Description#
You are given an empty 2D binary grid grid
of size m x n
. The grid represents a map where 0
's represent water and 1
's represent land. Initially, all the cells of grid
are water cells (i.e., all the cells are 0
's).
We may perform an add land operation which turns the water at position into a land. You are given an array positions
where positions[i] = [ri, ci]
is the position (ri, ci)
at which we should operate the ith
operation.
Return an array of integers answer
where answer[i]
is the number of islands after turning the cell (ri, ci)
into a land.
An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/0305.Number%20of%20Islands%20II/images/tmp-grid.jpg)
Input: m = 3, n = 3, positions = [[0,0],[0,1],[1,2],[2,1]]
Output: [1,1,2,3]
Explanation:
Initially, the 2d grid is filled with water.
- Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land. We have 1 island.
- Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land. We still have 1 island.
- Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land. We have 2 islands.
- Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land. We have 3 islands.
Example 2:
Input: m = 1, n = 1, positions = [[0,0]]
Output: [1]
Constraints:
1 <= m, n, positions.length <= 104
1 <= m * n <= 104
positions[i].length == 2
0 <= ri < m
0 <= ci < n
Follow up: Could you solve it in time complexity O(k log(mn))
, where k == positions.length
?
Solutions#
Solution 1: Union-Find#
We use a two-dimensional array $grid$ to represent a map, where $0$ and $1$ represent water and land respectively. Initially, all cells in $grid$ are water cells (i.e., all cells are $0$), and we use a variable $cnt$ to record the number of islands. The connectivity between islands can be maintained by a union-find set $uf$.
Next, we traverse each position $(i, j)$ in the array $positions$. If $grid[i][j]$ is $1$, it means that this position is already land, and we directly add $cnt$ to the answer; otherwise, we change the value of $grid[i][j]$ to $1$, and increase the value of $cnt$ by $1$. Then, we traverse the four directions of up, down, left, and right of this position. If a certain direction is $1$, and this position does not belong to the same connected component as $(i, j)$, then we merge this position with $(i, j)$, and decrease the value of $cnt$ by $1$. After traversing the four directions of up, down, left, and right of this position, we add $cnt$ to the answer.
The time complexity is $O(k \times \alpha(m \times n))$ or $O(k \times \log(m \times n))$, where $k$ is the length of $positions$, and $\alpha$ is the inverse function of the Ackermann function. In this problem, $\alpha(m \times n)$ can be considered as a very small constant.
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| class UnionFind:
def __init__(self, n: int):
self.p = list(range(n))
self.size = [1] * n
def find(self, x: int):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
def union(self, a: int, b: int) -> bool:
pa, pb = self.find(a - 1), self.find(b - 1)
if pa == pb:
return False
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]
return True
class Solution:
def numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:
uf = UnionFind(m * n)
grid = [[0] * n for _ in range(m)]
ans = []
dirs = (-1, 0, 1, 0, -1)
cnt = 0
for i, j in positions:
if grid[i][j]:
ans.append(cnt)
continue
grid[i][j] = 1
cnt += 1
for a, b in pairwise(dirs):
x, y = i + a, j + b
if (
0 <= x < m
and 0 <= y < n
and grid[x][y]
and uf.union(i * n + j, x * n + y)
):
cnt -= 1
ans.append(cnt)
return ans
|
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| class UnionFind {
private final int[] p;
private final int[] size;
public UnionFind(int n) {
p = new int[n];
size = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
size[i] = 1;
}
}
public int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
public boolean union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return false;
}
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
return true;
}
}
class Solution {
public List<Integer> numIslands2(int m, int n, int[][] positions) {
int[][] grid = new int[m][n];
UnionFind uf = new UnionFind(m * n);
int[] dirs = {-1, 0, 1, 0, -1};
int cnt = 0;
List<Integer> ans = new ArrayList<>();
for (var p : positions) {
int i = p[0], j = p[1];
if (grid[i][j] == 1) {
ans.add(cnt);
continue;
}
grid[i][j] = 1;
++cnt;
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1
&& uf.union(i * n + j, x * n + y)) {
--cnt;
}
}
ans.add(cnt);
}
return ans;
}
}
|
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| class UnionFind {
public:
UnionFind(int n) {
p = vector<int>(n);
size = vector<int>(n, 1);
iota(p.begin(), p.end(), 0);
}
bool unite(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return false;
}
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
return true;
}
int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
private:
vector<int> p, size;
};
class Solution {
public:
vector<int> numIslands2(int m, int n, vector<vector<int>>& positions) {
int grid[m][n];
memset(grid, 0, sizeof(grid));
UnionFind uf(m * n);
int dirs[5] = {-1, 0, 1, 0, -1};
int cnt = 0;
vector<int> ans;
for (auto& p : positions) {
int i = p[0], j = p[1];
if (grid[i][j]) {
ans.push_back(cnt);
continue;
}
grid[i][j] = 1;
++cnt;
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] && uf.unite(i * n + j, x * n + y)) {
--cnt;
}
}
ans.push_back(cnt);
}
return ans;
}
};
|
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| type unionFind struct {
p, size []int
}
func newUnionFind(n int) *unionFind {
p := make([]int, n)
size := make([]int, n)
for i := range p {
p[i] = i
size[i] = 1
}
return &unionFind{p, size}
}
func (uf *unionFind) find(x int) int {
if uf.p[x] != x {
uf.p[x] = uf.find(uf.p[x])
}
return uf.p[x]
}
func (uf *unionFind) union(a, b int) bool {
pa, pb := uf.find(a), uf.find(b)
if pa == pb {
return false
}
if uf.size[pa] > uf.size[pb] {
uf.p[pb] = pa
uf.size[pa] += uf.size[pb]
} else {
uf.p[pa] = pb
uf.size[pb] += uf.size[pa]
}
return true
}
func numIslands2(m int, n int, positions [][]int) (ans []int) {
uf := newUnionFind(m * n)
grid := make([][]int, m)
for i := range grid {
grid[i] = make([]int, n)
}
dirs := [5]int{-1, 0, 1, 0, -1}
cnt := 0
for _, p := range positions {
i, j := p[0], p[1]
if grid[i][j] == 1 {
ans = append(ans, cnt)
continue
}
grid[i][j] = 1
cnt++
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 && uf.union(i*n+j, x*n+y) {
cnt--
}
}
ans = append(ans, cnt)
}
return
}
|
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| class UnionFind {
p: number[];
size: number[];
constructor(n: number) {
this.p = Array(n)
.fill(0)
.map((_, i) => i);
this.size = Array(n).fill(1);
}
find(x: number): number {
if (this.p[x] !== x) {
this.p[x] = this.find(this.p[x]);
}
return this.p[x];
}
union(a: number, b: number): boolean {
const [pa, pb] = [this.find(a), this.find(b)];
if (pa === pb) {
return false;
}
if (this.size[pa] > this.size[pb]) {
this.p[pb] = pa;
this.size[pa] += this.size[pb];
} else {
this.p[pa] = pb;
this.size[pb] += this.size[pa];
}
return true;
}
}
function numIslands2(m: number, n: number, positions: number[][]): number[] {
const grid: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
const uf = new UnionFind(m * n);
const ans: number[] = [];
const dirs: number[] = [-1, 0, 1, 0, -1];
let cnt = 0;
for (const [i, j] of positions) {
if (grid[i][j]) {
ans.push(cnt);
continue;
}
grid[i][j] = 1;
++cnt;
for (let k = 0; k < 4; ++k) {
const [x, y] = [i + dirs[k], j + dirs[k + 1]];
if (x < 0 || x >= m || y < 0 || y >= n || !grid[x][y]) {
continue;
}
if (uf.union(i * n + j, x * n + y)) {
--cnt;
}
}
ans.push(cnt);
}
return ans;
}
|