"""
QUESTION:
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:

Given m = 3, n = 3, positions = [[0,0], [0,1], [1,2], [2,1]].
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
We return the result as an array: [1, 1, 2, 3]

Challenge:

Can you do it in time complexity O(k log mn), where k is the length of the positions?
ANSWER:
quick union
"""

class Solution(object):
    def numIslands2(self, m, n, positions):
        def find_root(p): #return root and height
            if connection[p] == p:
                return p, 1
            else:
                r, d = find_root(connection[p])
                return r, d+1

        ans = []
        last = 0
        connection = {}
        for p in positions:
            p = tuple(p)
            connection[p] = p
            last += 1
            root1 = None; d1 = 0
            for dx, dy in zip([1,0,-1,0],[0,1,0,-1]):
                q = tuple([p[0]+dx,p[1]+dy])
                if q in connection:
                    if not root1: root1, d1 = find_root(p)
                    root2, d2 = find_root(q)
                    if root1 != root2:
                        if d1<d2:
                            connection[root1] = root2
                        else:
                            connection[root2] = root1
                        last -= 1
            ans += [last]
        return ans


if __name__ == '__main__':
    print Solution().numIslands2(3, 3, [[0,0], [0,1], [1,2], [2,1], [0,2], [2,2]])