Making A Large Island

Problem Statement

You are given an n x n binary matrix grid. You are allowed to change at most one 0 to be 1.

Return the size of the largest island in grid after applying this operation.

An island is a 4-directionally connected group of 1s.

Leetcode Link

Example 1:

Input: grid = [[1,0],[0,1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3.

Example 2:

Input: grid = [[1,1],[1,0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger, only one island with area = 4.

Example 3:

Input: grid = [[1,1],[1,1]]
Output: 4
Explanation: Can't change any 0 to 1, only one island with area = 4.

Constraints:

• n == grid.length
• n == grid[i].length
• 1 <= n <= 500
• grid[i][j] is either 0 or 1.

Code

Python Code

class Solution(object):
    def largestIsland(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        if not grid or not grid[0]: return 0
        # First do a regular union find to find all the islands and their sizes, remember the biggest island's size in 'biggest'.
        # Also remember all zero positions
        uf={}
        size={}
        rank={}
        zeros=[]
        biggest=0
        def findparent(x):
            if x not in uf: return x
            if uf[x]!=x: return findparent(uf[x])
            return x
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]==0:
                    zeros.append((i,j))
                    continue
                if (i,j) not in uf:
                    uf[(i,j)]=(i,j)
                    rank[(i,j)]=0
                    size[(i,j)]=1
                    biggest=max(biggest,1)
                for ni,nj in [(i-1,j),(i,j-1)]:
                    if 0<=ni<len(grid) and 0<=nj<len(grid[0]) and (ni,nj) in uf:
                        parA=findparent((i,j))
                        parB=findparent((ni,nj))
                        if parA!=parB:
                            if rank[parA]>rank[parB]:
                                uf[parB]=parA
                                rank[parA]+=1
                                size[parA]+=size[parB]
                                biggest=max(biggest,size[parA])
                            else:
                                uf[parA]=parB
                                rank[parB]+=1
                                size[parB]+=size[parA]
                                biggest=max(biggest,size[parB])
        # Now loop over the zeros, and check all of its '1' neighbors, and tally their sizes in 'neisizes'
        # If this new '1' connects multiple previously separated islands, the new island's size would be sum(neisizes)+1
        for i,j in zeros:
            neisizes=[]
            visited=set()
            for ni,nj in [(i-1,j),(i+1,j),(i,j-1),(i,j+1)]:
                if 0<=ni<len(grid) and 0<=nj<len(grid[0]) and (ni,nj) in uf:
                    par=findparent((ni,nj))
                    if par not in visited:
                        neisizes.append(size[par])
                        visited.add(par)
            biggest=max(biggest,sum(neisizes)+1)
        return biggest