g1801_1900.s1861_rotating_the_box.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g1801_1900.s1861_rotating_the_box;

// #Medium #Array #Matrix #Two_Pointers #2022_05_09_Time_8_ms_(92.84%)_Space_145.6_MB_(5.03%)

/**
 * 1861 - Rotating the Box\.
 *
 * Medium
 *
 * 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/04/08/rotatingtheboxleetcodewithstones.png)
 *
 * **Input:**
 *
 *     box = \[\["#",".","#"]]
 *
 * **Output:**
 *
 *     [["."], 
 *      ["#"], 
 *      ["#"]]
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/04/08/rotatingtheboxleetcode2withstones.png)
 *
 * **Input:**
 *
 *     box = \[\["#",".","*","."], 
 *            ["#","#","*","."]]
 *
 * **Output:**
 *
 *     [["#","."], 
 *      ["#","#"], 
 *      ["*","*"], 
 *      [".","."]]
 *
 * **Example 3:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/04/08/rotatingtheboxleetcode3withstone.png)
 *
 * **Input:**
 *
 *     box = \[\["#","#","*",".","*","."], 
 *            ["#","#","#","*",".","."], 
 *            ["#","#","#",".","#","."]]
 *
 * **Output:**
 *
 *     [[".","#","#"], 
 *      [".","#","#"], 
 *      ["#","#","*"], 
 *      ["#","*","."], 
 *      ["#",".","*"], 
 *      ["#",".","."]]
 *
 * **Constraints:**
 *
 * *   `m == box.length`
 * *   `n == box[i].length`
 * *   `1 <= m, n <= 500`
 * *   `box[i][j]` is either `'#'`, `'*'`, or `'.'`.
**/
public class Solution {
    public char[][] rotateTheBox(char[][] box) {
        int n = box.length;
        int m = box[0].length;
        char[][] result = new char[m][n];
        for (int i = 0; i < n; i++) {
            int j = m - 1;
            int idx = m - 1;
            while (j >= 0) {
                if (box[i][j] == '#') {
                    result[j--][n - i - 1] = '.';
                    result[idx--][n - i - 1] = '#';
                } else {
                    char c = box[i][j];
                    result[j--][n - i - 1] = c;
                    if (c == '*') {
                        idx = j;
                    }
                }
            }
        }
        return result;
    }
}