<p>Given a 2D array of characters <code>grid</code> of size <code>m x n</code>, you need to find if there exists any cycle consisting of the <strong>same value</strong> in <code>grid</code>.</p>

<p>A cycle is a path of <strong>length 4 or more</strong> in the grid that starts and ends at the same cell. From a given cell, you can move to one of the cells adjacent to it - in one of the four directions (up, down, left, or right), if it has the <strong>same value</strong> of the current cell.</p>

<p>Also, you cannot move to the cell that you visited in your last move. For example, the cycle <code>(1, 1) -&gt; (1, 2) -&gt; (1, 1)</code> is invalid because from <code>(1, 2)</code> we visited <code>(1, 1)</code> which was the last visited cell.</p>

<p>Return <code>true</code> if any cycle of the same value exists in <code>grid</code>, otherwise, return <code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/1.png" style="width: 231px; height: 152px;" /></strong></p>

<pre>
<strong>Input:</strong> grid = [[&quot;a&quot;,&quot;a&quot;,&quot;a&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;b&quot;,&quot;b&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;b&quot;,&quot;b&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;a&quot;,&quot;a&quot;,&quot;a&quot;]]
<strong>Output:</strong> true
<strong>Explanation: </strong>There are two valid cycles shown in different colors in the image below:
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/11.png" style="width: 225px; height: 163px;" />
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/22.png" style="width: 236px; height: 154px;" /></strong></p>

<pre>
<strong>Input:</strong> grid = [[&quot;c&quot;,&quot;c&quot;,&quot;c&quot;,&quot;a&quot;],[&quot;c&quot;,&quot;d&quot;,&quot;c&quot;,&quot;c&quot;],[&quot;c&quot;,&quot;c&quot;,&quot;e&quot;,&quot;c&quot;],[&quot;f&quot;,&quot;c&quot;,&quot;c&quot;,&quot;c&quot;]]
<strong>Output:</strong> true
<strong>Explanation: </strong>There is only one valid cycle highlighted in the image below:
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/2.png" style="width: 229px; height: 157px;" />
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/3.png" style="width: 183px; height: 120px;" /></strong></p>

<pre>
<strong>Input:</strong> grid = [[&quot;a&quot;,&quot;b&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;z&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;b&quot;,&quot;a&quot;]]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>grid</code> consists only of lowercase English letters.</li>
</ul>

class Solution {
public:
    bool recurse(vector<vector<char>>& grid, vector<vector<int>>& dp, set<pair<int,int>>& visited, char currChar, int i, int j, int iLast, int jLast){
        if(i >= grid.size() || i < 0 || j >= grid[0].size() || j < 0 ){
            return false;
        }
        if(grid[i][j] != currChar){
            return false;
        }
        if(dp[i][j] != -1){
            return dp[i][j];
        }
        // if(iFirst == i && jFirst == j && visited[i][j] == 1 && idxToCount.find({i,j}) != idxToCount.end()){
        //     return true;
        // }
        // if(visited[i][j] != 0){
        //     return false;
        // }
        if(visited.find({i,j}) != visited.end()){
            return true;
        }
        // visited[i][j] = 1;
        // idxToCount[{i,j}] = idxToCount[{iLast, jLast}] + 1;
        visited.insert({i,j});
        vector<pair<int,int>> directions = {
            {1, 0},
            {0, 1},
            {-1, 0},
            {0, -1}
        };
        // bool found = false;
        for(int k = 0; k<4; k++){
            int newI = i + directions[k].first;
            int newJ = j + directions[k].second;
            if(newI == iLast && newJ == jLast){
                continue;
            }
            if(recurse(grid, dp, visited, currChar, newI, newJ, i, j)){
                dp[i][j] = 1;
                return true;
            }        
        }
        dp[i][j] = 0;
        return false;
    }
    bool containsCycle(vector<vector<char>>& grid) {
        vector<vector<int>> dp(grid.size(), vector<int>(grid[0].size(), -1));

        for(int i = 0; i<grid.size(); i++){
            for(int j = 0; j<grid[0].size(); j++){
                // visited[i][j] = 1;
                set<pair<int,int>> visited;
                bool found = recurse(grid, dp, visited, grid[i][j], i, j, -1, -1);
                if(found){
                    return true;
                }
            }
        }
        return false;
    }
};
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