<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:
    //to check if all the elements in the matrix are 0 or not
    bool check(vector<vector<int>>&matrix){
        for(int i=0;i<matrix.size();i++){
            for(int j=0;j<matrix[0].size();j++){
                if(matrix[i][j]==1)
                    return false;
            }
        }
        return true;
    }
    
    //to flip the matrix
    void flipMatrix(vector<vector<int>>&matrix,int row,int column){
        int rows=matrix.size();
        int columns=matrix[0].size();
        
        matrix[row][column]=1^matrix[row][column];
        
        vector<int>delRow={-1,0,1,0};
        vector<int>delColumn={0,1,0,-1};
        
        for(int i=0;i<4;i++){
            int nrow=row+delRow[i];
            int ncolumn=column+delColumn[i];
            
            if(nrow>=0 and nrow<rows and ncolumn>=0 and ncolumn<columns)
                matrix[nrow][ncolumn]=1^matrix[nrow][ncolumn];
        }
    }
    
    //bfs approach
    int minFlips(vector<vector<int>>& mat) {
       
        map<vector<vector<int>>,int> visited;
        queue<pair<vector<vector<int>>,int>>q;
        
        q.push({mat,0});
        visited[mat]=1;
        
        while(!q.empty()){
            vector<vector<int>>temp=q.front().first;
            int steps=q.front().second;
            q.pop();
                
            if(check(temp))
                return steps;
                
            for(int row=0;row<mat.size();row++){
                for(int column=0;column<mat[0].size();column++){
                    vector<vector<int>>v=temp;
                    flipMatrix(v,row,column);
                    if(visited[v]==0){
                        q.push({v,steps+1});
                        visited[v]=1;
                    }
                }
            }
        }
        return -1;
    }
};
16700