# LeetCode 305. Number of Islands II

## Description

https://leetcode.com/problems/number-of-islands-ii/

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:

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?

## Explanation

Use union-find data structure to help solve the problem. Whenever we make a position to island, we also use union-find to link its neighbors so that the next time if we visit any of neighbor positions, we know it belongs to the same island.

## Python Solution

class UnionFind:
def __init__(self):
self.father = {}
self.count = 0

def union(self, a, b):
root_a = self.find(a)
root_b = self.find(b)

if root_a == root_b:
return

self.father[root_b] = root_a
self.count -= 1

def find(self, point):
path = []

while point != self.father[point]:
path.append(point)

point = self.father[point]

for p in path:
self.father[p] = point

return point

class Solution:
def numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:

islands = set()

results = []

union_find = UnionFind()

directions = [[0, 1], [0, -1], [1, 0], [-1, 0]]

for position in positions:
x, y = position[0], position[1]

if (x, y) in islands:

results.append(union_find.count)

continue

union_find.father[(x, y)] = (x, y)

union_find.count += 1

for dx, dy in directions:
nx, ny = x + dx, y + dy

if (nx, ny) in islands:
union_find.union((x, y), (nx, ny))

results.append(union_find.count)

return results
• Time Complexity: O(m * n + L), where L is the number of operations.
• Space Complexity: O(m *n ).