<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>

#define ll long long
#define pb push_back
class Solution {
public:
    int minFlips(vector<vector<int>>& mat) {
        ll n = mat.size();
        ll m = mat[0].size();

        map<vector<vector<int>>,ll>mp;
        mp[mat]=0;
        ll delx[4]={-1,0,1,0};
        ll dely[4]={0,1,0,-1};
        priority_queue<pair<ll,vector<vector<int>>>,vector<pair<ll,vector<vector<int>>>>,greater<pair<ll,vector<vector<int>>>>>pq;
        pq.push({0,mat});
        while(!pq.empty()){
            vector<vector<int>>sam = pq.top().second;
            ll distance = pq.top().first;
            pq.pop();
            for(ll i=0;i<n;i++){
                for(ll j=0;j<m;j++){
                    sam[i][j]=1-sam[i][j];
                    for(ll k=0;k<4;k++){
                        ll newx = i+delx[k];
                        ll newy = j+dely[k];
                        if(newx>=0 && newx<n && newy>=0 && newy<m){
                            sam[newx][newy]=1-sam[newx][newy];
                        }
                    }
                    if(mp.find(sam)==mp.end()){
                        mp[sam]=distance+1;
                        pq.push({mp[sam],sam});
                    }else if(mp[sam]>distance+1){
                        mp[sam]=distance+1;
                        pq.push({mp[sam],sam});  
                    }
                     sam[i][j]=1-sam[i][j];

                    for(ll k=0;k<4;k++){
                        ll newx = i+delx[k];
                        ll newy = j+dely[k];
                        if(newx>=0 && newx<n && newy>=0 && newy<m){
                            sam[newx][newy]=1-sam[newx][newy];
                        }
                    }
                }
            }
        }
        vector<vector<int>>poke(n,vector<int>(m));
        if(mp.find(poke)==mp.end())return -1;
        else return mp[poke];
    }
};
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