Rotating the Box

Leetcode Problem:

• Rotating the Box

Summary

• Given an m x n matrix of characters boxGrid representing a side-view of a box, rotate the box 90 degrees clockwise and return the resulting matrix.

Approach

• The approach used is to first transpose the boxGrid and then reverse each row of the transposed matrix
• This is because the problem requires us to rotate the box 90 degrees clockwise, which can be achieved by first flipping the box horizontally and then flipping it vertically.

Complexity

• O(m*n) where m is the number of rows and n is the number of columns in the boxGrid.

Explanation

• The solution code first creates a new 2D vector rotateBox to store the resulting matrix
• It then iterates over each row in the boxGrid and performs the following operations: 1
• It starts from the rightmost column and moves towards the left
• If the current cell is a stone (#), it moves the stone to the bottom right corner of the row and marks the stone as empty (.) if it's not the last stone in the row
• 2
• If the current cell is an obstacle (*), it moves the obstacle to the left of the stone
• 3
• If the current cell is empty (.), it simply moves to the next cell
• After processing each row, the code transposes the boxGrid and reverses each row to achieve the rotation
• Finally, it returns the resulting matrix.

Solution Code:

class Solution {
public:
    vector> rotateTheBox(vector>& box) {
        vector> rotateBox;
        for(int i = 0 ; i< box.size(); i++){
            int r = box[0].size() - 1;
            int l = r;
            while(l >= 0){
                if(box[i][l] == '#'){
                    box[i][r] = box[i][l];
                    if(l != r){
                        box[i][l] = '.';
                    }
                    r--;
                } else if(box[i][l] == '*'){
                    r = l - 1;
                }
                l--;
            }

        }
        int R = box.size();
        int C = box[0].size();
        for(int j=0; j < C; j++){
            rotateBox.push_back(vector());
            for(int i = 0; i < R; i++){
                rotateBox[j].push_back(box[R-i-1][j]);
            }
        }
        
        


        return rotateBox;
    }
};