<p>You are given an <code>n x n</code> binary matrix <code>grid</code>. You are allowed to change <strong>at most one</strong> <code>0</code> to be <code>1</code>.</p>

<p>Return <em>the size of the largest <strong>island</strong> in</em> <code>grid</code> <em>after applying this operation</em>.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>s.</p>

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

<pre>
<strong>Input:</strong> grid = [[1,0],[0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1],[1,0]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>Change the 0 to 1 and make the island bigger, only one island with area = 4.</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1],[1,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> Can&#39;t change any 0 to 1, only one island with area = 4.
</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;= 500</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <set>
#include <algorithm>

using namespace std;

class Solution {
public:
    // Find with path compression
    int findpar(vector<int>& parent, int X) {
        if (parent[X] == X) return X;
        return parent[X] = findpar(parent, parent[X]);
    }

    // Union with size tracking
    void merge(vector<int>& parent, vector<int>& size, int X, int Y) {
        int rootX = findpar(parent, X);
        int rootY = findpar(parent, Y);
        if (rootX != rootY) {
            parent[rootY] = rootX;
            size[rootX] += size[rootY]; // Update the size of the root component
        }
    }

    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<int> parent(n * n);
        vector<int> size(n * n, 1); // Initialize size array to 1 for each cell

        // Initialize each node's parent to itself
        for (int i = 0; i < parent.size(); i++) {
            parent[i] = i;
        }

        vector<int> rdel = {-1, 0, 1, 0};
        vector<int> cdel = {0, 1, 0, -1};

        // First pass: Union all 1s
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    int a = i * n + j;
                    for (int k = 0; k < 4; k++) {
                        int nx = i + rdel[k];
                        int ny = j + cdel[k];
                        if (nx >= 0 && nx < n && ny >= 0 && ny < n && grid[nx][ny] == 1) {
                            int b = nx * n + ny;
                            merge(parent, size, a, b); // Merge adjacent 1s
                        }
                    }
                }
            }
        }

        int largestisland = 0;

        // Populate island sizes in a map after the first pass
        unordered_map<int, int> mp;
        for (int i = 0; i < n * n; ++i) {
            int root = findpar(parent, i);
            if (grid[i / n][i % n] == 1) {
                mp[root] = size[root];
            }
        }

        // Find the initial largest island
        for (auto& a : mp) {
            largestisland = max(largestisland, a.second);
        }

        // Second pass: Check each 0 to find the largest possible island
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    set<int> components;
                    int newSize = 1; // Size of the new island if this 0 is changed to 1
                    for (int k = 0; k < 4; k++) {
                        int nx = i + rdel[k];
                        int ny = j + cdel[k];
                        if (nx >= 0 && nx < n && ny >= 0 && ny < n && grid[nx][ny] == 1) {
                            int root = findpar(parent, nx * n + ny);
                            components.insert(root); // Add unique roots of adjacent islands
                        }
                    }
                    for (auto it : components) {
                        newSize += mp[it]; // Sum the sizes of adjacent islands
                    }
                    largestisland = max(largestisland, newSize);
                }
            }
        }

        return largestisland;
    }
};


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