#3240

Medium

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Minimum Number of Flips to Make Binary Grid Palindromic II

The problem asks to flip cells in a binary grid to make its rows and columns palindromic using the minimum number of flips.

array two pointers matrix

Problem Statement

You are given an m x n binary matrix grid. You can flip any number of cells in the grid from 0 to 1 or from 1 to 0.

A row or column is considered palindromic if its values read the same forward and backward. Your task is to find the minimum number of flips required to make all rows and columns palindromic, using an efficient approach.

Examples

Example 1

Input: grid = [[1,0,0],[0,1,0],[0,0,1]]

Output: 3

Example 2

Input: grid = [[0,1],[0,1],[0,0]]

Output: 2

Example 3

Input: grid = [[1],[1]]

Output: 2

Constraints

m == grid.length
n == grid[i].length
1 <= m * n <= 2 * 105
0 <= grid[i][j] <= 1

Solution Approach

Two-Pointer Scanning

Start with a two-pointer approach to check pairs of positions in each row and column. For each (x, y), verify that its mirrored counterparts, such as (m - 1 - x, y) and (x, n - 1 - y), hold the same value. Track how many flips are needed to achieve symmetry.

Invariant Tracking

Maintain invariant conditions for each mirrored pair of positions. If one position in a pair requires a flip, then the other must also be flipped to maintain palindromic symmetry. This tracking ensures the minimum number of flips is achieved.

Optimizing for Minimum Flips

After tracking the mirrored pairs, compute the flips for each segment and select the minimum required number of flips. This can be done efficiently by comparing the initial states of the grid cells and applying minimal changes.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on how efficiently the two-pointer scanning is implemented. For each cell in the grid, we are essentially comparing four positions, leading to an overall time complexity of O(m * n), where m and n are the grid dimensions. The space complexity depends on how the invariant tracking and flip counting are handled, generally O(1) additional space for the solution.

What Interviewers Usually Probe

Does the candidate understand two-pointer scanning?
Is the candidate tracking mirrored pairs correctly?
Can the candidate optimize for minimal flip calculations?

Common Pitfalls or Variants

Common pitfalls

Not correctly identifying mirrored pairs across the grid edges.
Failing to track the invariant condition of symmetry while counting flips.
Trying to solve the problem without recognizing that both row and column palindromes must be considered simultaneously.

Follow-up variants

Handling grids where all values are initially palindromic.
Optimizing for larger grids with minimal space overhead.
Dealing with cases where certain rows or columns are already palindromic.

FAQ

How do I minimize the number of flips for this problem?

Focus on mirrored pairs of grid positions and apply flips only where necessary to ensure both rows and columns are palindromic.

What is the primary approach for solving the Minimum Number of Flips to Make Binary Grid Palindromic II?

The primary approach involves using two-pointer scanning and invariant tracking to ensure minimal flips for symmetry.

Can I use a brute force method for this problem?

While a brute force method might work, it would be inefficient for large grids. The optimal solution uses a two-pointer approach with invariant tracking.

What are the edge cases I should consider for this problem?

Consider cases where the grid is already palindromic or where only a few cells need to be flipped.

What is the space complexity for solving this problem?

The space complexity can be O(1) if you track flips in place, though the implementation of the two-pointer scanning determines the exact space usage.

terminal

Solution

Solution 1: Case Analysis

If both rows and columns are palindromic, then for any $i \in [0, m / 2)$ and $j \in [0, n / 2)$, it must satisfy $\text{grid}[i][j] = \text{grid}[m - i - 1][j] = \text{grid}[i][n - j - 1] = \text{grid}[m - i - 1][n - j - 1]$. They must either all become $0$ or all become $1$. The number of changes to $0$ is $c_0 = \text{grid}[i][j] + \text{grid}[m - i - 1][j] + \text{grid}[i][n - j - 1] + \text{grid}[m - i - 1][n - j - 1]$, and the number of changes to $1$ is $c_1 = 4 - c_0$. We take the minimum of the two and add it to the answer.

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class Solution:
    def minFlips(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        ans = 0
        for i in range(m // 2):
            for j in range(n // 2):
                x, y = m - i - 1, n - j - 1
                cnt1 = grid[i][j] + grid[x][j] + grid[i][y] + grid[x][y]
                ans += min(cnt1, 4 - cnt1)
        if m % 2 and n % 2:
            ans += grid[m // 2][n // 2]
        diff = cnt1 = 0
        if m % 2:
            for j in range(n // 2):
                if grid[m // 2][j] == grid[m // 2][n - j - 1]:
                    cnt1 += grid[m // 2][j] * 2
                else:
                    diff += 1
        if n % 2:
            for i in range(m // 2):
                if grid[i][n // 2] == grid[m - i - 1][n // 2]:
                    cnt1 += grid[i][n // 2] * 2
                else:
                    diff += 1
        ans += diff if cnt1 % 4 == 0 or diff else 2
        return ans

Continue Practicing

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