Here is the blog I like to study about the algorithm which is locked on Leetcode.com.
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
Initially, the 2d grid
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
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.
C++ code in the study blog starts here:
C++ code in the study blog ends here:
Actionable Items
Plan to write C# solution as well in short future. The problem is called Leetcode 305, word ladder II.
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