<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>


int di[] = {-1, 0, 0, 1};
int dj[] = {0, -1, 1, 0};

class Solution {
public:
    bool isValid(int i, int j, int n, int m) {
        return i >= 0 && i < n && j >= 0 && j < m;
    }

    void getArea(vector<vector<int>>& grid, vector<vector<bool>>& vis, int i, int j, int& area, int n, int m) {
        if (!isValid(i, j, n, m) || grid[i][j] == 0 || vis[i][j]) return;
        vis[i][j] = true;
        area++;
        for (int k = 0; k < 4; k++) {
            int nextI = i + di[k];
            int nextJ = j + dj[k];
            getArea(grid, vis, nextI, nextJ, area, n, m);
        }
    }

    int findUltimateParent(vector<int>& ultimateParent, int node) {
        if (node == ultimateParent[node]) return node;
        return ultimateParent[node] = findUltimateParent(ultimateParent, ultimateParent[node]);
    }

    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<int> ultimateParent(n * m);

        for (int i = 0; i < n * m; i++) {
            ultimateParent[i] = i;
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    int currNode = i * m + j;
                    for (int k = 0; k < 4; k++) {
                        int nextI = i + di[k];
                        int nextJ = j + dj[k];
                        if (isValid(nextI, nextJ, n, m) && grid[nextI][nextJ] == 1) {
                            int nextNode = nextI * m + nextJ;
                            int p1 = findUltimateParent(ultimateParent, currNode);
                            int p2 = findUltimateParent(ultimateParent, nextNode);
                            if (p1 != p2) {
                                ultimateParent[p2] = p1;
                            }
                        }
                    }
                }
            }
        }

        vector<vector<bool>> vis(n, vector<bool>(m, false));
        unordered_map<int, int> mp;  // node -> area associated with it

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1 && !vis[i][j]) {
                    int area = 0;
                    getArea(grid, vis, i, j, area, n, m);
                    int root = findUltimateParent(ultimateParent, i * m + j);
                    mp[root] = area;
                }
            }
        }

        int ans = 0;
        unordered_set<int> st;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 0) {
                    st.clear();
                    int var = 1; // starting from 1 because we are converting this '0' to '1'
                    for (int k = 0; k < 4; k++) {
                        int nextI = i + di[k];
                        int nextJ = j + dj[k];
                        if (isValid(nextI, nextJ, n, m) && grid[nextI][nextJ] == 1) {
                            int root = findUltimateParent(ultimateParent, nextI * m + nextJ);
                            if (st.find(root) == st.end()) {
                                var += mp[root];
                                st.insert(root);
                            }
                        }
                    }
                    ans = max(ans, var);
                }
            }
        }

        // If there's no zero to flip, return the area of the largest island
        if (ans == 0) {
            for (auto& entry : mp) {
                ans = max(ans, entry.second);
            }
        }

        return ans;
    }
};

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