1263. Minimum Moves to Move a Box to Their Target Location

Description

A storekeeper is a game in which the player pushes boxes around in a warehouse trying to get them to target locations.

The game is represented by an m x n grid of characters grid where each element is a wall, floor, or box.

Your task is to move the box 'B' to the target position 'T' under the following rules:

• The character 'S' represents the player. The player can move up, down, left, right in grid if it is a floor (empty cell).
• The character '.' represents the floor which means a free cell to walk.
• The character '#' represents the wall which means an obstacle (impossible to walk there).
• There is only one box 'B' and one target cell 'T' in the grid.
• The box can be moved to an adjacent free cell by standing next to the box and then moving in the direction of the box. This is a push.
• The player cannot walk through the box.

Return the minimum number of pushes to move the box to the target. If there is no way to reach the target, return -1.

Example 1:

Input: grid = [["#","#","#","#","#","#"],
["#","T","#","#","#","#"],
["#",".",".","B",".","#"],
["#",".","#","#",".","#"],
["#",".",".",".","S","#"],
["#","#","#","#","#","#"]]
Output: 3
Explanation: We return only the number of times the box is pushed.

Example 2:

Input: grid = [["#","#","#","#","#","#"],
["#","T","#","#","#","#"],
["#",".",".","B",".","#"],
["#","#","#","#",".","#"],
["#",".",".",".","S","#"],
["#","#","#","#","#","#"]]
Output: -1


Example 3:

Input: grid = [["#","#","#","#","#","#"],
["#","T",".",".","#","#"],
["#",".","#","B",".","#"],
["#",".",".",".",".","#"],
["#",".",".",".","S","#"],
["#","#","#","#","#","#"]]
Output: 5
Explanation: push the box down, left, left, up and up.


Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 20
• grid contains only characters '.', '#', 'S', 'T', or 'B'.
• There is only one character 'S', 'B', and 'T' in the grid.

Solutions

Solution 1: Double-ended Queue + BFS

We consider the player’s position and the box’s position as a state, i.e., $(s_i, s_j, b_i, b_j)$, where $(s_i, s_j)$ is the player’s position, and $(b_i, b_j)$ is the box’s position. In the code implementation, we define a function $f(i, j)$, which maps the two-dimensional coordinates $(i, j)$ to a one-dimensional state number, i.e., $f(i, j) = i \times n + j$, where $n$ is the number of columns in the grid. So the player and the box’s state is $(f(s_i, s_j), f(b_i, b_j))$.

First, we traverse the grid to find the initial positions of the player and the box, denoted as $(s_i, s_j)$ and $(b_i, b_j)$.

Then, we define a double-ended queue $q$, where each element is a triplet $(f(s_i, s_j), f(b_i, b_j), d)$, indicating that the player is at $(s_i, s_j)$, the box is at $(b_i, b_j)$, and $d$ pushes have been made. Initially, we add $(f(s_i, s_j), f(b_i, b_j), 0)$ to the queue $q$.

Additionally, we use a two-dimensional array $vis$ to record whether each state has been visited. Initially, $vis[f(s_i, s_j), f(b_i, b_j)]$ is marked as visited.

Next, we start the breadth-first search.

In each step of the search, we take out the queue head element $(f(s_i, s_j), f(b_i, b_j), d)$, and check whether $grid[b_i][b_j] = ‘T’$ is satisfied. If it is, it means the box has been pushed to the target position, and now $d$ can be returned as the answer.

Otherwise, we enumerate the player’s next move direction. The player’s new position is denoted as $(s_x, s_y)$. If $(s_x, s_y)$ is a valid position, we judge whether $(s_x, s_y)$ is the same as the box’s position $(b_i, b_j)$:

• If they are the same, it means the player has reached the box’s position and pushed the box forward by one step. The box’s new position is $(b_x, b_y)$. If $(b_x, b_y)$ is a valid position, and the state $(f(s_x, s_y), f(b_x, b_y))$ has not been visited, then we add $(f(s_x, s_y), f(b_x, b_y), d + 1)$ to the end of the queue $q$, and mark $vis[f(s_x, s_y), f(b_x, b_y)]$ as visited.
• If they are different, it means the player has not pushed the box. Then we only need to judge whether the state $(f(s_x, s_y), f(b_i, b_j))$ has been visited. If it has not been visited, then we add $(f(s_x, s_y), f(b_i, b_j), d)$ to the head of the queue $q$, and mark $vis[f(s_x, s_y), f(b_i, b_j)]$ as visited.

We continue the breadth-first search until the queue is empty.

Note, if the box is pushed, the push count $d$ needs to be incremented by $1$, and the new state is added to the end of the queue $q$. If the box is not pushed, the push count $d$ remains unchanged, and the new state is added to the head of the queue $q$.

Finally, if no valid push scheme is found, then return $-1$.

The time complexity is $O(m^2 \times n^2)$, and the space complexity is $O(m^2 \times n^2)$. Where $m$ and $n$ are the number of rows and columns in the grid, respectively.

• class Solution {
private int m;
private int n;
private char[][] grid;

public int minPushBox(char[][] grid) {
m = grid.length;
n = grid[0].length;
this.grid = grid;
int si = 0, sj = 0, bi = 0, bj = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 'S') {
si = i;
sj = j;
} else if (grid[i][j] == 'B') {
bi = i;
bj = j;
}
}
}
int[] dirs = {-1, 0, 1, 0, -1};
Deque<int[]> q = new ArrayDeque<>();
boolean[][] vis = new boolean[m * n][m * n];
q.offer(new int[] {f(si, sj), f(bi, bj), 0});
vis[f(si, sj)][f(bi, bj)] = true;
while (!q.isEmpty()) {
var p = q.poll();
int d = p[2];
bi = p[1] / n;
bj = p[1] % n;
if (grid[bi][bj] == 'T') {
return d;
}
si = p[0] / n;
sj = p[0] % n;
for (int k = 0; k < 4; ++k) {
int sx = si + dirs[k], sy = sj + dirs[k + 1];
if (!check(sx, sy)) {
continue;
}
if (sx == bi && sy == bj) {
int bx = bi + dirs[k], by = bj + dirs[k + 1];
if (!check(bx, by) || vis[f(sx, sy)][f(bx, by)]) {
continue;
}
vis[f(sx, sy)][f(bx, by)] = true;
q.offer(new int[] {f(sx, sy), f(bx, by), d + 1});
} else if (!vis[f(sx, sy)][f(bi, bj)]) {
vis[f(sx, sy)][f(bi, bj)] = true;
q.offerFirst(new int[] {f(sx, sy), f(bi, bj), d});
}
}
}
return -1;
}

private int f(int i, int j) {
return i * n + j;
}

private boolean check(int i, int j) {
return i >= 0 && i < m && j >= 0 && j < n && grid[i][j] != '#';
}
}

• class Solution {
public:
int minPushBox(vector<vector<char>>& grid) {
int m = grid.size(), n = grid[0].size();
int si, sj, bi, bj;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 'S') {
si = i, sj = j;
} else if (grid[i][j] == 'B') {
bi = i, bj = j;
}
}
}
auto f = [&](int i, int j) {
return i * n + j;
};
auto check = [&](int i, int j) {
return i >= 0 && i < m && j >= 0 && j < n && grid[i][j] != '#';
};
int dirs[5] = {-1, 0, 1, 0, -1};
deque<tuple<int, int, int>> q;
q.emplace_back(f(si, sj), f(bi, bj), 0);
bool vis[m * n][m * n];
memset(vis, false, sizeof(vis));
vis[f(si, sj)][f(bi, bj)] = true;
while (!q.empty()) {
auto [s, b, d] = q.front();
q.pop_front();
si = s / n, sj = s % n;
bi = b / n, bj = b % n;
if (grid[bi][bj] == 'T') {
return d;
}
for (int k = 0; k < 4; ++k) {
int sx = si + dirs[k], sy = sj + dirs[k + 1];
if (!check(sx, sy)) {
continue;
}
if (sx == bi && sy == bj) {
int bx = bi + dirs[k], by = bj + dirs[k + 1];
if (!check(bx, by) || vis[f(sx, sy)][f(bx, by)]) {
continue;
}
vis[f(sx, sy)][f(bx, by)] = true;
q.emplace_back(f(sx, sy), f(bx, by), d + 1);
} else if (!vis[f(sx, sy)][f(bi, bj)]) {
vis[f(sx, sy)][f(bi, bj)] = true;
q.emplace_front(f(sx, sy), f(bi, bj), d);
}
}
}
return -1;
}
};

• class Solution:
def minPushBox(self, grid: List[List[str]]) -> int:
def f(i: int, j: int) -> int:
return i * n + j

def check(i: int, j: int) -> bool:
return 0 <= i < m and 0 <= j < n and grid[i][j] != "#"

for i, row in enumerate(grid):
for j, c in enumerate(row):
if c == "S":
si, sj = i, j
elif c == "B":
bi, bj = i, j
m, n = len(grid), len(grid[0])
dirs = (-1, 0, 1, 0, -1)
q = deque([(f(si, sj), f(bi, bj), 0)])
vis = [[False] * (m * n) for _ in range(m * n)]
vis[f(si, sj)][f(bi, bj)] = True
while q:
s, b, d = q.popleft()
bi, bj = b // n, b % n
if grid[bi][bj] == "T":
return d
si, sj = s // n, s % n
for a, b in pairwise(dirs):
sx, sy = si + a, sj + b
if not check(sx, sy):
continue
if sx == bi and sy == bj:
bx, by = bi + a, bj + b
if not check(bx, by) or vis[f(sx, sy)][f(bx, by)]:
continue
vis[f(sx, sy)][f(bx, by)] = True
q.append((f(sx, sy), f(bx, by), d + 1))
elif not vis[f(sx, sy)][f(bi, bj)]:
vis[f(sx, sy)][f(bi, bj)] = True
q.appendleft((f(sx, sy), f(bi, bj), d))
return -1


• func minPushBox(grid [][]byte) int {
m, n := len(grid), len(grid[0])
var si, sj, bi, bj int
for i, row := range grid {
for j, c := range row {
if c == 'S' {
si, sj = i, j
} else if c == 'B' {
bi, bj = i, j
}
}
}
f := func(i, j int) int {
return i*n + j
}
check := func(i, j int) bool {
return i >= 0 && i < m && j >= 0 && j < n && grid[i][j] != '#'
}
q := [][]int{[]int{f(si, sj), f(bi, bj), 0} }
vis := make([][]bool, m*n)
for i := range vis {
vis[i] = make([]bool, m*n)
}
vis[f(si, sj)][f(bi, bj)] = true
dirs := [5]int{-1, 0, 1, 0, -1}
for len(q) > 0 {
p := q[0]
q = q[1:]
si, sj, bi, bj = p[0]/n, p[0]%n, p[1]/n, p[1]%n
d := p[2]
if grid[bi][bj] == 'T' {
return d
}
for k := 0; k < 4; k++ {
sx, sy := si+dirs[k], sj+dirs[k+1]
if !check(sx, sy) {
continue
}
if sx == bi && sy == bj {
bx, by := bi+dirs[k], bj+dirs[k+1]
if !check(bx, by) || vis[f(sx, sy)][f(bx, by)] {
continue
}
vis[f(sx, sy)][f(bx, by)] = true
q = append(q, []int{f(sx, sy), f(bx, by), d + 1})
} else if !vis[f(sx, sy)][f(bi, bj)] {
vis[f(sx, sy)][f(bi, bj)] = true
q = append([][]int{[]int{f(sx, sy), f(bi, bj), d} }, q...)
}
}
}
return -1
}

• function minPushBox(grid: string[][]): number {
const [m, n] = [grid.length, grid[0].length];
let [si, sj, bi, bj] = [0, 0, 0, 0];
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (grid[i][j] === 'S') {
[si, sj] = [i, j];
} else if (grid[i][j] === 'B') {
[bi, bj] = [i, j];
}
}
}
const f = (i: number, j: number): number => i * n + j;
const check = (i: number, j: number): boolean =>
i >= 0 && i < m && j >= 0 && j < n && grid[i][j] !== '#';

const q: Deque<[number, number, number]> = new Deque();
const vis: boolean[][] = new Array(m * n).fill(0).map(() => new Array(m * n).fill(false));
q.push([f(si, sj), f(bi, bj), 0]);
vis[f(si, sj)][f(bi, bj)] = true;
const dirs: number[] = [-1, 0, 1, 0, -1];
while (q.size() > 0) {
const [s, b, d] = q.shift()!;
const [si, sj] = [Math.floor(s / n), s % n];
const [bi, bj] = [Math.floor(b / n), b % n];
if (grid[bi][bj] === 'T') {
return d;
}
for (let k = 0; k < 4; ++k) {
const [sx, sy] = [si + dirs[k], sj + dirs[k + 1]];
if (!check(sx, sy)) {
continue;
}
if (sx === bi && sy === bj) {
const [bx, by] = [bi + dirs[k], bj + dirs[k + 1]];
if (!check(bx, by) || vis[f(sx, sy)][f(bx, by)]) {
continue;
}
vis[f(sx, sy)][f(bx, by)] = true;
q.push([f(sx, sy), f(bx, by), d + 1]);
} else if (!vis[f(sx, sy)][f(bi, bj)]) {
vis[f(sx, sy)][f(bi, bj)] = true;
q.unshift([f(sx, sy), f(bi, bj), d]);
}
}
}
return -1;
}

/* 以下是双向列队模板类 */
class CircularDeque<T> {
prev: CircularDeque<T> | null;
next: CircularDeque<T> | null;
begin: number;
end: number;
empty: boolean;
data: T[];
constructor(N: number) {
this.prev = this.next = null;
this.begin = this.end = 0;
this.empty = true;
this.data = Array(N);
}

isFull(): boolean {
return this.end === this.begin && !this.empty;
}

isEmpty(): boolean {
return this.empty;
}

push(val: T): boolean {
if (this.isFull()) return false;
this.empty = false;
this.data[this.end] = val;
this.end = (this.end + 1) % this.data.length;
return true;
}

front(): T | undefined {
return this.isEmpty() ? undefined : this.data[this.begin];
}

back(): T | undefined {
return this.isEmpty() ? undefined : this.data[this.end - 1];
}

pop(): T | undefined {
if (this.isEmpty()) return undefined;
const value = this.data[this.end - 1];
this.end = (this.end - 1) % this.data.length;
if (this.end < 0) this.end += this.data.length;
if (this.end === this.begin) this.empty = true;
return value;
}

unshift(val: T): boolean {
if (this.isFull()) return false;
this.empty = false;
this.begin = (this.begin - 1) % this.data.length;
if (this.begin < 0) this.begin += this.data.length;
this.data[this.begin] = val;
return true;
}

shift(): T | undefined {
if (this.isEmpty()) return undefined;
const value = this.data[this.begin];
this.begin = (this.begin + 1) % this.data.length;
if (this.end === this.begin) this.empty = true;
return value;
}

*values(): Generator<T, void, undefined> {
if (this.isEmpty()) return undefined;
let i = this.begin;
do {
yield this.data[i];
i = (i + 1) % this.data.length;
} while (i !== this.end);
}
}

class Deque<T> {
tail: CircularDeque<T>;
_size: number;
constructor(collection: T[] = []) {
this.tail = new CircularDeque<T>(128);
this._size = 0;
for (const item of collection) this.push(item);
}

size(): number {
return this._size;
}

push(val: T): void {
let last = this.tail.prev!;
if (last.isFull()) {
const inserted = new CircularDeque<T>(128);

this.tail.prev = inserted;
inserted.next = this.tail;

last.next = inserted;
inserted.prev = last;

last = inserted;
}
last.push(val);
this._size++;
}

back(): T | undefined {
if (this._size === 0) return;
return this.tail.prev!.back();
}

pop(): T | undefined {
if (this.head.next === this.tail) return undefined;
const last = this.tail.prev!;
const value = last.pop();
if (last.isEmpty()) {
this.tail.prev = last.prev;
last.prev!.next = this.tail;
}
this._size--;
return value;
}

unshift(val: T): void {
if (first.isFull()) {
const inserted = new CircularDeque<T>(128);

inserted.next = first;
first.prev = inserted;

first = inserted;
}
first.unshift(val);
this._size++;
}

shift(): T | undefined {
if (this.head.next === this.tail) return undefined;
const value = first.shift();
if (first.isEmpty()) {
}
this._size--;
return value;
}

front(): T | undefined {
if (this._size === 0) return undefined;
}

*values(): Generator<T, void, undefined> {
while (node !== this.tail) {
for (const value of node.values()) yield value;
node = node.next!;
}
}
}