I encountered this question during an interview for a popular software company. I wasn’t able to reach a satisfactory solution to it during the course of the interview (and although I cleared that round, I didn’t get the role). However, I felt sufficiently close to the solution that I was tempted to work on it after the interview was over.

## Problem Statement#

You are given chess board (of arbitrary size $M \times N$) where each square has a certain non-negative value. A single rook is placed on the board. The rook can move horizontally or vertically as in regular chess. However, it can only move in this fashion to squares that have a strictly lower value than the square it starts from. Given the rook’s starting square, find the length of the longest path it can take on the board.

Here’s a sample board of size $6 \times 5$ –

23145
67921
35719
88876
94258
21479

Let’s say the rook is located at coordinates $(3, 0)$.

23145
67921
35719
8876
94258
21479

Given its starting square is of value $8$, it can only move (vertically or horizontally) to squares that are of value strictly lower than $8$.

3145
7921
5719
4258
1479

The objective is to find the longest path the rook can take on the board. It suffices to simply return the length.

## Approach#

This immediately felt like a dynamic programming problem. From a given state (of where the rook is located), you can call an imaginary routine $\text{longestPathFrom(rook)}$ to find the length of the longest path from that state. This is independent of how you arrived at that state. Thus, we can formulate a preliminary recurrence relation as –

$$\text{longestPathFrom(state)} = 1 + \text{longestPathFrom}(\text{state}')$$

for all states that are immediately reachable from the current state. Given how the rook moves, these immediately reachable squares are those on the same rank or file as the rook could be moved to in a single move (provided their value was strictly less than the value of the rook’s square).

However, there may be multiple ways of reaching the same square, and some may be longer than others. Hence, we need to modify the recurrence relation to –

$$\text{longestPathFrom(state)} = \max_{\text{state}'\ \in \text{ reachable}}\left(0, 1 + \text{longestPathFrom}(\text{state}')\right)$$

## Solution#

A simple algorithm to implement this is –

$M$: the number of rows in the board
$N$: the number of columns in the board
$\text{values}[M][N]$: the values of each cell in the board that govern reachability
$(i, j)$: the (row, column) of the cell from which we want to find the longest path
Initialize $\text{dp}[i][j] \leftarrow \varnothing, \forall i \in [1, M], j \in [1, N]$

$\text{longestPathFrom}(i, j)$
$\quad$if $\text{dp}[i][j] == \varnothing$ then
$\qquad d_{\max} \leftarrow \max_{i’ \in [1, M]}(0, 1 + \text{longestPathFrom}(i’, j))$
$\qquad d_{\max} \leftarrow \max_{j’ \in [1, N]} (0, 1 + \text{longestPathFrom}(i, j’))$
$\qquad \text{dp}[i][j] \leftarrow d_{\max}$
$\quad$endif
return $\text{dp}[i][j]$

Here’s the algorithm implemented as a full solution in Python –

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29  def print_matrix(matrix): print('\n'.join(['\t'.join([str(cell) for cell in row]) for row in matrix])) def max_dist(values, to_i, to_j, dp): """ max_dist(i, j) returns the max length path from coordinates (i, j) """ if dp[to_i][to_j] is None: curr_max_dist = 0 # default unexplored max distance (no reachable squares) for k in range(len(values[0])): if values[to_i][k] < values[to_i][to_j]: curr_max_dist = max(curr_max_dist, 1 + max_dist(values, to_i, k, dp)) for k in range(len(values)): if values[k][to_j] < values[to_i][to_j]: curr_max_dist = max(curr_max_dist, 1 + max_dist(values, k, to_j, dp)) dp[to_i][to_j] = curr_max_dist return dp[to_i][to_j] if __name__ == '__main__': with open('input.txt', 'r') as inputfile: n, m = map(int, inputfile.readline().split()) values = [[] for _ in range(n)] for i in range(n): values[i] = list(map(int, inputfile.readline().split())) u, v = map(int, inputfile.readline().split()) dp = [[None] * m for _ in range(n)] ans = max_dist(values, u, v, dp) # print_matrix(dp) print(ans) 

Here are the contents of a sample input.txt file you can use –

 1 2 3 4 5 6 7 8  6 5 2 3 1 4 5 6 7 9 2 1 3 5 7 1 9 8 8 8 7 6 9 4 2 5 8 2 1 4 7 9 3 0 

The correct answer is $7$. Here’s another one

 1 2 3 4 5 6 7 8  6 5 4 4 4 4 4 5 5 5 5 5 6 6 6 6 5 7 7 7 6 5 8 8 7 6 5 9 8 7 6 5 5 0 

The correct answer is $5$.