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; } } -
// 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; } }; -
print("Todo!")