[Leetcode] 0305. Number of Islands II

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內容目錄

Hard

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] = [r_i, c_i] is the position (r_i, c_i) at which we should operate the i^th operation.

Return an array of integers answer where answer[i] is the number of islands after turning the cell (r_i, c_i) 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:

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 <= r_i < m
0 <= c_i < n

Follow up: Could you solve it in time complexity O(k log(mn)), where k == positions.length?

Python

				
					# time complexity: O(m*n)
# space complexity: O(m*n)
from typing import List


class Solution:
    class UnionFind:
        def __init__(self, size):
            self.rep = [i for i in range(size)]
            self.rank = [0] * size
            self.count = 0

        def find(self, node):
            if node != self.rep[node]:
                self.rep[node] = self.find(self.rep[node])
            return self.rep[node]

        def union(self, node1, node2):
            rep1 = self.find(node1)
            rep2 = self.find(node2)
            if rep1 != rep2:
                if self.rank[rep1] > self.rank[rep2]:
                    self.rep[rep2] = rep1
                elif self.rank[rep2] < self.rank[rep1]:
                    self.rep[rep1] = rep2
                else:
                    self.rep[rep2] = rep1
                    self.rank[rep1] += 1
                self.count -= 1

    def numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:
        uf = self.UnionFind(m * n)
        res = []
        land_map = set()

        for row, col in positions:
            if (row, col) in land_map:
                res.append(uf.count)
                continue

            land_map.add((row, col))
            uf.count += 1

            for i, j in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
                r, c = row + i, col + j
                if r < 0 or c < 0 or r >= m or c >= n:
                    continue
                if (r, c) in land_map:
                    uf.union(row * n + col, r * n + c)

            res.append(uf.count)

        return res


m = 3
n = 3
positions = [[0, 0], [0, 1], [1, 2], [2, 1]]
print(Solution().numIslands2(m, n, positions))

Hard

Python

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