Number of Islands Ii
Created: March 6, 2020 by [lek-tin]
Last updated: March 6, 2020
A 2d grid map of m rows and n columns is initially filled with water. We may perform an addLand operation which turns the water at position (row, col) into a land. Given a list of positions to operate, count the number of islands after each addLand operation. 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
Input: m = 3, n = 3, positions = [[0,0], [0,1], [1,2], [2,1]]
Output: [1,1,2,3]
Explanation:
Initially, the 2d grid grid is filled with water. (Assume 0 represents water and 1 represents land).
0 0 0
0 0 0
0 0 0
Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land.
1 0 0
0 0 0 Number of islands = 1
0 0 0
Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land.
1 1 0
0 0 0 Number of islands = 1
0 0 0
Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land.
1 1 0
0 0 1 Number of islands = 2
0 0 0
Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land.
1 1 0
0 0 1 Number of islands = 3
0 1 0
Follow up
Can you do it in time complexity O(k log mn), where k is the length of the positions?
Solution
Time: O(m*n + L), L is the number of operations. Union operation takes essentially constant time when UnionFind is implemented with both path compression and union by rank.
Space: O(m*n)
class Solution:
def numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:
if m==0 or n==0 or len(positions)==0:
return 0
lookup = {}
counts = []
dirs = [ [0, 1], [0, -1], [1, 0], [-1, 0] ]
count = 0
for position in positions:
x, y = position
curr = (x, y)
if self.node_id(curr, n) in lookup:
# skip same coordinate. Update count with the previous count
counts.append(counts[-1])
count = counts[-1]
continue
count += 1
lookup[self.node_id(curr, n)] = self.node_id(curr, n)
for dir in dirs:
neighbor= (x+dir[0], y+dir[1])
if 0 <= neighbor[0] < m and 0 <= neighbor[1] < n and self.node_id(neighbor, n) in lookup:
if self.find(lookup, self.node_id(curr, n)) != self.find(lookup, self.node_id(neighbor, n)):
# Merge different islands, amortised time: O(log*k) ~= O(1)
self.union(lookup, self.node_id(curr, n), self.node_id(neighbor, n))
count -= 1
counts.append(count)
return counts
def node_id(self, coor, n):
return coor[0]*n + coor[1]
def find(self, lookup, x):
if lookup[x] != x:
# path compression
lookup[x] = self.find(lookup, lookup[x])
return lookup[x]
def union(self, lookup, x, y):
x_root, y_root = self.find(lookup, x), self.find(lookup, y)
lookup[min(x_root, y_root)] = max(x_root, y_root)