<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 {
    const bool check(const vector<vector<int>>& mat, const int rSize, const int cSize)
    {
        for(int r{}; r < rSize; ++r)
        {
            for(int c{}; c < cSize; ++c)
            {
                if(mat[r][c])
                    return false;
            }
        }

        return true;
    }

    void flip(vector<vector<int>>& mat, vector<vector<bool>>& flipped, const int rSize, const int cSize, 
        const int r, const int c)
    {
        flipped[r][c] = !flipped[r][c];

        mat[r][c] = !mat[r][c];

        if(r-1 >= 0)
            mat[r-1][c] = !mat[r-1][c];

        if(r+1 != rSize)
            mat[r+1][c] = !mat[r+1][c];

        if(c-1 >= 0)
            mat[r][c-1] = !mat[r][c-1];

        if(c+1 != cSize)
            mat[r][c+1] = !mat[r][c+1];
    }

    const int solve(vector<vector<int>>& mat, const int rSize, const int cSize,
        vector<vector<bool>>& flipped, const int maxFlip, const int currFlip)
    {
        if(maxFlip == currFlip)
        {
            if(check(mat,rSize,cSize))
                return currFlip;
            return -1;
        }

        for(int rr{}; rr < rSize; ++rr)
        {
            for(int cc{}; cc < cSize; ++cc)
            {
                if(flipped[rr][cc] == false)
                {
                    flip(mat,flipped,rSize,cSize,rr,cc);

                    const int innerResult = 
                        solve(mat,rSize,cSize,flipped,maxFlip,currFlip+1);

                    if(innerResult != -1)
                        return innerResult;

                    flip(mat,flipped,rSize,cSize,rr,cc);
                }
            }
        }

        return -1;
    }
public:
    int minFlips(vector<vector<int>>& mat) {
        const int rSize = mat.size();
        const int cSize = mat[0].size();
        const int totalCount = rSize * cSize;

        vector<vector<bool>> flipped(rSize, vector<bool>(cSize, false));
        
        // expected flip count
        for(int maxFlip{}; maxFlip <= totalCount; ++maxFlip)
        {
            // curr flip count, r, c
            const int result = solve(mat,rSize,cSize,flipped,maxFlip,0);

            if(result != -1)
                return result;
        }

        return -1;
    }
};
10600