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Problem solution · Java

Minimum Number of Flips to Make Binary Grid Palindromic II

Minimum Number of Flips to Make Binary Grid Palindromic II: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
52 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Minimum Number of Flips to Make Binary Grid Palindromic II, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 52 lines of Java from the credited upstream file 3240.java.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Number of Flips to Make Binary Grid Palindromic II · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minFlips(int[][] grid) {    final int m = grid.length;    final int n = grid[0].length;    int ans = 0;    int middleOnes = 0;    int mismatchedPairs = 0;     // Handle top-left, top-right, bottom-left, bottom-right cells.    for (int i = 0; i < m / 2; ++i) {      for (int j = 0; j < n / 2; ++j) {        final int ones =            grid[i][j] + grid[i][n - 1 - j] + grid[m - 1 - i][j] + grid[m - 1 - i][n - 1 - j];        ans += Math.min(ones, 4 - ones);      }    }     // Handle the middle row if the number of m is odd.    if (m % 2 == 1)      for (int j = 0; j < n / 2; ++j) {        final int leftCell = grid[m / 2][j];        final int rightCell = grid[m / 2][n - 1 - j];        mismatchedPairs += leftCell ^ rightCell;        middleOnes += leftCell + rightCell;      }     // Handle the middle column if the number of columns is odd.    if (n % 2 == 1)      for (int i = 0; i < m / 2; ++i) {        final int topCell = grid[i][n / 2];        final int bottomCell = grid[m - 1 - i][n / 2];        mismatchedPairs += topCell ^ bottomCell;        middleOnes += topCell + bottomCell;      }     if (mismatchedPairs == 0) {      // Since there's no mismatched pairs, middleOnes % 4 must be 0 or 2.      if (middleOnes % 4 == 2)        ans += 2; // Flip two 1s to 0s.    } else {      // Flip every mismatched pair 01 to 00 or 11. It doesn't matter.      ans += mismatchedPairs;    }     // Handle the center cell if both dimensions are odd.    if (m % 2 == 1 && n % 2 == 1)      ans += grid[m / 2][n / 2];     return ans;  }}

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