305. 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.

solution

The intuition to use union-find here is pretty obvious, but the difficulty is wrapping your head around the details of how it works.

A couple of points to note:

• Instead of using a list to store roots and sizes, use dictionaries, keyed by tuples of (x, y).
• We initially start with no nodes, and no sizes.
• When we add a node, it starts out being its own parent, with a size of 1.
  • We then check the 4 “edges”, running union on the neighboring node if it exists.

My main hiccup was trying to figure out what would happen if a new node caused two other components to now all become connected. How do we know which component this new node will choose to join?

After some thought, I realized that we union each edge in sequence. That means that we first connect the node to component 1. Then, now that the node is a part of component 1, it is unioned with component 2, as normal.

def numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:
	parents = defaultdict()
	sizes = defaultdict(int)
	dirs = [[0, 1], [1, 0], [0, -1], [-1, 0]]
	components = 0
	  
	def union(coord1, coord2):
		nonlocal components
	  
		p1, p2 = find(coord1), find(coord2)
	  
		if p1 == p2:
			return
	  
		# make p1 always smaller
		if sizes[p1] > sizes[p2]:
			p1, p2 = p2, p1
	  
		parents[p1] = p2
		sizes[p2] += sizes[p1]
		components -= 1
	  
	def find(coord):
		if coord != parents[coord]:
			parents[coord] = find(parents[coord])
		return parents[coord]
	  
	ans = []
	for x, y in positions:
		coord = (x, y)
 
		# duplicate operation
		if sizes[coord] > 0:
			ans.append(components)
			continue
 
		sizes[coord] = 1
		parents[coord] = coord
		components += 1
		  
		for dx, dy in dirs:
			nx, ny = x+dx, y+dy
			if 0 <= nx < m and 0 <= ny < n:
				if sizes[(nx, ny)] > 0:
					union((x,y), (nx, ny))
		ans.append(components)
	return ans