Leetcode 305: Island count II

Sept. 13, 2019

Introduction

It is so enjoyable to write a short algorithm called island count II. I like to post my solution somewhere so that I can track my progress of learning.

My practice

Here is my C# code from my Leetcode github algorithm folder.

Here is the folder in my repository called 100 hard level algorithms.

The problem statement can be looked up here:

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?

Highlights of solution using union find algorithms

1. Understand the union find algorithm, which can be implemented using the array with parent id
2. Union find algorithm original parent id is itself
3. Set all element with value not 0 as a new island
4. Check all its four neighbor, if the neighbor's node is not -1, then see if the parents are the same or not. If not, union two disjoint sets.
5. Only challenge job is to write function called findRoot using recursive function, path compression.
6. Two for loops, outside one is to loop each position, inside loop to check four directions.