Formatted question description: https://leetcode.ca/all/827.html
827. Making A Large Island (Hard)
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 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.
Notes:
1 <= grid.length = grid[0].length <= 50.0 <= grid[i][j] <= 1.
Related Topics: Depth-first Search
Solution 1. Union Find
// OJ: https://leetcode.com/problems/making-a-large-island/
// Time: O(MN)
// Space: O(MN)
class UnionFind {
vector<int> id, size;
public:
UnionFind(int N) : id(N), size(N, 1) {
iota(begin(id), end(id), 0);
}
int find(int x) {
return id[x] == x ? x : (id[x] = find(id[x]));
}
void connect(int x, int y) {
int p = find(x), q = find(y);
if (p == q) return;
id[p] = q;
size[q] += size[p];
}
int getSize(int x) { return size[find(x)]; }
};
class Solution {
public:
int largestIsland(vector<vector<int>>& G) {
int M = G.size(), N = G[0].size(), dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} }, ans = 0;
UnionFind uf(M * N);
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (G[i][j] == 0) continue;
for (auto &[dx, dy] : dirs) {
int x = i + dx, y = j + dy;
if (x < 0 || y < 0 || x >= M || y >= N || G[x][y] == 0) continue;
uf.connect(i * N + j, x * N + y);
}
ans = max(ans, uf.getSize(i * N + j));
}
}
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (G[i][j] == 1) continue;
unordered_map<int, int> m;
for (auto &[dx, dy] : dirs) {
int x = i + dx, y = j + dy;
if (x < 0 || y < 0 || x >= M || y >= N || G[x][y] == 0) continue;
m[uf.find(x * N + y)] = uf.getSize(x * N + y);
}
int size = 1;
for (auto &p : m) size += p.second;
ans = max(ans, size);
}
}
return ans;
}
};
Java
class Solution {
int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
public int largestIsland(int[][] grid) {
int side = grid.length;
int[][] islandNumbers = new int[side][side];
int islandCount = 0;
int maxSize = 0;
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
List<int[]> zeros = new ArrayList<int[]>();
for (int i = 0; i < side; i++) {
for (int j = 0; j < side; j++) {
if (grid[i][j] == 0)
zeros.add(new int[]{i, j});
else if (islandNumbers[i][j] == 0) {
islandCount++;
int size = breadthFirstSearch(grid, islandNumbers, islandCount, i, j);
maxSize = Math.max(maxSize, size);
map.put(islandCount, size);
}
}
}
for (int[] zero : zeros) {
int size = 1;
Set<Integer> islandsSet = new HashSet<Integer>();
int row = zero[0], column = zero[1];
for (int[] direction : directions) {
int newRow = row + direction[0], newColumn = column + direction[1];
if (newRow >= 0 && newRow < side && newColumn >= 0 && newColumn < side) {
int islandNumber = islandNumbers[newRow][newColumn];
if (islandNumber > 0)
islandsSet.add(islandNumber);
}
}
for (int island : islandsSet)
size += map.get(island);
maxSize = Math.max(maxSize, size);
}
return maxSize;
}
public int breadthFirstSearch(int[][] grid, int[][] islandNumbers, int islandCount, int row, int column) {
int size = 0;
islandNumbers[row][column] = islandCount;
int side = grid.length;
Queue<int[]> queue = new LinkedList<int[]>();
queue.offer(new int[]{row, column});
while (!queue.isEmpty()) {
int[] cell = queue.poll();
size++;
int curRow = cell[0], curColumn = cell[1];
for (int[] direction : directions) {
int newRow = curRow + direction[0], newColumn = curColumn + direction[1];
if (newRow >= 0 && newRow < side && newColumn >= 0 && newColumn < side && grid[newRow][newColumn] == 1 && islandNumbers[newRow][newColumn] == 0) {
islandNumbers[newRow][newColumn] = islandCount;
queue.offer(new int[]{newRow, newColumn});
}
}
}
return size;
}
}