Daily Challenge #105 - High-Sum Matrix Drop

You have a square matrix of unsigned integer values. Say, 5 + 5.
Each row of that matrix contains numbers from 1 to 5 (or to whatever length of the matrix), randomly sorted.

You can move through the matrix only one step at the time - either straight down, left-down or right-down.

Your challenge is to write a code that will start at the middle of the top row, and find the path down with the highest sum.

Example:
For the following matrix, you start in c/0. For this matrix the best path would be: c/0, d/1, d/2, e/3, d/5 which scores 4+4+5+4+5 = 22.

+---+---+---+---+---+---+
|   | a | b | c | d | e |
+---+---+---+---+---+---+
| 0 | 2 | 1 | 4 | 5 | 3 |
+---+---+---+---+---+---+
| 1 | 5 | 2 | 1 | 4 | 3 |
+---+---+---+---+---+---+
| 2 | 1 | 3 | 2 | 5 | 4 |
+---+---+---+---+---+---+
| 3 | 1 | 5 | 2 | 3 | 4 |
+---+---+---+---+---+---+
| 4 | 3 | 2 | 1 | 5 | 4 |
+---+---+---+---+---+---+

Good luck!

---

This challenge comes from peledzohar here on DEV. Want to propose a challenge idea for a future post? Email [e-mail] with your suggestions!

Comments

[Craig McIlwrath]

I didn't do #90, so I'll put my answer here. This is a really stupid, slow, "just works" algorithm. It checks every single possible path. I guess it's around O(n^3)?

(Language is Haskell).

data Move = D | DL | DR deriving (Show) 

type Size = Int
type Row = Int
type Col = Int
type Pos = (Row, Col) 

isInSquare :: Size -> Pos -> Bool
isInSquare n (r, c) = all id
  [ r >= 0
  , r < n
  , c >= 0
  , c < n
  ]

executeMove :: Move -> Pos -> Pos
executeMove D  (r, c) = (r + 1, c) 
executeMove DL (r, c) = (r + 1, c - 1)
executeMove DR (r, c) = (r + 1, c + 1)

validMove :: Size -> Pos -> Move -> Bool
validMove n p m = isInSquare n $ executeMove m p

possibleMoves :: Size -> Pos -> [Move]
possibleMoves n p = filter (validMove n p) [D, DL, DR]

generatePaths :: Size -> Pos -> [[Move]]
generatePaths n p
  | fst p == n - 1 = [[]]
  | otherwise      = possibleMoves n p >>= nextMoves
  where nextMoves :: Move -> [[Move]] 
        nextMoves m = generatePaths n (executeMove m p) >>= \ms -> [m : ms] 

-- Middle spot of an even array is defined as the right middle for simplicity. 
middle n
  | odd n     = n `div` 2
  | otherwise = n `div` 2

access :: [[a]] -> Pos -> a
access xss (r, c) = xss !! r !! c

sumIn :: (Num a) => [[a]] -> Pos -> [Move] -> a 
sumIn xss start path = let ps = scanl (flip executeMove) start path
                       in  sum $ map (access xss) ps 

maxBy :: (Ord b) => (a -> b) -> [a] -> a
maxBy f = foldl1 maxOf 
  where maxOf m x = if (f x > f m) then x else m

bestPath :: (Ord a, Num a) => Size -> [[a]] -> ([Move], a)
bestPath n xss = let start = (0, middle n)
                     paths = generatePaths n start 
                 in  maxBy snd $ zip paths $ map (sumIn xss start) paths

[sinx]

My solution using queue. Should be o(n).

import string

def get_steps(i, j, row, col):
    left = j - 1
    right = j + 1
    down = i + 1

    if down >= row:
        return []

    steps = []
    if left >= 0:
        steps.append((down, left))

    steps.append((down, j))

    if right < col:
        steps.append((down, right))

    return steps

def make_and_fill(row, col, fill):
    ret = []
    for _ in range(row):
        row_ = [fill] * col
        ret.append(row_)

    return ret

def find_max_index(xs):
    max_ = max(xs)
    return xs.index(max_)

def get_path(i, j, backtraces):
    ret = []
    queue = [(i, j)]

    while queue:
        pos = queue.pop(0)
        if pos:
            ret.append(pos)
            i, j = pos
            backtrace = backtraces[i][j]
            queue.append(backtrace)

    ret.reverse()
    ret = [f'{string.ascii_letters[j]}/{i}' for (i, j) in ret]
    return ', '.join(ret)


def solve(xs):
    row, col = len(xs), len(xs[0])
    sums = make_and_fill(row, col, 0)
    sums[0][:] = xs[0]
    backtraces = make_and_fill(row, col, 0)
    queue = [(0, col//2)]

    while queue:
        pos = queue.pop(0)
        i, j = pos
        for step in get_steps(i, j, row, col):
            step_i, step_j = step
            acc = sums[i][j] + xs[step_i][step_j]
            if sums[step_i][step_j] < acc:
                queue.append((step_i, step_j))
                sums[step_i][step_j] = acc
                backtraces[step_i][step_j] = pos

    max_index = find_max_index(sums[-1])
    print(sums[-1][max_index])

    path = get_path(row-1, max_index, backtraces)
    print(path)


if __name__ == '__main__':
    matrix = [
        [2, 1, 4, 5, 3],
        [5, 2, 1, 4, 3],
        [1, 3, 2, 5, 4],
        [1, 5, 2, 3, 4],
        [3, 2, 1, 5, 4],
    ]
    solve(matrix)

[Michael Tharrington]

Thanks for pointing this out, David! This was a mistake.

[Zohar Peled]

Hey guys, Someone got mixed up a bit here. That was the 90# challenge, this one was suppose to be the one I've sent you later, with the bottles... :-)

[Michael Tharrington]

Ah jeez... my bad here. 🤦‍♂️

You are absolutely right, Zohar!