<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>

int dx[] = {1,0,0,-1};
int dy[] = {0,1,-1,0};
class Solution {
    int getIndex(int x,int y,int n,int m)
    {
        return ((x*m) + y);
    }


    vector<string> helper(string s1,int n,int m)
    {
        vector<string> res;
        for(int i = 0;i < s1.length();i++)
        {
            string s2 = s1;
            s2[i] = (s2[i] == '0') ? '1' : '0';
            int x = i/m;
            int y = i%m;
            for(int j = 0;j < 4;j++)
            {
                int i1 = x + dx[j];
                int i2 = y + dy[j];
                if(i1 >= 0 && i1 < n && i2 >= 0 && i2 < m)
                {
                    int ind = getIndex(i1,i2,n,m);
                    s2[ind] = (s2[ind] == '0') ? '1' : '0';
                }
            }
            res.push_back(s2);
        }
        return res;
    }
public:
    int minFlips(vector<vector<int>>& mat) {
        int n = mat.size();
        int m = mat[0].size();

        string s1 = "";
        for(int i= 0;i < n;i++)
        {
            for(int j = 0;j < m;j++)
            {
                if(mat[i][j])
                    s1 += '1';
                else
                    s1 += '0';
//                s1 += (mat[i][j] + '0');
            }
        }

        string target = string(n*m,'0');
        if(s1 == target)
            return 0;
        
        queue<string> q;
        q.push(s1);
        unordered_map<string,int> mp;
        mp[s1]++;
        int cou = 0;
        while(!q.empty())
        {
            int x = q.size();
            while(x--)
            {
                string tt = q.front();
                q.pop();
                if(tt == target)
                    return cou;
                
                vector<string> v = helper(tt,n,m);
                for(int i = 0;i < v.size();i++)
                {
                    if(mp.count(v[i]) == 0)
                    {
                        mp[v[i]]++;
                        q.push(v[i]);
                    }
                }
            }
            cou++;
        }
        return -1;
    }
};
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