euler018.js

// By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

  //  3
  // 7 4
 // 2 4 6
// 8 5 9 3

// That is, 3 + 7 + 4 + 9 = 23.

// Find the maximum total from top to bottom of the triangle below:

              // 75
             // 95 64
            // 17 47 82
           // 18 35 87 10
          // 20 04 82 47 65
         // 19 01 23 75 03 34
        // 88 02 77 73 07 63 67
       // 99 65 04 28 06 16 70 92
      // 41 41 26 56 83 40 80 70 33
     // 41 48 72 33 47 32 37 16 94 29
    // 53 71 44 65 25 43 91 52 97 51 14
   // 70 11 33 28 77 73 17 78 39 68 17 57
  // 91 71 52 38 17 14 91 43 58 50 27 29 48
 // 63 66 04 68 89 53 67 30 73 16 69 87 40 31
// 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

// NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

var fs = require('fs');
var triangle = fs.readFileSync('euler018.txt').toString().split("\n");

for(i in triangle) {
    triangle[i] = triangle[i].match(/\d{2}/g)
    for (j in triangle[i]) {
        triangle[i][j] = parseInt(triangle[i][j]) // not entirely necessary in JS, but tidier.
    }
}

for (i=triangle.length-2; i>=0; i--) {
    for (j=0; j<=triangle[i].length-1; j++) {
        var value = triangle[i][j] + Math.max(triangle[i+1][j],triangle[i+1][j+1])
        triangle[i][j] = value;
    }
}

console.log(triangle[0][0])

// Since this problem mentions another, which can have a common solution, I decided to figure out an algorithm that works for both.
// I used this one: https://www.youtube.com/watch?v=6zcFB1nIoq8
// The idea is to take the values on the 2nd row, and check both the values below it, and add the largest of the two to the value.
// continue to do this all the way to the top, and the final value is the largest sum.

// If you want to track the path, you can use a binary number, where 0 represents taking the left number, and 1 represents the right.