<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:
    bool check(vector<vector<int>> pairs,  vector<vector<int>> a) {
        int n =  a.size();
        int m  = a[0].size();
        for(auto pair : pairs){
            vector<int> dir = {1, 0, -1, 0, 1};
            a[pair[0]][pair[1]] = (a[pair[0]][pair[1]] ? 0 : 1);
            for(int d = 0 ; d < 4; d++){
                int x  = pair[0] + dir[d];
                int  y =  pair[1] + dir[d+1];
                if(x>=0 && x < n && y>=0 && y < m){
                    a[x][y] = (a[x][y] ? 0 : 1);
                }
            }
        }
        for(int i = 0 ; i <  n; i++){
            for(int j = 0 ; j < m; j++){
                if(a[i][j]) return false;
            }
        }
        return true;
    }
    int minFlips(vector<vector<int>>& a) {
        int n =  a.size();
        int m  = a[0].size();

        int ans =  INT_MAX;
        
        for(int i = 0 ; i < (1 << (m * n)); i++){
            int bit =  i;
            
            int index = 0;
            vector<vector<int>> pairs;
            while(bit){
                // cout << bit << endl;
                if(bit & 1){
                    int c = index % m;
                    int r = index / m;
                    pairs.push_back({r, c});
                }
                index++;
                // cout << index << endl;
                bit >>= 1;
            }
            // for(auto  i : pairs)cout << i[0] << " " << i[1]<< endl;
            if(check(pairs, a)){
                
                ans =  min(ans, (int)pairs.size());
            }
        }
        return ans == INT_MAX ? -1 : ans;
    }
};
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