<p>You are given an <code>n x n</code> integer matrix <code>grid</code> where each value <code>grid[i][j]</code> represents the elevation at that point <code>(i, j)</code>.</p>

<p>The rain starts to fall. At time <code>t</code>, the depth of the water everywhere is <code>t</code>. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most <code>t</code>. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.</p>

<p>Return <em>the least time until you can reach the bottom right square </em><code>(n - 1, n - 1)</code><em> if you start at the top left square </em><code>(0, 0)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim1-grid.jpg" style="width: 164px; height: 165px;" />
<pre>
<strong>Input:</strong> grid = [[0,2],[1,3]]
<strong>Output:</strong> 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim2-grid-1.jpg" style="width: 404px; height: 405px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
<strong>Output:</strong> 16
<strong>Explanation:</strong> The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;&nbsp;n<sup>2</sup></code></li>
	<li>Each value <code>grid[i][j]</code> is <strong>unique</strong>.</li>
</ul>

// class Solution {
// public:
//     int swimInWater(vector<vector<int>>& grid) {
//         int n = grid.size();
//         int vis[n][n];
//         memset(vis, 0, sizeof(vis));
//         priority_queue<pair<int, pair<int, int>>,
//                        vector<pair<int, pair<int, int>>>,
//                        greater<pair<int, pair<int, int>>>>
//             pq;

//         pq.push({grid[0][0], {0, 0}});
//         vis[0][0] = 1;
//         int max_time = 0;
//         while (!pq.empty()) {
//             auto it = pq.top();
//             int time = it.first;
//             int row = it.second.first;
//             int col = it.second.second;
//             pq.pop();
//              max_time = max(max_time,time);
//             if (row == n - 1 && col == n - 1)
//                 return max_time;

//             int dr[] = {-1, 1, 0, 0};
//             int dc[] = {0, 0, -1, 1};

//             for (int i = 0; i < 4; i++) {
//                 int newr = row + dr[i];
//                 int newc = col + dc[i];
//                 if (newr >= 0 && newc >= 0 && newr < n && newc < n &&
//                     !vis[newr][newc]) {
                    
//                     vis[newr][newc] = 1;
//                     pq.push({grid[newr][newc], {newr, newc}});
//                 }
//             }
//         }
//         return 0;
//     }
// };
class Disjointset
{
private:
    vector<int> parent, size; // parent will store the parents '
                              // size will be taken to rank or compare the parent order while doing union.
public:
    Disjointset(int n)
    {
        parent.resize(n + 1);
        size.resize(n, 1);
        for (int i = 0; i < n + 1; i++)
        {
            parent[i] = i; // element will be it's parent only
        }
    }
    int find(int X)
    {
        if (parent[X] == X)
            return X;

        return parent[X] = find(parent[X]); // path compression
    }
    void UNION(int X, int Y)
    {
        int parent1 = find(X);
        int parent2 = find(Y);
        if (parent1 == parent2)
            return; // CYCLE DETECTED.
        if (size[parent1] < size[parent2])
        {
            parent[parent1] = parent2;
            size[parent2] += size[parent1];
        }
        else
        {
            parent[parent2] = parent1;
            size[parent1] += size[parent2];
        }
    }
    void printArrays()
    {
        cout << "PARENT" << endl;
        for (auto it : parent)
        {
            cout << it << " ";
        }
        cout << endl;
        cout << "SIZE" << endl;
        for (auto it : size)
        {
            cout << it << " ";
        }
    }
    int SIZE(int vertex) // returns the size of the whole disjoint set of which the current vertex is a part of.
    {
        return size[find(vertex)]; // returns the size of disjoint set that it is a part of.
    }

    int noOfdisjointSets(int n) // pass n as indexing. , gives all disjoint sets and not particular
    {
        // find the parent of all nodes.

        for (int i = 0; i < n; i++)
        {
            parent[i] = find(i);
        }

        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
            if (parent[i] == i)
            {
                cnt++;
            }
        }
        return cnt;

        /*
      //   METHOD 2
        set<int> s;
        for(int i=0;i<n;i++)
        {
            s.insert(parent[i]);
        }
        return s.size();
        */
    }
};

class Solution
{
public:
    int swimInWater(vector<vector<int>> &grid)
    {
        return UnionAndFind_swimInWater(grid);
    }
    int UnionAndFind_swimInWater(vector<vector<int>> &grid)
    {
        int n = grid.size();
        Disjointset ds(n * n);
        pair<int, int> locations[n * n];                // store the location of the element reachable exactly at time t.
        vector<vector<int>> vis(n, vector<int>(n, -1)); // can use grid for this purpose also.
        for (int i = 0; i < n; i++)                     // storing the locations of elements
        {
            for (int j = 0; j < n; j++)
            {
                locations[grid[i][j]] = {i, j};
            }
        }

        int delRow[] = {0, 0, 1, -1};
        int delCol[] = {1, -1, 0, 0};
        // now the time will be in range from 0 to n*n -1
        for (int time = 0; time < n * n; time++)
        {

            // find the loc of element reachable exactly at  this time {x,y}
            int x = locations[time].first;
            int y = locations[time].second;
            // mark the current as visited
            vis[x][y] = 1;
            // check if all it's neighbours, if they have already been reached then merge with it.
            for (int k = 0; k < 4; k++)
            {
                int nrow = x + delRow[k];
                int ncol = y + delCol[k];
                if (underBoundaries(nrow, ncol, n, n) && vis[nrow][ncol] == 1)
                {
                    ds.UNION(x*n+y,nrow*n+ncol);//join the locations here and not the heights.
                }
            }
            if (ds.find(0) == ds.find(n*n-1)) // if last element location in the connected component
            {
                return time;
            }
        }

        return n * n - 1; // since in max time it will be possible to reach the desired
        // the code will itself cover this situation and does not require it here only.
        /*
         some intutive improvements in the problem:
          since time is increasing linearly,
          so any element which has lesser value than current time is already visited
          so we can say element<time then already visited.
          we can save ourselves of vis extra n square space.
        */
    }
    bool underBoundaries(int nrow, int ncol, int n, int m)
    {
        if (nrow >= 0 && nrow < n && ncol >= 0 && ncol < m)
        {
            return true;
        }
        return false;
    }
};
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