<p>Given a 2D&nbsp;<code>grid</code> consists of <code>0s</code> (land)&nbsp;and <code>1s</code> (water).&nbsp; An <em>island</em> is a maximal 4-directionally connected group of <code><font face="monospace">0</font>s</code> and a <em>closed island</em>&nbsp;is an island <strong>totally</strong>&nbsp;(all left, top, right, bottom) surrounded by <code>1s.</code></p>

<p>Return the number of <em>closed islands</em>.</p>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png" style="width: 240px; height: 120px;" /></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
Islands in gray are closed because they are completely surrounded by water (group of 1s).</pre>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png" style="width: 160px; height: 80px;" /></p>

<pre>
<strong>Input:</strong> grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1],
&nbsp;              [1,0,0,0,0,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,1,0,1,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
<strong>Output:</strong> 2
</pre>

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

<ul>
	<li><code>1 &lt;= grid.length, grid[0].length &lt;= 100</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;=1</code></li>
</ul>

#include<bits/stdc++.h>
using namespace std;

class disjoint {
    int* parent;
    int* size;
public:
    disjoint(int n) {
        parent = new int[n];
        size = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            size[i] = 1;
        }
    }

    ~disjoint() {
        delete[] parent;
        delete[] size;
    }

    int findPar(int i) {
        if (parent[i] == i) {
            return i;
        }
        return parent[i] = findPar(parent[i]);
    }

    void unionBySize(int x, int y) {
        int ulpx = findPar(x);
        int ulpy = findPar(y);
        if (ulpx != ulpy) {
            if (size[ulpx] > size[ulpy]) {
                parent[ulpy] = ulpx;
                size[ulpx] += size[ulpy];
            } else {
                parent[ulpx] = ulpy;
                size[ulpy] += size[ulpx];
            }
        }
    }
};

class Solution {
public:
    bool isValid(int n, int m, int a, int b) {
        return a >= 0 && a < n && b >= 0 && b < m;
    }
    int closedIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int drow[] = {0, 0, 1, -1};
        int dcol[] = {1, -1, 0, 0};
        
        // Handle border connections
        disjoint ds(n * m);
        // for (int i = 0; i < n; ++i) {
        //     for (int j = 0; j < m; ++j) {
        //         if ((i == 0 || i == n-1 || j == 0 || j == m-1) && grid[i][j] == 0) {
        //             int val1 = i * m + j;
        //             for (int k = 0; k < 4; ++k) {
        //                 int a = i + drow[k];
        //                 int b = j + dcol[k];
        //                 if (isValid(n, m, a, b) && grid[a][b] == 0) {
        //                     int val2 = a * m + b;
        //                     ds.unionBySize(val1, val2);
        //                 }
        //             }
        //         }
        //     }
        // }

        // // Count non-closed islands by iterating over the border again
        // int cnt;
        // unordered_set<int> s;
        // for (int i = 0; i < n; ++i) {
        //     if (grid[i][0] == 0 && ds.findPar(i * m) == i * m) {
        //         s.insert(i*m);
        //     }
        //     if (grid[i][m-1] == 0 && ds.findPar(i * m +( m-1)) == i * m + m-1) {
        //         s.insert(i * m +( m-1));
        //     }
        // }
        // for (int j = 0; j < m; ++j) {
        //     if (grid[0][j] == 0 && ds.findPar(j) == j){
        //         s.insert(j);
        //     }
        //     if (grid[n-1][j] == 0 && ds.findPar((n-1) * m + j) == (n-1) * m + j) {
        //         s.insert((n-1)*m+j);
        //     }
        // }
        // Union all interior zero cells
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (grid[i][j] == 0) {
                    int val1 = i * m + j;
                    for (int k = 0; k < 4; ++k) {
                        int a = i + drow[k];
                        int b = j + dcol[k];
                        if (isValid(n, m, a, b) && grid[a][b] == 0) {
                            int val2 = a * m + b;
                            ds.unionBySize(val1, val2);
                        }
                    }
                }
            }
        }
        unordered_set<int> s1,s2;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==0){
                if((i==0 || i==n-1 || j==0 || j==m-1 )) s1.insert(ds.findPar(i*m+j));
                s2.insert(ds.findPar(i*m+j));
                }
            }
        }
        return s2.size()-s1.size();
    }
};

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