Trebuchet?! | Yuvrajsinh Gohil

Problem Overview

The newly-improved trebuchet calibration document has been "amended" by a young Elf. Each line contains a calibration value that we need to extract by combining the first and last digit to form a two-digit number. Our task is to sum all these calibration values.

Part 1: Finding Digits

Starting simple - I need to find the first and last digit in each line and combine them. The approach is straightforward: scan from the beginning to find the first digit, scan from the end to find the last digit, then combine them as a two-digit number.

What if there's only one digit? Well, it becomes both the first and last digit, so "a7b" gives us 77. Nice and clean!

#include<bits/stdc++.h>
using namespace std;

#define int long long

signed main() {
  string ch;
  fstream my_file;
  my_file.open("input.txt", ios::in);

  int ans = 0;

  while (true) {
      my_file >> ch;

      int first = 0, last = 0;

      for (int i = 0; i < ch.size(); ++i) {
          if (ch[i] - '0' <= 9 && ch[i] - '0' > 0) {
              first = ch[i] - '0';
              break;
          }
      }

      for (int i = ch.size() - 1; i >= 0; --i) {
          if (ch[i] - '0' <= 9 && ch[i] - '0' > 0) {
              last = ch[i] - '0';
              break;
          }
      }

      ans += (first * 10) + last;
      if (my_file.eof()) break;
  }

  cout << ans << endl;

  my_file.close();
  return 0;
}

Part 2: Spelled Out Numbers

Oh, the Elf was being creative! Turns out digits can also be spelled out: "one", "two", "three", etc. Now I need to handle both numeric digits and written numbers.

The clever bit here is handling overlapping patterns like "twone" (is it "two" + "one" or just looking for first and last?). My strategy: for the first digit, scan forward checking both numeric digits and spelled numbers. For the last digit, I reverse the search - and here's the trick: I also reverse the spelled numbers ("one" becomes "eno") to make the backward search elegant!

#include<bits/stdc++.h>
using namespace std;

#define int long long

signed main() {
  string ch;
  fstream my_file;
  my_file.open("input.txt", ios::in);

  map<string, int> conv = {
      {"one", 1},
      {"two", 2},
      {"three", 3},
      {"four", 4},
      {"five", 5},
      {"six", 6},
      {"seven", 7},
      {"eight", 8},
      {"nine", 9},
  };

  map<string, int> vnoc = {
      {"eno", 1},
      {"owt", 2},
      {"eerht", 3},
      {"ruof", 4},
      {"evif", 5},
      {"xis", 6},
      {"neves", 7},
      {"thgie", 8},
      {"enin", 9},
  };

  int ans = 0;

  while (true) {
      my_file >> ch;
      int first = 0, last = 0;

      for (int i = 0; i < ch.size(); ++i) {
          string f = "";
          bool found = false;
          if (ch[i] - '0' <= 9 && ch[i] - '0' > 0) {
              first = ch[i] - '0';
              break;
          }

          for (int j = i; j < i + 5; ++j) {
              f += ch[j];
              if (conv.find(f) != conv.end()) {
                  first = conv[f];
                  found = true;
                  break;
              }
          }
          if (found) break;
      }

      for (int i = ch.size() - 1; i >= 0; --i) {
          bool found = false;
          string l = "";
          if (ch[i] - '0' <= 9 && ch[i] - '0' > 0) {
              last = ch[i] - '0';
              break;
          }

          for (int j = i; j >= i - 5; --j) {
              l += ch[j];
              if (vnoc.find(l) != vnoc.end()) {
                  last = vnoc[l];
                  found = true;
                  break;
              }
          }
          if (found) break;
      }

      ans += (first * 10) + last;
      if (my_file.eof()) break;
  }

  cout << ans << endl;

  my_file.close();
  return 0;
}

The reversed words map (vnoc = conv backwards) is peak programming humor. Who needs reverse() when you can just hardcode "thgie" for 8?