<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:
    bool isValid(int i, int j, int n, int m) {
        return i >= 0 && i < n && j >= 0 && j < m;
    }
    bool isFinal(vector<vector<int>> &grid) {
        bool ans = true;
        for (auto row : grid) {
            for (auto cell : row) {
                if (cell == 1) ans = false;
            }
        }
        return ans;
    }
    void flipGrid(int i, int j, vector<vector<int>> &grid) {
        int dr[4] = {-1, 0, 1, 0};
        int dc[4] = {0, 1, 0, -1};
        int n = grid.size();
        int m = grid[0].size();
        grid[i][j] = 1 - grid[i][j];
        for (int k = 0; k < 4; k++) {
            int adj_i = i + dr[k];
            int adj_j = j + dc[k];
            if (isValid(adj_i, adj_j, n, m)) {
                grid[adj_i][adj_j] = 1 - grid[adj_i][adj_j];
            }
        }
    }

    void showGrid(vector<vector<int>> &grid) {
        int n = grid.size();
        int m = grid[0].size();

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cout<<grid[i][j]<<"   ";
            }
            cout<<endl;
        }
    }
    string getString(vector<vector<int>> &grid) {
        string grid_str = "";
        for (auto row : grid) {
            for (auto cell : row) {
                grid_str = grid_str + (char)(cell + 48);
            }
        }
        return grid_str;
    }
    int minPos(vector<vector<int>> &grid, int cur_step,
    int &limit, int l_i, int l_j, vector<unordered_map<string, int>> &dp) {
        if (cur_step > limit) return 10000;
        string grid_str = getString(grid);
        if (dp[cur_step][grid_str] != 0) return dp[cur_step][grid_str];
        bool isFinal = true;
        int n = grid.size();
        int m = grid[0].size();
        int ans = 10000;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) isFinal = false;
                if (i != l_i || j != l_j) {
                    flipGrid(i, j, grid);
                    ans = min(ans, 1 + minPos(grid, cur_step + 1, limit, i, j, dp));
                    flipGrid(i, j, grid);
                }
            }
        }
        if (isFinal) return dp[cur_step][grid_str] = 0;
        return dp[cur_step][grid_str] = ans;
    }
    int minFlips(vector<vector<int>>& grid) {
        int limit = 9;
        vector<unordered_map<string, int>> dp(11);
        int ans = minPos(grid, 0, limit, -1, -1, dp);
        return ans == 10000 ? -1 : ans;
    }
};
18600