LeetCode #1861 — MEDIUM

Rotating the Box

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an m x n matrix of characters boxGrid 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 boxGrid 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: boxGrid = [["#",".","#"]]
Output: [["."],
         ["#"],
         ["#"]]

Example 2:

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

Example 3:

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

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given an m x n matrix of characters boxGrid 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 boxGrid 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers

Example 1

[["#",".","#"]]

Example 2

[["#",".","*","."],["#","#","*","."]]

Example 3

[["#","#","*",".","*","."],["#","#","#","*",".","."],["#","#","#",".","#","."]]

Step 02

Core Insight

What unlocks the optimal approach

Rotate the box using the relation rotatedBox[i][j] = box[m - 1 - j][i].
Start iterating from the bottom of the box and for each empty cell check if there is any stone above it with no obstacles between them.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1861: Rotating the Box
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;
    }
}

// Accepted solution for LeetCode #1861: Rotating the Box
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
}

# Accepted solution for LeetCode #1861: Rotating the Box
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

// Accepted solution for LeetCode #1861: Rotating the Box
struct Solution;

impl Solution {
    pub fn rotate_the_box(b: Vec<Vec<char>>) -> Vec<Vec<char>> {
        let n = b.len();
        let m = b[0].len();
        let mut a = vec![];
        for _ in 0..m {
            a.push(vec!['.'; n]);
        }
        for i in 0..n {
            let mut x = m;
            let mut y = m;
            for _j in 0..m {
                match b[i][x - 1] {
                    '.' => {}
                    '*' => {
                        a[x - 1][n - 1 - i] = '*';
                        y = x - 1;
                    }
                    '#' => {
                        a[y - 1][n - 1 - i] = '#';
                        y -= 1;
                    }
                    _ => {}
                }
                x -= 1;
            }
        }
        a
    }
}

#[test]
fn test() {
    let b: Vec<Vec<char>> = vec_vec_char![['#', '.', '#']];
    let a: Vec<Vec<char>> = vec_vec_char![['.'], ['#'], ['#']];
    assert_eq!(Solution::rotate_the_box(b), a);
    let b: Vec<Vec<char>> = vec_vec_char![['#', '.', '*', '.'], ['#', '#', '*', '.']];
    let a: Vec<Vec<char>> = vec_vec_char![['#', '.'], ['#', '#'], ['*', '*'], ['.', '.']];
    assert_eq!(Solution::rotate_the_box(b), a)
}

// Accepted solution for LeetCode #1861: Rotating the Box
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1861: Rotating the Box
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.