/**
 * P18-Maximum Path Sum
 * 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)
 */
const val str = "75\n" +
        "95 64\n" +
        "17 47 82\n" +
        "18 35 87 10\n" +
        "20 04 82 47 65\n" +
        "19 01 23 75 03 34\n" +
        "88 02 77 73 07 63 67\n" +
        "99 65 04 28 06 16 70 92\n" +
        "41 41 26 56 83 40 80 70 33\n" +
        "41 48 72 33 47 32 37 16 94 29\n" +
        "53 71 44 65 25 43 91 52 97 51 14\n" +
        "70 11 33 28 77 73 17 78 39 68 17 57\n" +
        "91 71 52 38 17 14 91 43 58 50 27 29 48\n" +
        "63 66 04 68 89 53 67 30 73 16 69 87 40 31\n" +
        "04 62 98 27 23 09 70 98 73 93 38 53 60 04 23"
val rows: List<List<Int>> = str.split("\n").map { x -> x.split(" ").map { it.toInt() } }

fun p18() {
    print(getMax(0, 0, 0))
}

fun getMax(sum: Int, i: Int, j: Int): Int {
    if (i > rows.lastIndex) {
        println("Total: $sum")
        return sum
    }
    println("-> ${rows[i][j]} [$i,$j] sum: ${sum + rows[i][j]}")
    return maxOf(getMax(sum + rows[i][j], i+1, j), getMax(sum + rows[i][j], i+1,j+1))
}



