What is the brute force solution for Minimum Number of Flips to Make Binary Grid Palindromic II?

The brute force approach for Minimum Number of Flips to Make Binary Grid Palindromic II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each cell in the grid and flipping it to achieve the desired palindromic structure. This method is straightforward but inefficient, as it requires examining every possible combination of flips. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Flips to Make Binary Grid Palindromic II?

Minimum Number of Flips to Make Binary Grid Palindromic II uses the following concepts: Array, Two Pointers, Matrix. The recommended patterns to study are: Two Pointers, Matrix Manipulation.

What are the interview tips for Minimum Number of Flips to Make Binary Grid Palindromic II?

Clearly explain your thought process while coding. Discuss edge cases, such as grids with only one row or column. Practice similar problems to recognize patterns in symmetry.

What are common mistakes when solving Minimum Number of Flips to Make Binary Grid Palindromic II?

Forgetting to handle the middle row/column in odd dimensions. Not considering all symmetric pairs when counting flips.

#3240

Minimum Number of Flips to Make Binary Grid Palindromic II

Medium

ArrayTwo PointersMatrixTwo PointersMatrix Manipulation

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution leverages symmetry in the grid to minimize the number of flips. By focusing on pairs of cells that must match, we can efficiently calculate the required flips without checking every combination.

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Algorithm

4 steps

1Step 1: Initialize a flip counter and iterate through the first half of the grid.
2Step 2: For each cell, identify its symmetric counterparts and count the number of 1's and 0's.
3Step 3: Calculate the minimum flips needed to make all symmetric cells the same.
4Step 4: After processing the grid, ensure the total number of 1's is divisible by 4.

solution.py14 lines

1def minFlips(grid):
2    m, n = len(grid), len(grid[0])
3    flips = 0
4    for i in range((m + 1) // 2):
5        for j in range((n + 1) // 2):
6            count = [0, 0]
7            for x in [i, m - 1 - i]:
8                for y in [j, n - 1 - j]:
9                    count[grid[x][y]] += 1
10            flips += min(count[0], count[1])
11    total_ones = sum(sum(row) for row in grid)
12    if total_ones % 4 != 0:
13        flips += 1
14    return flips

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Complexity note: The time complexity is O(n) because we are effectively processing each cell in the grid a limited number of times, focusing on symmetry. The space complexity is O(1) as we are using a constant amount of extra space.

1Understanding symmetry is crucial for minimizing flips.
2The total number of 1's must be adjusted to meet divisibility requirements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.