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Problem018.js
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/**
* @file Provides solution for Project Euler Problem 18 - Maximum path sum I
* @author Eric Lavault {@link https://github.com/lvlte}
* @license MIT
*/
/**
* Problem 18 - Maximum path sum I
*
* @see {@link https://projecteuler.net/problem=18}
*
* 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 triangle = `
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
`
export const maxPathSum = function (grid = triangle) {
/**
* If we reduce the problem to its simplest form, considering :
*
* 7 -> The max sum depends on the two adjacent numbers below 7,
* 2 4 not 7 itself.
*
* obviously 4 > 2 therefore the max sum is 7 + 4 = 11
*
* 6
* Likewise, with : 4 6 6 > 4 therefore the max sum is 6 + 6 = 12
*
* Now, let's say we are given :
*
* 3
* 7 6
* 2 4 6
*
* and we decompose it into sub-problems such that each one fits the simple
* case above, we got :
*
* . . 3
* 7 . . 6 ? ?
* 2 4 . . 4 6 . . .
*
* Again, considering any number, the best path depends on the two adjacent
* numbers below it, not the number itself. That's why we have to compute
* the max sum from bottom to top, replacing each number with the sum of
* that number plus the greatest of the two adjacent numbers computed from
* the previous row.
*
* . . 3 15
* 11 . . 12 -> 11 12 -> x x
* x x . . x x x x x x x x
*
* We are simplifying a complicated problem by breaking it down into simpler
* sub-problems in a recursive manner, this is called Dynamic Programming.
*/
grid = grid.split(/\r\n|\n/).filter(l => l).map(r => r.split(' ').map(n => +n))
for (let i = grid.length - 2; i >= 0; i--) {
for (let j = 0; j < grid[i].length; j++) {
grid[i][j] += Math.max(grid[i + 1][j], grid[i + 1][j + 1])
}
}
return grid[0][0]
}