Last Updated: 14 Sep, 2022
COUNT ISLANDS
Moderate
You are given an empty 2D binary 'GRID' of size ‘N’ x 'M' The 'GRID' represents a map where 0's represent water and 1's represent land. Initially, all the cells of the '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' of size ‘K’ 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.
Example:
Input: ‘M’ = 3, 'N' = 3, 'POSITIONS' = [ [ 0,0 ], [ 0,1 ], [ 1,2 ], [ 2,1 ] ]
Output: [1, 1, 2, 3]
Initially, the 2d 'GRID' is filled with water.
- Operation #1: addLand(0, 0) turns the water at ‘GRID[0][0]’ into a land. We have 1 island.
- Operation #2: addLand(0, 1) turns the water at ‘GRID[0][1]’ into a land. We still have 1 island.
- Operation #3: addLand(1, 2) turns the water at 'GRID[1][2]’ into a land. We have 2 islands.
- Operation #4: addLand(2, 1) turns the water at 'GRID[2][1]’ into a land. We have 3 islands.
Input Format :
The first line contains ‘T’, denoting the number of test cases.
The first line of each test case contains two integers, ‘N’ and ‘M’, denoting the number of rows and columns in the 'GRID'.
The second line of each test case contains an integer ‘K’ denoting the size of ‘POSITION’ array.
Each of the next ‘K’ lines contains two space-separated integers denoting the (‘Ri’, ‘Ci’) of ‘POSITION[i]’.
Output format :
For each test case, return the array of integers as described in the problem statement.
Note :
You don't need to print anything. It has already been taken care of. Just implement the given function.
Constraints :
1 <= T <= 10
1 <= M, N, K <= 100
0 <= Ri< m
0 <= Ci < n
Time Limit: 1 sec
Approaches
Linear scan the 2d grid map, if a node contains a '1', then it is a root node that triggers a Depth First Search. During DFS, every visited node should be set as '0' to mark it as a visited node. Count the number of root nodes that trigger DFS, this number would be the number of islands since each DFS starting at some root identifies an island.
The steps are as follows:
- In the function ‘numIslands’, for each cell in ‘POSITIONS’ array, starting from the beginning, we set the value of the corresponding cell as ‘1’ and call the ‘numIslandsUtil’ function passing the current configuration of ‘GRID’.
- We initially have an empty array ‘numIslandsArr’ in which we append the values returned by ‘numIslandsUtil’ function and return it at the last.
- In the function numIslandsUtil, we scan the 2-D ‘GRID’ from top to bottom if we find a cell with value ‘1’ then we can call the dfs function and pass the cell as one of it’s parameters.
- In dfs function, we set the value of current cell as ‘0’ then, we iterate through it’s 4-directional neighbours.
- During iteration, if any of the neighbour cell is found as ‘1’ then, we call the dfs function on it.
- We have a counter variable in numIslandsUtil function, which we increment every time we call dfs function and return it at the last.
Treat the 2d 'GRID' map as an undirected graph (formatted as adjacency matrix) and there is an edge between two horizontally or vertically adjacent nodes of value 1, then the problem reduces to finding the number of connected components in the graph after each ‘addLand’ operation.
The steps are as follows:
- Make use of a Union Find data structure of size M*N to store all the nodes in the graph and initially each node's parent value is set to -1 to represent an empty graph. Our goal is to update Union Find with lands added by ‘addLand’ operation and union lands belongs to the same island.
- For each ‘addLand’ operation at position (row, col), union it with its adjacent neighbors if they belongs to some islands, if none of its neighbors belong to any islands, then initialize the new position as a new island (set parent value to itself) within Union Find.