row <= endRow; ++row) {
            std::cout << row << "," << (sum-row) << " ";
        }
        std::cout << '\n';
    }
}

---

Day 5: Print Queue

Solution Overview

Part 1: Check Page Order

The first part of the input contains a list of rules X|Y which states that page X must be printed at some point before page Y. Next part have list page sequences, identify which one of those are in right order.

• Iterates over all pairs (i,j) in the given page sequence where i<j.

• For each pair, checking if it violates any of the rule.

• If no rule violated, the order is valid.

bool isOrderValid(const std::vector<int>& o, const std::vector<std::pair<int, int>>& rule)
{
    for(auto i=0; i < o.size()-1; ++i) 
    {
        for(auto j=i+1; j < o.size(); ++j) 
        {
            for(auto r: rule)
            {
                if(o[i] == r.second && o[j] == r.first) // incorrect order
                {
                    return false;
                }
            }
        }
    }
    return true;
}

This is simply brute force and horrible time complexity of \(O(n^2\times m)\) where \(n\) is length of the page sequence (o.size()) and \(m\) is a number of rules(rule.size()).

Optimization

A faster solution (aoc2024/day05/solution.cpp at main · UnicycleBloke/aoc2024) (\(O(m\times n)\)) I found on Reddit:

• Iterate over rules \(O(m)\)

• For each rule find its first and second elements in the sequence (Linear search \(O(2n)\), might short-circuit if first not found \(O(n + n/2)\))

• check if both finds follow the rule, based on position they appear in the sequence. Else order is not valid.

More optimized solution would require a Hash Map:

bool isOrderValid(const std::vector<int>& o, const std::vector<std::pair<int, int>>& rule) {
    // Map elements in 'o' to their indices for fast lookup
    std::unordered_map<int, int> indexMap;
    for (int i = 0; i < o.size(); ++i) {
        indexMap[o[i]] = i;
    }

    // Check if any rule is violated
    for (const auto& r : rule) {
        if (indexMap.find(r.first) != indexMap.end() && indexMap.find(r.second) != indexMap.end()) {
            if (indexMap[r.first] > indexMap[r.second]) {
                // Order is invalid
                return false;
            }
        }
    }
    // Order is valid
    return true;
}

Bringing time complexity to \(O(m + n)\) as find in Hash Map is \(O(1)\) operation.

// Before
[100%] Built target day5
Part 1: 4790
Elapsed time: 7330 us
Part 2: 6319
Elapsed time: 9174 us
Elapsed time: 45189 us
[100%] Built target run5

// After Optimization
[100%] Built target day5
Part 1: 4790
Elapsed time: 2959 us
Part 2: 6319
Elapsed time: 3437 us
Elapsed time: 38453 us
[100%] Built target run5

Part 2: Sort pages as per the Rules

For each of the incorrectly-ordered updates, use the page ordering rules to put the page numbers in the right order.

Implemented a bubble sort such that while not sorted keep swapping the elements in incorrect order.

    for(std::vector<int>& o: incorrect_orders)
    {
        while(true)
        {
            bool order_valid = true;

            for(auto r: rule)
            {
                auto i=0,j=0;
                for(; i < o.size(); ++i)
                {
                    if(o[i] == r.first) break;
                }
                for(; j < o.size(); ++j) 
                {
                    if(o[j] == r.second) break;
                }

                if(i==o.size() || j==o.size()) continue;
                if(j < i)
                {
                    std::swap(o[i], o[j]);
                    order_valid = false;
                    break;
                }
            }

            if (order_valid) break;
        }
    }

Found a more optimized solution on the internet, using \(m\times m\) matrix data, where data[a][b] == true means a must come before b in any valid sequence.

Key Takeaways

std::move

Example std::move

    while (std::getline(file, line))
    {
        cout << line << endl;
        std::istringstream iss(line);
        std::vector<int> pages;
        std::string page;
        while (std::getline(iss, page, ','))
        {
            pages.push_back(std::stoi(page));
        }
        order.push_back(std::move(pages));
        /**
        order.push_back(pages);           - Copies the content of pages into order
        order.push_back(std::move(pages)) - Moves the contents, no copying, ownership of interal is transferred
        When use std::move??
        -> Don't need the original vector
        -> performance optimization
		*/
    }

for loop with pass by reference

1. for(auto o: orders): The copy constructor of std::vector<int> is called for each element of orders.

    for(std::vector<int> o: incorrect_orders)
    {
	    // some condition
        std::swap(o[i], o[j]);
	}
	// Does NOT change the original
	// Instead do
	for(std::vector<int>& o: incorrect_orders) 

Fun

Ref: 2024 Day 5 (part 2) Non-Transitivity, non-schmansitivity : r/adventofcode

References and Resources

• C++ Weekly - Ep 278 - emplace_back vs push_back

---

This marks the first part of my Advent of Code 2024 journey. More updates soon. If you’re also learning C++ or participating in Advent of Code, I’d love to hear about your experiences! Share your thoughts, tips, or solutions in the comments below. Feel free to check out my solutions on GitHub.”

Hello World Advent of Code 2024 [Days 6 to 8]

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