Number of Islands II Python Solution

Tested Python solution for LeetCode 305 with 15 pytest cases. Generate a practice environment with lcpy.

LeetCode 305, Hard. Topics: Union Find, Array, Hash Table. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 15 parametrized pytest cases, and a playground notebook:

lcpy gen -n 305   # by problem number
lcpy gen -s number_of_islands_ii   # by problem name

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Problem

You are given an empty 2D binary grid grid of size m x n. The grid represents a map where 0s represent water and 1s represent land. Initially, all the cells of grid are water cells (i.e., all the cells are 0s).

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.

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Examples

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.

Input: m = 1, n = 1, positions = [[0,0]]
Output: [1]

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Constraints

• 1 <= m, n, positions.length <= 10^4
• 1 <= m * n <= 10^4
• 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?

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class UnionFind:
    def __init__(self, n: int) -> None:
        self.p = list(range(n))
        self.size = [1] * n

    def find(self, x: int) -> 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), self.find(b)
        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:
    # Time: O(k * alpha(m * n)) — one union-find pass over positions
    # Space: O(m * n) — parent and size arrays
    def num_islands2(self, m: int, n: int, positions: list[list[int]]) -> list[int]:
        uf = UnionFind(m * n)
        land = set()
        count = 0
        answer: list[int] = []
        for i, j in positions:
            if (i, j) in land:
                answer.append(count)
                continue
            land.add((i, j))
            count += 1
            for x, y in ((i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)):
                if (x, y) in land and uf.union(i * n + j, x * n + y):
                    count -= 1
            answer.append(count)
        return answer

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Complexity

Time	Space
O(k * alpha(m * n)) — one union-find pass over positions	O(m * n) — parent and size arrays

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Tags

NeetCode All.