Rotating the box

Array Two Pointers Matrix

79.2% acceptance

Feb 25, 2026

1625

You are given an m x n matrix of characters boxGrid representing a side-view of a box. Each cell is '#' (stone), '*' (obstacle), or '.' (empty). The box is rotated 90 degrees clockwise, causing stones to fall due to gravity. Return the n x m matrix representing the box after rotation.

Solution

Rust

Time O(n * m)

Space O(n * m)

LeetCode

solution.rs

impl Solution {
  pub fn rotate_the_box(mut box_grid: Vec<Vec<char>>) -> Vec<Vec<char>> {
    let m = box_grid.len();
    let n = box_grid[0].len();

    // Apply gravity: stones fall to the right in each row
    for row in box_grid.iter_mut() {
      let mut empty = n - 1; // rightmost available position
      for col in (0..n).rev() {
        match row[col] {
          '#' => {
            row[col] = '.';
            row[empty] = '#';
            if empty > 0 { empty -= 1; }
          }
          '*' => {
            if col > 0 { empty = col - 1; }
          }
          _ => {}
        }
      }
    }

    // Rotate 90 degrees clockwise: result[j][m-1-i] = box[i][j]
    // Result is n rows x m cols
    let mut result = vec![vec!['.'; m]; n];
    for i in 0..m {
      for j in 0..n {
        result[j][m - 1 - i] = box_grid[i][j];
      }
    }
    result
  }
}