Minimum Number of Flips to Make Binary Grid Palindromic I

Problem Description

Ah, the joys of binary grids! You know, those delightful little squares of 0s and 1s that make you feel like a computer scientist every time you look at them. The problem at hand is to transform a binary grid into a palindromic structure. Yes, you heard it right! We want our grid to look the same when viewed from either end.

Imagine you’re trying to arrange your bookshelf. You want it to look symmetrical, but your books are all over the place. You could flip some books around (or maybe just your sanity) to achieve that perfect symmetry. In this problem, we need to flip the bits in the grid to achieve that same level of perfection.

The task is to find the minimum number of flips required to make the binary grid palindromic.

Code Solution


class Solution:
    def minFlips(self, grid: list[list[int]]) -> int:
        rowFlips = sum(row[i] != row[-1 - i]
                       for row in grid for i in range(len(row) // 2))
        colFlips = sum(col[i] != col[-1 - i] for col in zip(*grid)
                       for i in range(len(col) // 2))
        return min(rowFlips, colFlips)

Approach

The approach taken in the code is quite straightforward. It calculates the number of flips required for both rows and columns to achieve a palindromic structure. For each row, it checks how many bits differ from their corresponding bits on the opposite end. The same logic applies to columns. Finally, it returns the minimum number of flips needed between the two.

Time and Space Complexity

Time Complexity: O(n), where n is the number of elements in the grid. We traverse each row and column once.
Space Complexity: O(1), as we are using a constant amount of space for our calculations.

Real-World Example

Let’s say you’re hosting a dinner party and you want your table to look symmetrical. You have a mix of plates and cups, and you want them to be arranged in a way that they mirror each other. If you have to flip some plates or cups to achieve that perfect look, you’re essentially performing the same task as the binary grid problem. The fewer flips you make, the better!