# 1861. Rotating the Box

## Description

You are given an m x n matrix of characters box representing a side-view of a box. Each cell of the box is one of the following:

• A stone '#'
• A stationary obstacle '*'
• Empty '.'

The box is rotated 90 degrees clockwise, causing some of the stones to fall due to gravity. Each stone falls down until it lands on an obstacle, another stone, or the bottom of the box. Gravity does not affect the obstacles' positions, and the inertia from the box's rotation does not affect the stones' horizontal positions.

It is guaranteed that each stone in box rests on an obstacle, another stone, or the bottom of the box.

Return an n x m matrix representing the box after the rotation described above.

Example 1:

Input: box = [["#",".","#"]]
Output: [["."],
["#"],
["#"]]

Example 2:

Input: box = [["#",".","*","."],
["#","#","*","."]]
Output: [["#","."],
["#","#"],
["*","*"],
[".","."]]

Example 3:

Input: box = [["#","#","*",".","*","."],
["#","#","#","*",".","."],
["#","#","#",".","#","."]]
Output: [[".","#","#"],
[".","#","#"],
["#","#","*"],
["#","*","."],
["#",".","*"],
["#",".","."]]

Constraints:

• m == box.length
• n == box[i].length
• 1 <= m, n <= 500
• box[i][j] is either '#', '*', or '.'.

## Solutions

Solution 1: Queue Simulation

First, we rotate the matrix 90 degrees clockwise, then simulate the falling process of the stones in each column.

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

• class Solution {
public char[][] rotateTheBox(char[][] box) {
int m = box.length, n = box[0].length;
char[][] ans = new char[n][m];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans[j][m - i - 1] = box[i][j];
}
}
for (int j = 0; j < m; ++j) {
Deque<Integer> q = new ArrayDeque<>();
for (int i = n - 1; i >= 0; --i) {
if (ans[i][j] == '*') {
q.clear();
} else if (ans[i][j] == '.') {
q.offer(i);
} else if (!q.isEmpty()) {
ans[q.pollFirst()][j] = '#';
ans[i][j] = '.';
q.offer(i);
}
}
}
return ans;
}
}

• class Solution {
public:
vector<vector<char>> rotateTheBox(vector<vector<char>>& box) {
int m = box.size(), n = box[0].size();
vector<vector<char>> ans(n, vector<char>(m));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans[j][m - i - 1] = box[i][j];
}
}
for (int j = 0; j < m; ++j) {
queue<int> q;
for (int i = n - 1; ~i; --i) {
if (ans[i][j] == '*') {
queue<int> t;
swap(t, q);
} else if (ans[i][j] == '.') {
q.push(i);
} else if (!q.empty()) {
ans[q.front()][j] = '#';
q.pop();
ans[i][j] = '.';
q.push(i);
}
}
}
return ans;
}
};

• class Solution:
def rotateTheBox(self, box: List[List[str]]) -> List[List[str]]:
m, n = len(box), len(box[0])
ans = [[None] * m for _ in range(n)]
for i in range(m):
for j in range(n):
ans[j][m - i - 1] = box[i][j]
for j in range(m):
q = deque()
for i in range(n - 1, -1, -1):
if ans[i][j] == '*':
q.clear()
elif ans[i][j] == '.':
q.append(i)
elif q:
ans[q.popleft()][j] = '#'
ans[i][j] = '.'
q.append(i)
return ans

• func rotateTheBox(box [][]byte) [][]byte {
m, n := len(box), len(box[0])
ans := make([][]byte, n)
for i := range ans {
ans[i] = make([]byte, m)
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
ans[j][m-i-1] = box[i][j]
}
}
for j := 0; j < m; j++ {
q := []int{}
for i := n - 1; i >= 0; i-- {
if ans[i][j] == '*' {
q = []int{}
} else if ans[i][j] == '.' {
q = append(q, i)
} else if len(q) > 0 {
ans[q[0]][j] = '#'
q = q[1:]
ans[i][j] = '.'
q = append(q, i)
}
}
}
return ans
}