C

R

```
/*
* Problem Statement:
* - Given a NxN grid, find whether rat in cell [0, 0] can reach the target in cell [N-1, N-1]
* - The grid is represented as an array of rows. Each row is represented as an array of 0 or 1 values.
* - A cell with value 0 can not be moved through. Value 1 means the rat can move here.
* - The rat can not move diagonally.
*
* Reference for this problem: https://www.geeksforgeeks.org/rat-in-a-maze-backtracking-2/
*
* Based on the original implementation contributed by Chiranjeev Thapliyal (https://github.com/chiranjeev-thapliyal).
*/
/**
* Checks if the given grid is valid.
*
* A grid needs to satisfy these conditions:
* - must not be empty
* - must be a square
* - must not contain values other than {@code 0} and {@code 1}
*
* @param grid The grid to check.
* @throws TypeError When the given grid is invalid.
*/
function validateGrid (grid) {
if (!Array.isArray(grid) || grid.length === 0) throw new TypeError('Grid must be a non-empty array')
const allRowsHaveCorrectLength = grid.every(row => row.length === grid.length)
if (!allRowsHaveCorrectLength) throw new TypeError('Grid must be a square')
const allCellsHaveValidValues = grid.every(row => {
return row.every(cell => cell === 0 || cell === 1)
})
if (!allCellsHaveValidValues) throw new TypeError('Grid must only contain 0s and 1s')
}
function isSafe (grid, x, y) {
const n = grid.length
return x >= 0 && x < n && y >= 0 && y < n && grid[y][x] === 1
}
/**
* Attempts to calculate the remaining path to the target.
*
* @param grid The full grid.
* @param x The current X coordinate.
* @param y The current Y coordinate.
* @param solution The current solution matrix.
* @param path The path we took to get from the source cell to the current location.
* @returns {string|boolean} Either the path to the target cell or false.
*/
function getPathPart (grid, x, y, solution, path) {
const n = grid.length
// are we there yet?
if (x === n - 1 && y === n - 1 && grid[y][x] === 1) {
solution[y][x] = 1
return path
}
// did we step on a 0 cell or outside the grid?
if (!isSafe(grid, x, y)) return false
// are we walking onto an already-marked solution coordinate?
if (solution[y][x] === 1) return false
// none of the above? let's dig deeper!
// mark the current coordinates on the solution matrix
solution[y][x] = 1
// attempt to move right
const right = getPathPart(grid, x + 1, y, solution, path + 'R')
if (right) return right
// right didn't work: attempt to move down
const down = getPathPart(grid, x, y + 1, solution, path + 'D')
if (down) return down
// down didn't work: attempt to move up
const up = getPathPart(grid, x, y - 1, solution, path + 'U')
if (up) return up
// up didn't work: attempt to move left
const left = getPathPart(grid, x - 1, y, solution, path + 'L')
if (left) return left
// no direction was successful: remove this cell from the solution matrix and backtrack
solution[y][x] = 0
return false
}
function getPath (grid) {
// grid dimensions
const n = grid.length
// prepare solution matrix
const solution = []
for (let i = 0; i < n; i++) {
const row = Array(n)
row.fill(0)
solution[i] = row
}
return getPathPart(grid, 0, 0, solution, '')
}
/**
* Creates an instance of the "rat in a maze" based on a given grid (maze).
*/
export class RatInAMaze {
constructor (grid) {
// first, let's do some error checking on the input
validateGrid(grid)
// attempt to solve the maze now - all public methods only query the result state later
const solution = getPath(grid)
if (solution !== false) {
this.path = solution
this.solved = true
} else {
this.path = ''
this.solved = false
}
}
}
```