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

The brute force approach for Minimum Number of Flips to Make Binary Grid Palindromic I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each row and column to see how many flips are needed to make them palindromic. This is straightforward but inefficient for larger matrices. 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 I?

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

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

Always clarify whether you need to check rows, columns, or both. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Don’t rush; take your time to think through the problem and outline your approach before coding.

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

Students often forget to check both dimensions (rows and columns) separately. Not considering edge cases like already palindromic rows or columns can lead to incorrect counts.

#3239

Minimum Number of Flips to Make Binary Grid Palindromic I

Medium

ArrayTwo PointersMatrixTwo PointersArray

LeetCode ↗

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 the symmetry of palindromes and uses a two-pointer technique to count the flips needed in a more efficient manner. This reduces the number of checks significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to count flips for rows and another for columns.
2Step 2: For each row, use two pointers to compare elements from the start and end, counting mismatches.
3Step 3: Repeat the same for columns using two pointers.
4Step 4: Return the minimum flips needed for either rows or columns.

solution.py26 lines

1# Full working Python code
2
3def minFlips(grid):
4    m, n = len(grid), len(grid[0])
5    row_flips = 0
6    col_flips = 0
7
8    # Check rows
9    for i in range(m):
10        left, right = 0, n - 1
11        while left < right:
12            if grid[i][left] != grid[i][right]:
13                row_flips += 1
14            left += 1
15            right -= 1
16
17    # Check columns
18    for j in range(n):
19        top, bottom = 0, m - 1
20        while top < bottom:
21            if grid[top][j] != grid[bottom][j]:
22                col_flips += 1
23            top += 1
24            bottom -= 1
25
26    return min(row_flips, col_flips)

ℹ

Complexity note: The time complexity is O(n) because we only need to check each element once per row and column. The space complexity is O(1) since we use a constant amount of space for counting flips.

1Flipping a cell affects both its row and column, so we need to consider both dimensions.
2Using two pointers allows us to efficiently count mismatches without redundant checks.

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