Lintcode 1391. Making A Large Island

In a 2D grid of 0s and 1s, we change at most one 0 to a 1.

After, what is the size of the largest island? (An island is a 4-directionally connected group of 1s).

Example

Example 1:

Input:[[1, 0], [0, 1]]
Output:3

Explanation:
Change one 0 to 1 and connect two 1s, then we get an island with area = 3.

Example 2:

Input: [[1, 1], [1, 0]]
Output:4

Explanation:
 Change the 0 to 1 and make the island bigger, only one island with area = 4.

Example 3:

Input:[[1, 1], [1, 1]]
Output:4

Explanation:
 Can't change any 0 to 1, only one island with area = 4.

Notice

• 1 <= grid.length = grid[0].length <= 50.
• 0 <= grid[i][j] <= 1.

Code (Java):

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public class Solution {     /**      * @param grid:      * @return: nothing      */     public int largestIsland(int[][] grid) {         //         if (grid == null || grid.length == 0) {             return 0;         }                   int m = grid.length;         int n = grid[0].length;                   UF uf = new UF(m, n, grid);                   int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};                   // step 1: connect all 1s         for (int i = 0; i < m; i++) {             for (int j = 0; j < n; j++) {                 for (int[] dir : dirs) {                     int nx = i + dir[0];                     int ny = j + dir[1];                                           if (grid[i][j] == 1 && nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] == 1) {                         uf.union(i * n + j, nx * n + ny);                     }                 }             }         }                   int maxSize = 0;                   // step 2: find all 0s and try to connect         for (int i = 0; i < m; i++) {             for (int j = 0; j < n; j++) {                 int size = 1;                 Set<Integer> parents = new HashSet<>();                 for (int[] dir : dirs) {                     int nx = i + dir[0];                     int ny = j + dir[1];                                           if (grid[i][j] == 0 && nx >= 0 && nx < m && ny >=0 && ny < n && grid[nx][ny] == 1) {                         int parent = uf.getParent(nx * n + ny);                         if (!parents.contains(parent)) {                             parents.add(parent);                             size += uf.getSize(nx * n + ny);                         }                     }                 }                                   maxSize = Math.max(size, maxSize);             }         }                   return maxSize;     } }   class UF {     int[] parents;     int[] size;     int m;     int n;     int[][] grid;           public UF (int m, int n, int[][] grid) {         this.m = m;         this.n = n;         this.grid = grid;                   parents = new int[m * n];         size = new int[m * n];                   for (int i = 0; i < m; i++) {             for (int j = 0; j < n; j++) {                 parents[i * n + j] = i * n + j;                 size[i * n + j] = 1;             }         }     }           public void union(int a, int b) {         int pa = find(a);         int pb = find(b);                   if (pa != pb) {             parents[pa] = pb;             size[pb] += size[pa];         }     }           private int find(int a) {         int root = a;                   while (root != parents[root]) {             root = parents[root];         }                   // path compression         while (root != a) {             int parent = parents[a];             parents[a] = root;             a = parent;         }                   return root;     }           public int getSize(int a) {         int pa = find(a);         return size[pa];     }           public int getParent(int a) {         return find(a);     } }