<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:
    vector<vector<int>> intToMat(int num, int m, int n){
        vector<vector<int>> ans(m, vector<int>(n, 0));
        for(int i=m-1;i>=0;i--){
            for(int j=n-1;j>=0;j--){
                ans[i][j] = num%2;
                num /= 2;
            }
        }
        return ans;
    }

    int matToInteger(vector<vector<int>> & mat){
        int ans = 0;
        for(int i=0;i<mat.size();i++){
            for(int j=0;j<mat[i].size();j++){
                ans = ans*2 + mat[i][j];
            }
        }
        return ans;
    }

    vector<vector<int>> flip(vector<vector<int>> mat, int i, int j){
        int m = mat.size();
        int n = mat[0].size();
        mat[i][j] = (mat[i][j] + 1)%2;
        if(i-1>=0){
            mat[i-1][j] = (mat[i-1][j] + 1)%2;
        }
        if(i+1<m){
            mat[i+1][j] = (mat[i+1][j] + 1)%2;
        }
        if(j-1>=0){
            mat[i][j-1] = (mat[i][j-1] + 1)%2;
        }
        if(j+1<n){
            mat[i][j+1] = (mat[i][j+1] + 1)%2;
        }
        return mat;
    }

    bool isFound(vector<vector<int>>& mat, vector<bool> & isVisited){
        int num = matToInteger(mat);
        if(num == 0){
            return 1;
        }
        if(isVisited[num]){
            return 0;
        }
        isVisited[num] = 1;
        for(int i=0;i<mat.size();i++){
            for(int j=0;j<mat[i].size();j++){
                vector<vector<int>> fliped_map = flip(mat, i, j);
                if(isFound(fliped_map, isVisited)){
                    return 1;
                }
            }
        }
        return 0;
    }

    int minFlips(vector<vector<int>>& mat) {
        vector<bool> isVisited(512, 0);
        queue<vector<vector<int>>> Q;
        Q.push(mat);
        queue<int> Q2;
        Q2.push(0);
        while(!Q.empty()){
            vector<vector<int>> newMat = Q.front();
            Q.pop();
            int level = Q2.front();
            Q2.pop();
            int num = matToInteger(newMat);
            if(num == 0){
                return level;
            }
            if(isVisited[num]){
                continue;
            }
            isVisited[num] = 1;

            for(int i=0;i<newMat.size();i++){
                for(int j=0;j<newMat[i].size();j++){
                    vector<vector<int>> fliped_map = flip(newMat, i, j);
                    Q.push(fliped_map);
                    Q2.push(level + 1);
                }
            }
        }
        return -1;

    }
};
21000