How should I practice Binary Maze Reachability?

Start by reading the full statement, solve the visible examples, then compare the pattern against related Amazon phone screen problems in FastPrep.

Is Binary Maze Reachability associated with Amazon?

FastPrep tags this as a Amazon phone screen coding practice problem when the problem catalog has that company and stage context.

Binary Maze Reachability

Amazon● EasyFULLTIMEPHONE SCREEN

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Problem statement

You are given a nonempty rectangular binary matrix maze. A cell containing 1 is passable, and a cell containing 0 is blocked.

The two arrays start and destination contain zero-based coordinates in the form [row, column]. From a passable cell, each move goes exactly one cell up, down, left or right to another passable cell inside the matrix.

Return true if a route from start to destination exists; otherwise return false. If either endpoint is blocked, return false. A passable cell can reach itself using zero moves.

Function

canReachMaze(maze: int[][], start: int[], destination: int[]) → boolean

Examples

Example 1

maze = [[1,1,0],[0,1,1],[1,0,1]]start = [0,0]destination = [2,2]return = true

One route is (0,0) -> (0,1) -> (1,1) -> (1,2) -> (2,2). Every visited cell contains 1, and each move joins orthogonally adjacent cells.

Example 2

maze = [[1,0,1],[0,1,0],[1,0,1]]start = [0,0]destination = [2,2]return = false

The start has no passable orthogonal neighbor. The diagonal passable cells do not connect it to the destination.

Example 3

maze = [[1,0,1],[1,1,1]]start = [0,0]destination = [0,2]return = true

A route goes down to (1,0), right through (1,1) and (1,2), then up to (0,2). Moving only right and down would miss this valid route.

Constraints

1 <= maze.length <= 40.
1 <= maze[row].length <= 40; every row has the same length.
Every cell is 0 or 1.
start.length == destination.length == 2.
Both coordinates name cells inside the matrix: 0 <= row < maze.length and 0 <= column < maze[0].length.