Longest Common Subsequence in C++

Longest Common

Subsequence in C++

Welcome to the Longest Common Subsequence in C++ page! Here, you'll find the source code for this program as well as a description of how the program works.

Current Solution

#include <algorithm>
#include <charconv>
#include <concepts>
#include <iostream>
#include <optional>
#include <ranges>
#include <string_view>
#include <utility>
#include <vector>

namespace views = std::views;
namespace ranges = std::ranges;

[[noreturn]] void usage() {
    std::cerr
        << R"(Usage: please provide two lists in the format "1, 2, 3, 4, 5")"
        << '\n';
    std::exit(1);
}

static constexpr std::string_view ws = " \t\n\r\f\v";
constexpr std::string_view trim(std::string_view s) {
    const auto start = s.find_first_not_of(ws);
    if (start == std::string_view::npos) return "";
    s.remove_prefix(start);

    const auto end = s.find_last_not_of(ws);
    s.remove_suffix(s.size() - 1 - end);
    return s;
}

std::optional<int> to_int(std::string_view s) {
    int value{};
    auto [ptr, ec] = std::from_chars(s.data(), s.data() + s.size(), value);
    return (ec == std::errc{} && ptr == s.data() + s.size())
               ? std::make_optional(value)
               : std::nullopt;
}

std::optional<std::vector<int>> parse_vec(std::string_view s) {
    auto pipe = s | views::split(',') | views::transform([](auto&& r) {
                    return std::string_view{
                        std::addressof(*ranges::begin(r)),
                        static_cast<std::size_t>(ranges::distance(r))};
                }) |
                views::transform(trim) | views::transform(to_int);

    std::vector<int> out;
    for (auto&& opt : pipe) {
        if (!opt) return std::nullopt;
        out.push_back(*opt);
    }
    return out.empty() ? std::nullopt : std::make_optional(out);
}

template <typename T>
std::vector<T> find_lcs(const std::vector<T>& a, const std::vector<T>& b) {
    const std::size_t m = a.size();
    const std::size_t n = b.size();

    auto idx = [n](std::size_t i, std::size_t j) { return i * (n + 1) + j; };

    std::vector<int> dp((m + 1) * (n + 1), 0);

    for (std::size_t i = 1; i <= m; ++i) {
        for (std::size_t j = 1; j <= n; ++j) {
            if (a[i - 1] == b[j - 1])
                dp[idx(i, j)] = dp[idx(i - 1, j - 1)] + 1;
            else
                dp[idx(i, j)] = std::max(dp[idx(i - 1, j)], dp[idx(i, j - 1)]);
        }
    }

    std::vector<T> result;
    result.reserve(dp.back());

    std::size_t i = m, j = n;

    while (i > 0 && j > 0) {
        if (a[i - 1] == b[j - 1]) {
            result.push_back(a[i - 1]);
            --i;
            --j;
        } else if (dp[idx(i - 1, j)] > dp[idx(i, j - 1)]) {
            --i;
        } else {
            --j;
        }
    }

    ranges::reverse(result);
    return result;
}

int main(int argc, char* argv[]) {
    if (argc != 3) usage();

    const auto a = parse_vec(argv[1]);
    const auto b = parse_vec(argv[2]);
    if (!a || !b) usage();

    const auto result = find_lcs(*a, *b);

    for (const char* sep = ""; int val : result) {
        std::cout << std::exchange(sep, ", ") << val;
    }
    std::cout << "\n";
}

Longest Common Subsequence in C++ was written by:

• Jeremy Grifski
• Shubham Tiwari
• Ștefan-Iulian Alecu

If you see anything you'd like to change or update, please consider contributing.

How to Implement the Solution

No 'How to Implement the Solution' section available. Please consider contributing.

How to Run the Solution

No 'How to Run the Solution' section available. Please consider contributing.