<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>

vector<vector<int>>dir{{0,-1},{0,1},{1,0},{-1,0}};
class Solution {
public:
    int flip(int num,int i,int n,int m) {
        int x=i/m,y=i%m;
        num^=(1<<i);
        for(auto it:dir) {
            int a=it[0]+x;
            int b =it[1]+y;
            if(a>=0&&a<n&&b>=0&&b<m) {
                int z = a*m+b;
                num^=(1<<z);
            }
        }
        return num;
    }
    int minFlips(vector<vector<int>>& mat) {
        int n=mat.size(),m=mat[0].size();
        queue<pair<int,int>>q;
        int num=0;
        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                if(mat[i][j]) {
                    int a = i*m+j;
                    num+=(1<<a);
                }
            }
        }
        q.push({num,0});
        unordered_map<int,int>mp;
        mp[num]=1;
        while(!q.empty()) {
            int u = q.front().first;
            int d = q.front().second;
            q.pop();
            if(u==0) return d;
            for(int i=0;i<n*m;i++) {
                int x = flip(u,i,n,m);
                if(!mp.count(x)) {
                    mp[x]=1;
                    q.push({x,d+1});
                }
            }
        }
        return -1;
    }
};
12200