<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>

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

        string s = "", zero = "";
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                s += to_string(mat[i][j]);
                zero += "0";
            }
        }
        if (s == zero) 
            return 0;

        unordered_map<string, bool> vis;
        queue<string> q;
        q.push(s);
        vis[s] = true;
        int flips = 0;

        while (!q.empty()) {
            int size = q.size();
            flips++;
            while (size--) {
                string curr = q.front();
                q.pop();

                for (int k = 0; k < curr.size(); k++) {
                    int i = k / n;
                    int j = k % n;

                    string t = curr;
                    t[k] = (t[k] == '0') ? '1' : '0';

                    // Flip adjacent cells
                    // bottom
                    if (i + 1 < m) {
                        t[(i + 1) * n + j] = (t[(i + 1) * n + j] == '0') ? '1' : '0';
                    }

                    // top
                    if (i - 1 >= 0) {
                        t[(i - 1) * n + j] = (t[(i - 1) * n + j] == '0') ? '1' : '0';
                    }

                    // right
                    if (j + 1 < n) {
                        t[i * n + (j + 1)] = (t[i * n + (j + 1)] == '0') ? '1' : '0';
                    }
                    
                    // left
                    if (j - 1 >= 0) {
                        t[i * n + (j - 1)] = (t[i * n + (j - 1)] == '0') ? '1' : '0';
                    }

                    if (t == zero)
                        return flips;

                    if (!vis[t]) {
                        q.push(t);
                        vis[t] = true;
                    }
                }
            }
        }

        return -1;
    }
};

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