LeetCode #1284 — HARD

Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighbors if they share one edge.

Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot.

A binary matrix is a matrix with all cells equal to 0 or 1 only.

A zero matrix is a matrix with all cells equal to 0.

Example 1:

Input: mat = [[0,0],[0,1]]
Output: 3
Explanation: One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.

Example 2:

Input: mat = [[0]]
Output: 0
Explanation: Given matrix is a zero matrix. We do not need to change it.

Example 3:

Input: mat = [[1,0,0],[1,0,0]]
Output: -1
Explanation: Given matrix cannot be a zero matrix.

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 3
mat[i][j] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighbors if they share one edge. Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot. A binary matrix is a matrix with all cells equal to 0 or 1 only. A zero matrix is a matrix with all cells equal to 0.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

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

Example 2

[[0]]

Example 3

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

Core Insight

What unlocks the optimal approach

Flipping same index two times is like not flipping it at all. Each index can be flipped one time. Try all possible combinations. O(2^(n*m)).

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

Largest constraint values

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 #1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
class Solution {
    public int minFlips(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        int state = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (mat[i][j] == 1) {
                    state |= 1 << (i * n + j);
                }
            }
        }
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(state);
        Set<Integer> vis = new HashSet<>();
        vis.add(state);
        int ans = 0;
        int[] dirs = {0, -1, 0, 1, 0, 0};
        while (!q.isEmpty()) {
            for (int t = q.size(); t > 0; --t) {
                state = q.poll();
                if (state == 0) {
                    return ans;
                }
                for (int i = 0; i < m; ++i) {
                    for (int j = 0; j < n; ++j) {
                        int nxt = state;
                        for (int k = 0; k < 5; ++k) {
                            int x = i + dirs[k], y = j + dirs[k + 1];
                            if (x < 0 || x >= m || y < 0 || y >= n) {
                                continue;
                            }
                            if ((nxt & (1 << (x * n + y))) != 0) {
                                nxt -= 1 << (x * n + y);
                            } else {
                                nxt |= 1 << (x * n + y);
                            }
                        }
                        if (!vis.contains(nxt)) {
                            vis.add(nxt);
                            q.offer(nxt);
                        }
                    }
                }
            }
            ++ans;
        }
        return -1;
    }
}

// Accepted solution for LeetCode #1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
func minFlips(mat [][]int) int {
	m, n := len(mat), len(mat[0])
	state := 0
	for i, row := range mat {
		for j, v := range row {
			if v == 1 {
				state |= 1 << (i*n + j)
			}
		}
	}
	q := []int{state}
	vis := map[int]bool{state: true}
	ans := 0
	dirs := []int{0, -1, 0, 1, 0, 0}
	for len(q) > 0 {
		for t := len(q); t > 0; t-- {
			state = q[0]
			if state == 0 {
				return ans
			}
			q = q[1:]
			for i := 0; i < m; i++ {
				for j := 0; j < n; j++ {
					nxt := state
					for k := 0; k < 5; k++ {
						x, y := i+dirs[k], j+dirs[k+1]
						if x < 0 || x >= m || y < 0 || y >= n {
							continue
						}
						if (nxt & (1 << (x*n + y))) != 0 {
							nxt -= 1 << (x*n + y)
						} else {
							nxt |= 1 << (x*n + y)
						}
					}
					if !vis[nxt] {
						vis[nxt] = true
						q = append(q, nxt)
					}
				}
			}
		}
		ans++
	}
	return -1
}

# Accepted solution for LeetCode #1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
class Solution:
    def minFlips(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        state = sum(1 << (i * n + j) for i in range(m) for j in range(n) if mat[i][j])
        q = deque([state])
        vis = {state}
        ans = 0
        dirs = [0, -1, 0, 1, 0, 0]
        while q:
            for _ in range(len(q)):
                state = q.popleft()
                if state == 0:
                    return ans
                for i in range(m):
                    for j in range(n):
                        nxt = state
                        for k in range(5):
                            x, y = i + dirs[k], j + dirs[k + 1]
                            if not 0 <= x < m or not 0 <= y < n:
                                continue
                            if nxt & (1 << (x * n + y)):
                                nxt -= 1 << (x * n + y)
                            else:
                                nxt |= 1 << (x * n + y)
                        if nxt not in vis:
                            vis.add(nxt)
                            q.append(nxt)
            ans += 1
        return -1

// Accepted solution for LeetCode #1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
struct Solution;

impl Solution {
    fn min_flips(mut mat: Vec<Vec<i32>>) -> i32 {
        let n = mat.len();
        let m = mat[0].len();
        let mut res = std::usize::MAX;
        Self::dfs(0, 0, &mut mat, &mut res, n, m);
        if res == std::usize::MAX {
            -1
        } else {
            res as i32
        }
    }

    fn dfs(start: usize, k: usize, mat: &mut Vec<Vec<i32>>, min: &mut usize, n: usize, m: usize) {
        if start == n * m {
            if Self::ones(mat, n, m) == 0 {
                *min = (*min).min(k);
            }
        } else {
            let r = start / m;
            let c = start % m;
            Self::flip(r, c, mat, n, m);
            Self::dfs(start + 1, k + 1, mat, min, n, m);
            Self::flip(r, c, mat, n, m);
            Self::dfs(start + 1, k, mat, min, n, m);
        }
    }

    fn flip(i: usize, j: usize, mat: &mut Vec<Vec<i32>>, n: usize, m: usize) {
        mat[i][j] = 1 - mat[i][j];
        if i > 0 {
            mat[i - 1][j] = 1 - mat[i - 1][j];
        }
        if j > 0 {
            mat[i][j - 1] = 1 - mat[i][j - 1];
        }
        if i + 1 < n {
            mat[i + 1][j] = 1 - mat[i + 1][j];
        }
        if j + 1 < m {
            mat[i][j + 1] = 1 - mat[i][j + 1];
        }
    }

    fn ones(mat: &[Vec<i32>], n: usize, m: usize) -> usize {
        let mut res = 0;
        for i in 0..n {
            for j in 0..m {
                if mat[i][j] == 1 {
                    res += 1;
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let mat = vec_vec_i32![[0, 0], [0, 1]];
    let res = 3;
    assert_eq!(Solution::min_flips(mat), res);
    let mat = vec_vec_i32![[0]];
    let res = 0;
    assert_eq!(Solution::min_flips(mat), res);
    let mat = vec_vec_i32![[1, 1, 1], [1, 0, 1], [0, 0, 0]];
    let res = 6;
    assert_eq!(Solution::min_flips(mat), res);
    let mat = vec_vec_i32![[1, 0, 0], [1, 0, 0]];
    let res = -1;
    assert_eq!(Solution::min_flips(mat), res);
}

// Accepted solution for LeetCode #1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
// class Solution {
//     public int minFlips(int[][] mat) {
//         int m = mat.length, n = mat[0].length;
//         int state = 0;
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 if (mat[i][j] == 1) {
//                     state |= 1 << (i * n + j);
//                 }
//             }
//         }
//         Deque<Integer> q = new ArrayDeque<>();
//         q.offer(state);
//         Set<Integer> vis = new HashSet<>();
//         vis.add(state);
//         int ans = 0;
//         int[] dirs = {0, -1, 0, 1, 0, 0};
//         while (!q.isEmpty()) {
//             for (int t = q.size(); t > 0; --t) {
//                 state = q.poll();
//                 if (state == 0) {
//                     return ans;
//                 }
//                 for (int i = 0; i < m; ++i) {
//                     for (int j = 0; j < n; ++j) {
//                         int nxt = state;
//                         for (int k = 0; k < 5; ++k) {
//                             int x = i + dirs[k], y = j + dirs[k + 1];
//                             if (x < 0 || x >= m || y < 0 || y >= n) {
//                                 continue;
//                             }
//                             if ((nxt & (1 << (x * n + y))) != 0) {
//                                 nxt -= 1 << (x * n + y);
//                             } else {
//                                 nxt |= 1 << (x * n + y);
//                             }
//                         }
//                         if (!vis.contains(nxt)) {
//                             vis.add(nxt);
//                             q.offer(nxt);
//                         }
//                     }
//                 }
//             }
//             ++ans;
//         }
//         return -1;
//     }
// }

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

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → 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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.