# Project Euler in PL/SQL: Problem 18

Solving problem 18 of Project Euler in PL/SQL.

## Problem 18:

### Maximum Path Sum 1

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

To get a feel of how to intelligently solve this problem without brute force, I found this method from another programmer and decided it would be the best way to go about it in PL/SQL as well.

Unfortunately, PL/SQL isn’t quite as succinct or pretty as Ruby, especially when it comes to loops and enumerables, so I had to put a function and procedure into my `EULER_PKG` to help keep things readable. Here are those new entries –

```FUNCTION get_num_from_list(line_in VARCHAR2, num_in INTEGER)
RETURN VARCHAR2
AS
--grabs a number from a line of numbers delimited by spaces
BEGIN
RETURN regexp_substr(line_in, '\S+', 1, num_in);
END;

PROCEDURE set_num_from_list(line_in IN OUT VARCHAR2, num_in INTEGER, new_val VARCHAR2)
AS
BEGIN
line_in := REGEXP_REPLACE(line_in, '\S+', new_val, 1, num_in);
END;

```

Pretty simple, just an easy way to retrieve a value from these lines of numbers, as well as update one number with another within the line.

For our solution, we’ll start at the 2nd to bottom row of the triangle. For each number, we’ll check one of the two “adjacent” numbers below it, take the larger number of the two, and add it to our original number. If we do this for every number and every row working up the pyramid, we’ll be left with the maximum possible value.

```DECLARE
TYPE triangle_array IS VARRAY(20) OF VARCHAR2(60);
triangle triangle_array := triangle_array();
new_num INTEGER;
BEGIN
triangle.EXTEND(15);
triangle(1) := '75';
triangle(2) := '95 64';
triangle(3) := '17 47 82';
triangle(4) := '18 35 87 10';
triangle(5) := '20 04 82 47 65';
triangle(6) := '19 01 23 75 03 34';
triangle(7) := '88 02 77 73 07 63 67';
triangle(8) := '99 65 04 28 06 16 70 92';
triangle(9) := '41 41 26 56 83 40 80 70 33';
triangle(10) := '41 48 72 33 47 32 37 16 94 29';
triangle(11) := '53 71 44 65 25 43 91 52 97 51 14';
triangle(12) := '70 11 33 28 77 73 17 78 39 68 17 57';
triangle(13) := '91 71 52 38 17 14 91 43 58 50 27 29 48';
triangle(14) := '63 66 04 68 89 53 67 30 73 16 69 87 40 31';
triangle(15) := '04 62 98 27 23 09 70 98 73 93 38 53 60 04 23';

FOR i IN REVERSE 1..triangle.count-1 LOOP --starting at the 2nd to bottom line
FOR x IN 1..REGEXP_COUNT(triangle(i),' ')+1 LOOP

--get the max value of the two numbers below current number in the triangle
SELECT MAX(column_value)
FROM table(sys.odcinumberlist(EULER_PKG.get_num_from_list(triangle(i+1),x),EULER_PKG.get_num_from_list(triangle(i+1),x+1)));

EULER_PKG.set_num_from_list(triangle(i), x,new_num);

END LOOP;

dbms_output.put_line(triangle(i));

END LOOP;
END;

/*output:
125 164 102 95 112 123 165 128 166 109 122 147 100 54
255 235 154 150 140 179 256 209 224 172 174 176 148
325 246 187 178 256 329 273 302 263 242 193 233
378 317 231 321 354 372 393 354 360 293 247
419 365 393 387 419 425 430 376 454 322
460 434 419 475 508 470 510 524 487
559 499 479 536 514 526 594 616
647 501 613 609 533 657 683
666 614 636 684 660 717
686 640 766 731 782
704 801 853 792
818 900 935
995 999
1074 */
```

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