<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 isValid(int n, int m, int i, int j){
        if(i<0 || i>=n || j<0 || j>=m)
            return 0;
        return 1;
    }

    int bfs(vector<vector<int>>&mat, map<vector<vector<int>>, int> &vis, map<vector<vector<int>>, int> &ans){
        int n = mat.size();
        int m = mat[0].size();
        queue<pair< vector<vector<int>>, int>>q;

        vector<pair<int, int>>side{{-1, 0}, {1, 0}, {0, 1}, {0, -1}, {0,0}};

        q.push({mat, 0});
        vis[mat]=1;
        ans[mat]=0;
        while(!q.empty()){
            vector<vector<int>> tp = q.front().first;
            int dis = q.front().second;
            q.pop();
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    vector<vector<int>>ngr = tp;
                    for(int k=0;k<side.size();k++){
                        if(isValid(n, m, i+side[k].first, j+side[k].second)){
                            ngr[i+side[k].first][j+side[k].second] = ngr[i+side[k].first][j+side[k].second]==0?1:0;
                        }      
                    }
                    if(!vis[ngr]){
                            vis[ngr]=1;
                            ans[ngr]=dis+1;
                            q.push({ngr, dis+1});
                    }
                }
            }
        }
        vector<vector<int>>zeroMat = mat;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                zeroMat[i][j]=0;
            }
        }

        return vis[zeroMat]==0 ? -1 : ans[zeroMat];
    }

    int minFlips(vector<vector<int>>& mat) {
        map<vector<vector<int>>, int>isVisited;
        map<vector<vector<int>>, int>ans;

        return bfs(mat, isVisited, ans);

        //return ans[mat];
    }
};
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