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

<p>Return the <em>minimum number of steps</em> required to convert <code>mat</code> to a zero matrix or <code>-1</code> if you cannot.</p>

<p>A <strong>binary matrix</strong> is a matrix with all cells equal to <code>0</code> or <code>1</code> only.</p>

<p>A <strong>zero matrix</strong> is a matrix with all cells equal to <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/11/28/matrix.png" style="width: 409px; height: 86px;" />
<pre>
<strong>Input:</strong> mat = [[0,0],[0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,0,0],[1,0,0]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> Given matrix cannot be a zero matrix.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 3</code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

#include <vector>
#include <queue>
#include <bitset>
#include <unordered_map>

using namespace std;

class Solution {
public:
    int minFlips(vector<vector<int>>& mat) {
        int m = mat.size();
        int n = mat[0].size();
        int totalCells = m * n;

        auto toBitmask = [&](const vector<vector<int>>& matrix) -> int {
            int bitmask = 0;
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (matrix[i][j] == 1) {
                        bitmask |= (1 << (i * n + j));
                    }
                }
            }
            return bitmask;
        };

        auto isZeroMatrix = [&](int bitmask) -> bool {
            return bitmask == 0;
        };

        vector<pair<int, int>> neighbors = {{0, 0}, {1, 0}, {0, 1}, {-1, 0}, {0, -1}};
        
        auto getFlipEffect = [&](int x, int y) -> int {
            int effect = 0;
            for (const auto& [dx, dy] : neighbors) {
                int nx = x + dx, ny = y + dy;
                if (nx >= 0 && nx < m && ny >= 0 && ny < n) {
                    effect |= (1 << (nx * n + ny));
                }
            }
            return effect;
        };
        
        int startBitmask = toBitmask(mat);
        queue<int> q;
        unordered_map<int, int> dist;
        
        q.push(startBitmask);
        dist[startBitmask] = 0;
        
        while (!q.empty()) {
            int current = q.front();
            q.pop();
            
            if (isZeroMatrix(current)) {
                return dist[current];
            }
            
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    int newBitmask = current ^ getFlipEffect(i, j);
                    if (dist.find(newBitmask) == dist.end()) {
                        dist[newBitmask] = dist[current] + 1;
                        q.push(newBitmask);
                    }
                }
            }
        }
        
        return -1;
    }
};
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