Formatted question description: https://leetcode.ca/all/1263.html

# 1263. Minimum Moves to Move a Box to Their Target Location (Hard)

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 a grid of size m x n, where each element is a wall, floor, or a box.

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

• Player is represented by character 'S' and can move up, down, left, right in the grid if it is a floor (empy cell).
• Floor is represented by character '.' that means free cell to walk.
• Wall is represented by character '#' that means 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.


Example 4:

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


Constraints:

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

## Solution 1. BFS + DP

dp[sx][sy][bx][by] is the minimal number of pushes needed to make S at (sx, sy) and B at (bx, by).

// OJ: https://leetcode.com/problems/minimum-moves-to-move-a-box-to-their-target-location/
// Time: O((MN)^2)
// Space: O((MN)^2)
class Solution {
public:
int minPushBox(vector<vector<char>>& G) {
int dp = {}, M = G.size(), N = G.size(), sx, sy, bx, by, tx, ty, dirs = { {0,1},{0,-1},{1,0},{-1,0} }, ans = INT_MAX;
memset(dp, 0x3f, sizeof(dp));
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (G[i][j] == 'S') sx = i, sy = j;
else if (G[i][j] == 'B') bx = i, by = j;
else if (G[i][j] == 'T') tx = i, ty = j;
}
}
queue<vector<int>> q;
q.push({ sx, sy, bx, by, 0 });
dp[sx][sy][bx][by] = 0;
while (q.size()) {
auto v = q.front();
q.pop();
int sx = v, sy = v, bx = v, by = v, push = v;
if (bx == tx && by == ty) ans = min(ans, push);
for (auto [dx, dy] : dirs) {
int x = sx + dx, y = sy + dy;
if (x < 0 || y < 0 || x >= M || y >= N || G[x][y] == '#') continue;
int bx2 = bx, by2 = by, push2 = push;
if (x == bx && y == by) bx2 += dx, by2 += dy, ++push2;
if (bx2 < 0 || by2 < 0 || bx2 >= M || by2 >= N || G[bx2][by2] == '#' || push2 >= dp[x][y][bx2][by2]) continue;
dp[x][y][bx2][by2] = push2;
q.push({ x, y, bx2, by2, push2 });
}
}
return ans == INT_MAX ? -1 : ans;
}
};


Java

• class Solution {
public int minPushBox(char[][] grid) {
final int BLOCK = -1;
final int WHITE = 0;
final int GRAY = 1;
final int BLACK = 2;
int rows = grid.length, columns = grid.length;
int[][][][] colors = new int[rows][columns][rows][columns];
int[][][][] distances = new int[rows][columns][rows][columns];
for (int bRow = 0; bRow < rows; bRow++) {
for (int bCol = 0; bCol < columns; bCol++) {
for (int pRow = 0; pRow < rows; pRow++) {
for (int pCol = 0; pCol < columns; pCol++)
distances[bRow][bCol][pRow][pCol] = Integer.MAX_VALUE;
}
}
}
for (int bRow = 0; bRow < rows; bRow++) {
for (int bCol = 0; bCol < columns; bCol++) {
for (int pRow = 0; pRow < rows; pRow++) {
for (int pCol = 0; pCol < columns; pCol++) {
if (grid[bRow][bCol] == '#' || grid[pRow][pCol] == '#') {
colors[bRow][bCol][pRow][pCol] = BLOCK;
distances[bRow][bCol][pRow][pCol] = -1;
}
}
}
}
}
int initialBoxRow = -1, initialBoxColumn = -1, initialPlayerRow = -1, initialPlayerColumn = -1, targetRow = -1, targetColumn = -1;
int count = 0;
outer:
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
char c = grid[i][j];
if (c == 'B') {
initialBoxRow = i;
initialBoxColumn = j;
count++;
} else if (c == 'S') {
initialPlayerRow = i;
initialPlayerColumn = j;
count++;
} else if (c == 'T') {
targetRow = i;
targetColumn = j;
count++;
}
if (count == 3)
break outer;
}
}
int[][] directions = { {-1, 0}, {0, 1}, {1, 0}, {0, -1} };
distances[initialBoxRow][initialBoxColumn][initialPlayerRow][initialPlayerColumn] = 0;
PriorityQueue<Status> queue = new PriorityQueue<Status>();
queue.offer(new Status(initialBoxRow, initialBoxColumn, initialPlayerRow, initialPlayerColumn, 0));
while (!queue.isEmpty()) {
Status status = queue.poll();
int boxRow = status.boxRow, boxColumn = status.boxColumn, playerRow = status.playerRow, playerColumn = status.playerColumn, distance = status.distance;
for (int[] direction : directions) {
int playerNewRow = playerRow + direction, playerNewColumn = playerColumn + direction;
if (playerNewRow < 0 || playerNewRow >= rows || playerNewColumn < 0 || playerNewColumn >= columns || grid[playerNewRow][playerNewColumn] == '#')
continue;
if (playerNewRow == boxRow && playerNewColumn == boxColumn) {
int boxNewRow = boxRow + direction, boxNewColumn = boxColumn + direction;
if (boxNewRow < 0 || boxNewRow >= rows || boxNewColumn < 0 || boxNewColumn >= columns || grid[boxNewRow][boxNewColumn] == '#')
continue;
if (boxNewRow == targetRow && boxNewColumn == targetColumn)
return distance + 1;
else if (colors[boxNewRow][boxNewColumn][playerNewRow][playerNewColumn] == WHITE) {
colors[boxNewRow][boxNewColumn][playerNewRow][playerNewColumn] = GRAY;
distances[boxNewRow][boxNewColumn][playerNewRow][playerNewColumn] = distance + 1;
queue.offer(new Status(boxNewRow, boxNewColumn, playerNewRow, playerNewColumn, distance + 1));
}
} else {
if (colors[boxRow][boxColumn][playerNewRow][playerNewColumn] == WHITE) {
colors[boxRow][boxColumn][playerNewRow][playerNewColumn] = GRAY;
distances[boxRow][boxColumn][playerNewRow][playerNewColumn] = distance;
queue.offer(new Status(boxRow, boxColumn, playerNewRow, playerNewColumn, distance));
}

}
}
colors[boxRow][boxColumn][playerRow][playerColumn] = BLACK;
}
int totalDistance = Integer.MAX_VALUE;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
int distance = distances[targetRow][targetColumn][i][j];
totalDistance = Math.min(totalDistance, distance);
}
}
}
}

class Status implements Comparable<Status> {
int boxRow;
int boxColumn;
int playerRow;
int playerColumn;
int distance;

public Status() {

}

public Status(int boxRow, int boxColumn, int playerRow, int playerColumn, int distance) {
this.boxRow = boxRow;
this.boxColumn = boxColumn;
this.playerRow = playerRow;
this.playerColumn = playerColumn;
this.distance = distance;
}

public int compareTo(Status status2) {
return this.distance - status2.distance;
}
}

• // OJ: https://leetcode.com/problems/minimum-moves-to-move-a-box-to-their-target-location/
// Time: O((MN)^2)
// Space: O((MN)^2)
class Solution {
public:
int minPushBox(vector<vector<char>>& A) {
int M = A.size(), N = A.size(), tx, ty, sx, sy, bx, by, dirs = { {0,1},{0,-1},{1,0},{-1,0} }, dp = {}, ans = INT_MAX;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] == 'S') sx = i, sy = j;
else if (A[i][j] == 'B') bx = i, by = j;
else if (A[i][j] == 'T') tx = i, ty = j;
}
}
memset(dp, 0x3f, sizeof(dp));
dp[sx][sy][bx][by] = 0;
queue<array<int, 4>> q{ { {sx, sy, bx, by} } };
while (q.size()) {
auto [sx, sy, bx, by] = q.front();
q.pop();
int step = dp[sx][sy][bx][by];
if (bx == tx && by == ty) ans = min(ans, step);
for (auto &[dx, dy] : dirs) {
int a = sx + dx, b = sy + dy, bx2 = bx, by2 = by, step2 = step;
if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] == '#') continue;
if (a == bx && b == by) bx2 += dx, by2 += dy, step2++;
if (bx2 < 0 || bx2 >= M || by2 < 0 || by2 >= N || A[bx2][by2] == '#' || step2 >= dp[a][b][bx2][by2]) continue;
dp[a][b][bx2][by2] = step2;
q.push({a, b, bx2, by2});
}
}
return ans == INT_MAX ? -1 : ans;
}
};

• print("Todo!")