<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 DisJointSet{
    vector<int> parent,rank,size;
public:
    //constructor function
    DisJointSet(int n){
        parent.resize(n+1);
        rank.resize(n+1,0);
        size.resize(n+1,1);
        for(int i=0;i<=n;i++){
            parent[i] = i;
        }
    }
    //find parent of node with compression
    //compression means making the ultimate parent for each node
    // example - 1->2->3->4 everyone parent will be 1;
    int findUParent(int node){
        if(parent[node] == node){
            return node;
        }
        return parent[node] = findUParent(parent[node]);
    }

//Note in both rank/size we attach small to large
//if same then any to any

    //by Rank method
    void unionByRank(int u, int v){
        int parent_u = findUParent(u);
        int parent_v = findUParent(v);
        if(parent_u == parent_v) return;
        if(rank[parent_u] < rank[parent_v]){
            parent[parent_u] = parent_v;
        }else if(rank[parent_u] > rank[parent_v]){
            parent[parent_v] = parent_u;
        }else{
            parent[parent_u] = parent_v;
            //increasing the rank because both have same rank
            rank[parent_v]++;
        }
    }

    //by Size method
    void unionBySize(int u, int v){
        int parent_u = findUParent(u);
        int parent_v = findUParent(v);
        if(parent_u == parent_v) return;
        if(size[parent_u] < size[parent_v]){
            parent[parent_u] = parent_v;
            //since now parent_v will have all node inside which are of parent_u
            size[parent_v] += size[parent_u];
        }else{
            parent[parent_v] = parent_u;
            //vice-versa(above comment)
            size[parent_u] += size[parent_v];
        }
    }
};

class Solution {
public:
    int drow[4] = {1,0,-1,0};
    int dcol[4] = {0,1,0,-1};
    int swimInWater(vector<vector<int>>& grid) {
        // vector<pair<int,int>> edges();
        int n = grid.size();
        vector<pair<int,int>> edges(n*n);
        DisJointSet ds(n*n);
        vector<vector<int>> vis(n,vector<int> (n, -1));
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                int wt = grid[i][j];
                edges[wt] = {i,j};
            }
        }
        for(int i=0;i<n*n;i++){
            int x = edges[i].first;
            int y = edges[i].second;
            int currId = x*n + y;
            vis[x][y] = 1;
            for(int j=0;j<4;j++){
                int nx = x + drow[j];
                int ny = y + dcol[j];
                int adjId = nx*n + ny;
                if(nx >=0 && nx < n && ny >=0 && ny < n &&
                vis[nx][ny] == 1){
                    ds.unionBySize(currId, adjId);
                }
            }
            if(ds.findUParent(0) == ds.findUParent(n*n-1)){
                return i;
            }
        }
        //this will never reach since when i reaches n*n it will get the answer and return from inside loop.
        return 0;
    }
};
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