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🎁 - 2023 DAY 5 SOLUTIONS -🎁

Day 5: If You Give a Seed a Fertilizer


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  • [JavaScript] Well that was by far the hardest out of all of the days, part 1 was relatively fine but part 2 took me awhile of trying different things

    Ended up solving it by working backwards by trying different location values and seeing if that can become a valid seed. Takes around 3 secs to compute the answer.

    Link to code

    Part 1 Code Block
    // Part 1
    // ======
    
    function part1(input) {
      const split = input.split("\r\n\r\n");
    
      let pastValues = split[0].match(/\d+/g).map((x) => parseInt(x));
      let currentValues = [];
    
      for (const section of split.slice(1)) {
        for (const line of section.split("\r\n")) {
          const values = line.match(/\d+/g)?.map((x) => parseInt(x));
    
          if (!values) {
            continue;
          }
    
          const sourceStart = values[1];
          const destinationStart = values[0];
          const length = values[2];
    
          for (let i = 0; i < pastValues.length; i++) {
            if (
              pastValues[i] >= sourceStart &&
              pastValues[i] < sourceStart + length
            ) {
              currentValues.push(destinationStart + pastValues[i] - sourceStart);
              pastValues.splice(i, 1);
              i--;
            }
          }
        }
    
        for (let i = 0; i < pastValues.length; i++) {
          currentValues.push(pastValues[i]);
        }
    
        pastValues = [...currentValues];
        currentValues = [];
      }
    
      return Math.min(...pastValues);
    }
    
    Part 2 Code Block
    // Part 2
    // ======
    
    function part2(input) {
      const split = input.split("\r\n\r\n");
    
      let seeds = split[0].match(/\d+/g).map((x) => parseInt(x));
      seeds = seeds
        .filter((x, i) => i % 2 == 0)
        .map((x, i) => [x, seeds[i * 2 + 1]]);
    
      const maps = split
        .slice(1)
        .map((x) => {
          const lines = x.split("\r\n");
          return lines
            .map((x) => x.match(/\d+/g)?.map((x) => parseInt(x)))
            .filter((x) => x);
        })
        .reverse();
    
      for (let i = 0; true; i++) {
        let curValue = i;
    
        for (const map of maps) {
          for (const line of map) {
            const sourceStart = line[1];
            const destinationStart = line[0];
            const length = line[2];
    
            if (
              curValue >= destinationStart &&
              curValue < destinationStart + length
            ) {
              curValue = sourceStart + curValue - destinationStart;
              break;
            }
          }
        }
    
        for (const [seedRangeStart, seedRangeLength] of seeds) {
          if (
            curValue >= seedRangeStart &&
            curValue < seedRangeStart + seedRangeLength
          ) {
            return i;
          }
        }
      }
    }
    
    • Ended up solving it by working backwards by trying different location values and seeing if that can become a valid seed.

      Huh, that's clever.

      • Turns out I got really lucky and my location value is much lower than most peoples which is why it can be solved relatively quickly

    • Torn between doing the problem backwards and implementing a more general case -- glad to know both approaches work out in the end!

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