Rotating the Box | Brute Force | Optimal | Leetcode 1861 | codestorywithMIK
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This is the 125th Video of our Playlist "Array 1D/2D : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good practice 2D Array based problem : Rotating the Box | Brute Force | Optimal | Leetcode 1861 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Rotating the Box | Brute Force | Optimal | Leetcode 1861 | codestorywithMIK
Company Tags : will update
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Arrays/2-D Array/Rotating the Box.cpp
Leetcode Link : https://leetcode.com/problems/rotating-the-box/description
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Summary :
Approach 1 (First Code)
Steps:
Transpose the 2D array to swap rows with columns.
Reverse each row in the transposed matrix to rotate the box 90 degrees clockwise.
Apply gravity by moving stones (#) to the lowest available position, stopping when encountering obstacles (*) or other stones.
Gravity Logic:
For each empty space (.), scan upward in the column to find the nearest stone and move it to the current position.
Efficiency:
Gravity is applied in O(n²) due to upward scanning for each empty space.
Approach 2 (Second Code)
Steps:
Transpose and reverse rows to rotate the box 90 degrees clockwise, as in the first approach.
Apply gravity with an optimized approach.
Gravity Logic:
Use a spaceBottomRow pointer to track the lowest available position for stones in each column.
Iterate from the bottom to the top, moving stones directly without scanning upward.
Efficiency:
Gravity is applied in O(n × m), avoiding redundant upward scans.
✨ Timelines✨
00:00 - Introduction
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