LeetCode #3240 — MEDIUM

Minimum Number of Flips to Make Binary Grid Palindromic II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an m x n binary matrix grid.

A row or column is considered palindromic if its values read the same forward and backward.

You can flip any number of cells in grid from 0 to 1, or from 1 to 0.

Return the minimum number of cells that need to be flipped to make all rows and columns palindromic, and the total number of 1's in grid divisible by 4.

Example 1:

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

Output: 3

Explanation:

Example 2:

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

Output: 2

Explanation:

Example 3:

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

Output: 2

Explanation:

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given an m x n binary matrix grid. A row or column is considered palindromic if its values read the same forward and backward. You can flip any number of cells in grid from 0 to 1, or from 1 to 0. Return the minimum number of cells that need to be flipped to make all rows and columns palindromic, and the total number of 1's in grid divisible by 4.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers

Example 1

[[1,0,0],[0,1,0],[0,0,1]]

Example 2

[[0,1],[0,1],[0,0]]

Example 3

[[1],[1]]

Step 02

Core Insight

What unlocks the optimal approach

For each <code>(x, y)</code>, find <code>(m - 1 - x, y)</code>, <code>(m - 1 - x, n - 1 - y)</code>, and <code>(x, n - 1 - y)</code>; they should be the same.
Note that we need to specially handle the middle row (column) if the number of rows (columns) is odd.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3240: Minimum Number of Flips to Make Binary Grid Palindromic II
class Solution {
    public int minFlips(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int ans = 0;
        for (int i = 0; i < m / 2; ++i) {
            for (int j = 0; j < n / 2; ++j) {
                int x = m - i - 1, y = n - j - 1;
                int cnt1 = grid[i][j] + grid[x][j] + grid[i][y] + grid[x][y];
                ans += Math.min(cnt1, 4 - cnt1);
            }
        }
        if (m % 2 == 1 && n % 2 == 1) {
            ans += grid[m / 2][n / 2];
        }

        int diff = 0, cnt1 = 0;
        if (m % 2 == 1) {
            for (int j = 0; j < n / 2; ++j) {
                if (grid[m / 2][j] == grid[m / 2][n - j - 1]) {
                    cnt1 += grid[m / 2][j] * 2;
                } else {
                    diff += 1;
                }
            }
        }
        if (n % 2 == 1) {
            for (int i = 0; i < m / 2; ++i) {
                if (grid[i][n / 2] == grid[m - i - 1][n / 2]) {
                    cnt1 += grid[i][n / 2] * 2;
                } else {
                    diff += 1;
                }
            }
        }
        ans += cnt1 % 4 == 0 || diff > 0 ? diff : 2;
        return ans;
    }
}

// Accepted solution for LeetCode #3240: Minimum Number of Flips to Make Binary Grid Palindromic II
func minFlips(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	ans := 0

	for i := 0; i < m/2; i++ {
		for j := 0; j < n/2; j++ {
			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 == 1 && n%2 == 1 {
		ans += grid[m/2][n/2]
	}

	diff, cnt1 := 0, 0

	if m%2 == 1 {
		for j := 0; j < n/2; j++ {
			if grid[m/2][j] == grid[m/2][n-j-1] {
				cnt1 += grid[m/2][j] * 2
			} else {
				diff += 1
			}
		}
	}

	if n%2 == 1 {
		for i := 0; i < m/2; i++ {
			if grid[i][n/2] == grid[m-i-1][n/2] {
				cnt1 += grid[i][n/2] * 2
			} else {
				diff += 1
			}
		}
	}

	if cnt1%4 == 0 || diff > 0 {
		ans += diff
	} else {
		ans += 2
	}

	return ans
}

# Accepted solution for LeetCode #3240: Minimum Number of Flips to Make Binary Grid Palindromic II
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

// Accepted solution for LeetCode #3240: Minimum Number of Flips to Make Binary Grid Palindromic II
/**
 * [3240] Minimum Number of Flips to Make Binary Grid Palindromic II
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn min_flips(grid: Vec<Vec<i32>>) -> i32 {
        let (m, n) = (grid.len(), grid[0].len());
        let mut result = 0;

        // For the kernel
        for i in 0..m / 2 {
            for j in 0..n / 2 {
                let mut one_count = 0;

                if grid[i][j] == 1 {
                    one_count += 1;
                }

                if grid[m - 1 - i][j] == 1 {
                    one_count += 1;
                }

                if grid[i][n - 1 - j] == 1 {
                    one_count += 1;
                }

                if grid[m - 1 - i][n - 1 - j] == 1 {
                    one_count += 1;
                }

                result += one_count.min(4 - one_count);
            }
        }

        let mut one_count = 0;
        let mut flip_count = 0;

        if m % 2 == 1 {
            for i in 0..n / 2 {
                if grid[m / 2][i] != grid[m / 2][n - 1 - i] {
                    flip_count += 1;
                } else {
                    if grid[m / 2][i] == 1 {
                        one_count += 2;
                    }
                }
            }
        }

        if n % 2 == 1 {
            for i in 0..m / 2 {
                if grid[i][n / 2] != grid[m - 1 - i][n / 2] {
                    flip_count += 1;
                } else {
                    if grid[i][n / 2] == 1 {
                        one_count += 2;
                    }
                }
            }
        }

        if one_count % 4 == 2 {
            if flip_count == 0 {
                result += 2;
            }
        }
        result += flip_count;

        if m % 2 == 1 && n % 2 == 1 {
            if grid[m / 2][n / 2] == 1 {
                result += 1;
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3240() {
        assert_eq!(2, Solution::min_flips(vec![vec![0, 0, 1], vec![1, 1, 1]]));
        assert_eq!(
            2,
            Solution::min_flips(vec![vec![0, 1], vec![0, 1], vec![0, 0]])
        );
        assert_eq!(
            3,
            Solution::min_flips(vec![vec![1, 0, 0], vec![0, 1, 0], vec![0, 0, 1]])
        );
        assert_eq!(2, Solution::min_flips(vec![vec![1], vec![1]]));
    }
}

// Accepted solution for LeetCode #3240: Minimum Number of Flips to Make Binary Grid Palindromic II
function minFlips(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let ans = 0;

    for (let i = 0; i < Math.floor(m / 2); i++) {
        for (let j = 0; j < Math.floor(n / 2); j++) {
            const x = m - i - 1;
            const y = n - j - 1;
            const cnt1 = grid[i][j] + grid[x][j] + grid[i][y] + grid[x][y];
            ans += Math.min(cnt1, 4 - cnt1);
        }
    }

    if (m % 2 === 1 && n % 2 === 1) {
        ans += grid[Math.floor(m / 2)][Math.floor(n / 2)];
    }

    let diff = 0,
        cnt1 = 0;

    if (m % 2 === 1) {
        for (let j = 0; j < Math.floor(n / 2); j++) {
            if (grid[Math.floor(m / 2)][j] === grid[Math.floor(m / 2)][n - j - 1]) {
                cnt1 += grid[Math.floor(m / 2)][j] * 2;
            } else {
                diff += 1;
            }
        }
    }

    if (n % 2 === 1) {
        for (let i = 0; i < Math.floor(m / 2); i++) {
            if (grid[i][Math.floor(n / 2)] === grid[m - i - 1][Math.floor(n / 2)]) {
                cnt1 += grid[i][Math.floor(n / 2)] * 2;
            } else {
                diff += 1;
            }
        }
    }

    ans += cnt1 % 4 === 0 || diff > 0 ? diff : 2;
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.