<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 flip_bit(int s, int idx) {
        s = s ^ (1 << idx);
        return s;
    }
    int flip(string s, int bin, int& mask, int n, int m) {
        for (int i = 0; i < n * m; i++) {
            if ((mask >> i) & 1) {// it is one make it as zero and flip other bits also
                bin = flip_bit(bin, i);
                if (i - m >= 0) //yop
                    bin = flip_bit(bin, i - m);
                if (i + m < n * m) // bottom
                    bin = flip_bit(bin, i + m);
                if (i % m < m - 1)          // left
                    bin = flip_bit(bin, i + 1);
                if (i % m > 0)        // right
                    bin = flip_bit(bin, i - 1);
            }
        }
        if (bin == 0)
            return __builtin_popcount(mask);// total set bits
        else
            return -1;// bin!=0 means all the elm in matrox are not zero
    }
    int minFlips(vector<vector<int>>& mat) {
        int n = mat.size();
        int m = mat[0].size();
        int ans = n * m + 1;
        string s = converMatrixToString(mat);
        // cout<<s;
        int bin = stoi(s, 0, 2);//statring pos and base 2
        
        cout <<bin;// if bin==0 atat would be the answer 
        for (int mask = 0; mask < (1 << (n * m)); mask++) {// tis has produced all the possible sub sequence to shift the bist there is one there
            int f = flip(s, bin, mask, n, m);//  
            if (f != -1) {// it measn bin is 0
                ans = min(ans, f);
            }
        }
        return ((ans == n * m + 1) ? -1 : ans);
    }
    string converMatrixToString(vector<vector<int>>& mat) {
        string ans = "";
        for (int i = 0; i < mat.size(); i++) {
            for (int j = 0; j < mat[0].size(); j++) {
                ans += to_string(mat[i][j]);
            }
        }
        return ans;
    }
};
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