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FAANG interview question in Rust

I recently got this question on a FAANG interview and decided to try and solve it in Rust:
// Given an MxN array of “boxes”, where each box contains some number of coins C[i][j], // you want to maximize the number of coins you can take. You take coins by traversing // row by row, taking all of the coins from ONE box in each row. However, any time you // change the index of the box you take coins from, you must pay a “change fee” equal // to ABS(x - y) where x and y are the previous and new row indices. Write a function // that can determine the optimal set of boxes to take coins from in order to maximize // your profit after change fees 
My first though is that there are a few ways we could attack this problem:
  • Naively find all paths and the value of each path (slow)
  • DFS to find possible solutions (not sure about improvement over the above)
  • Greedily determine the next possible best step (not forward thinking enough)
  • Try and determine best profit at each index and back out the path to get there
I implemented a (faulty) greedy version of the algorithm first:
fn find_max_coins(grid: Vec>) -> Vec { let mut solution: Vec = vec![]; let mut prev_idx = &mut 0usize; // Iterate through the grid for row in 0..grid.len() { // For readability let current_row = &grid[row]; // For the initial setup, find the best option in the first row if row == 0 { let mut best_final_index: usize = 0; let mut max_found: i16 = 0; for possible_best_index in 0..current_row.len() { *prev_idx = possible_best_index; // Reset after each loop because `possible_option` writes over it let possible_option = prev_row_max( &grid[row + 1], current_row[possible_best_index], &mut prev_idx, ); if possible_option + current_row[possible_best_index] > max_found { max_found = possible_option + current_row[possible_best_index]; best_final_index = possible_best_index; } } *prev_idx = best_final_index; solution.push(current_row[best_final_index]); continue; } // For all subsequent rows, greedily pick the best path solution.push(prev_row_max( current_row, grid[row - 1][*prev_idx], &mut prev_idx, )); } solution } fn prev_row_max(current_row: &Vec, target: i16, prev_idx: &mut usize) -> i16 { let mut current_max: i16 = 0; // Don't modify the heap value until we have a value to replace it with let mut local_max_idx: usize = *prev_idx; for col in 0..current_row.len() { // Determine value less change fee let value = current_row[col] + target - (*prev_idx as i16 - col as i16).abs(); if value > current_max { current_max = value; local_max_idx = col; } } // Copy the final value back *prev_idx = local_max_idx; return current_row[*prev_idx]; } 
This should not work but it surprisingly passed all the test cases they gave me. I bought up that it did not actually work, as it does not solve this matrix:
vec![ vec![1, 1, 1, 1, 1, 1, 1], vec![1, 1, 1, 1, 1, 1, 1], vec![1, 1, 1, 1, 1, 1, 1], vec![1, 1, 1, 1, 1, 1, 2], vec![1, 1, 1, 1, 1, 1, 1000], ]; 
My next thought was to try and calculate the best profit we can get at each spot, which is a simpler function:
fn find_max_coins(grid: Vec>) -> Vec { let rows = grid.len(); let cols = grid[0].len(); let fill_zero = || -> Vec> { (0..rows).map(|_| (0..cols).map(|_| 0).collect()).collect() }; let mut step_profit: Vec> = fill_zero(); let mut total_profit: Vec> = grid.to_vec(); for row in 1..rows { // Set seed values for possible solutions step_profit[row][0] = total_profit[row - 1][0]; // from left-hand side step_profit[row][cols - 1] = total_profit[row - 1][cols - 1]; // from right-hand side // Find left max profit for col in 1..cols { step_profit[row][col] = max(step_profit[row][col - 1] - 1, total_profit[row - 1][col]); } // Find right max profit for col in (0..cols - 1).rev() { step_profit[row][col] = max(step_profit[row][col], max(step_profit[row][col + 1] - 1, total_profit[row - 1][col])); } // Update profit for col in 0..cols { total_profit[row][col] = grid[row][col] + step_profit[row][col]; } } println!("Profit: {:?}", total_profit); // no solution yet... } 
Given the following matrix:
let grid = vec![ vec![0, 0, 10, 0, 0], vec![0, 2, 0, 3, 0], vec![5, 0, 0, 0, 3], ]; 
This generates the best profit at each cell correctly:
[ [ 0, 0, 10, 0, 0], [ 8, 11, 10, 12, 8], [15, 11, 11, 12, 14] ] 
However, I cannot think of a way to back out the path from this result without running into the same problem as my greedy algorithm, as knowing the profits at each step does not tell you the path needed to get there.
The solution should be 10 -> 2 -> 5, but I do not see how an algorithm would back that out of this set of optimums. Without knowing path to the profit at a given spot, there is not a way to determine what path led to the 15.
I wish I could get closure on this problem, but they would not tell me what the correct solution was. What would the solution be here? It felt way, way too complex to solve for a 20 minute whiteboard.
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[S] [USA-MD] Expired 35mm film cartridges — mixed color print C-41, Kodachrome K-14, and mixed chrome E-6 films — all kept in cold storage

I am getting out of photography — both film and digital — and am selling off everything that I own... cameras, film, lenses, strobes, hand meters, camera bags, etc. Everything must go.
Reason for selling: I have had zero income since December of last year. As a result, I need to liquidate the remainder of my personal items to cover our upcoming mortgage payments. I’ve been selling on-and-off previously here on Photomarket and WatchExchange, and have positive feedback.
In this post, I’m selling off my cache of expired 35mm color print C-41, Kodachrome K-14, and chrome E-6 film cartridges; I will have a separate posting of expired B&W cartridges and bulk films later after I get this collection out the door. I’ve bought my film from a number of different sources, and all of it has been kept refrigerated for the past many years.
Please note: With just one exception, none of this film is in a box with the expiration date; as such, some of the film is relatively recent (less than 6-years old), while the bulk of it is 10+ years and older. The color film that I’ve shot over the years is still good; no color shifts and only moderate fading, which can be addressed in post-processing.
My pricing for each individual cartridge is as follows:
  • 12 exposure cartridges — $3
  • 24 exposure cartridges — $4
  • 27 exposure cartridges — $4
  • 36 exposure cartridges — $5
  • Kodachrome cartridges — $10
Shipping is CONUS only via USPS Priority Mail flat rate box, sized appropriately for the amount of film ordered. Multiple batches can be combined for a savings on shipping. Insurance is extra. Please do not ask me to ship outside of the US; I’ve gotten burned by international buyers in the past and will not do that again. PayPal only — no trades, exchanges, or returns.
Proof photos here: https://imgur.com/gallery/hBEVIls
Mixed B&W print film, C-41 processing batch — $69 + shipping
  • 1 — Ilford XP2 Super 400 C-41 24exp
  • 1 — Ilford XP2 Super 400 C-41 36exp
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  • 6 — Kodak Black & White 400 C-41 24exp (assorted packaging)
Kodachrome slide film, K-14 processing batch — $30 + shipping You can get it processed in COLOR here: https://www.shootfilmco.com/blogs/shoot-film-co/kodachrome-processed-in-color-seriously
  • 2 — Kodak Kodachrome 64 K-14 36exp
  • 1 — Kodak Kodachrome 25 K-14 36exp
SOLD Mixed chrome slide film, E-6 processing batch — $55 + shipping
  • 5 — Kodak Elite Chrome 100 E-6 36exp
  • 6 — Fujichrome 100 E-6 36exp
SOLD Mixed color print film, C-41 processing batch — $85 + shipping
  • 5 — Fujicolor Superia 200 C-41 36exp
  • 12 — Kodak Ektar 100 C-41 36exp
SOLD Kodak color print film, C-41 processing batch — $100 + shipping
  • 2 — Kodak Portra 400 C-41 36exp
  • 10 — Kodak Portra 160 C-41 36exp
  • 8 — Kodak Ektar 100 C-41 36exp
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Mixed off-brand color print film, C-41 processing batch — $66 + shipping
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Kodak 200 color print film, C-41 processing batch — $156 + shipping
  • 21 — Kodak Gold 200 C-41 24exp
  • 3 — Kodak Gold 200 C-41 24 (sealed box) Expiration 01/2008
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  • 7 — Kodak 400 C-41 24exp
  • 11 — Kodak MAX 400 C-41 24exp
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  • 2 — Kodak MAX 400 C-41 36exp
  • 1 — Kodak Kodacolor 400 C-41 36exp
Kodak 800 color print film, C-41 processing batch — $44 + shipping
  • 11 — Kodak Gold MAX 800 C-41 24exp
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