Number of Islands

Problem Highlights

🔗 Leetcode Link: https://leetcode.com/problems/number-of-islands

⏰ Time to complete: 15 mins

1. U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

- What are the constraints?
    - `m` == grid.length (number of rows)
    - `n` == grid[i].length (number of columns)
    - 1 <= `m`, `n` <= 300
    - `grid[i][j]` is '0' or '1'
- Can we simply iterate through the grid and count each time that there isn’t a 1 to the left or above another 1? 
That won’t actually work though because there can be islands with isolated cells that have 0s above and to the left, but are still part of the island.
- What is the definition of an island?
    - One or more pieces of land ('1') connected horizontally or vertically

```markdown
HAPPY CASE
Input: [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]
Output: 1

Input: [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]]
Output: 3

EDGE CASE
Input: [["0","0","1","0","0"],["0","0","0","0","0"],["0","0","0","0","0"],["0","0","1","0","0"]]
Output: 2
```

2. M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For this graph problem, some things we want to consider are:

- Depth-First Search
    - The idea is to consider the given matrix as a graph, where each cell is a node of the given graph. Two nodes contain an edge if and only if there is a ‘1’ either horizontally or vertically.
- Find-Union
    - Each set is a single island
    - Two pieces of land should be merged if they are adjacent

3. P-lan

Plan the solution with appropriate visualizations and pseudocode.

1. Create the initial parent array
    1. Number of entries: nrows * ncols
    2. Set parent[i] = i for all elements
2. For each element of the grid
    1. If this element is '1'
        1. If the left neighbor is '1', call union
        2. If the upper neighbor is '1', call union
3. Return the number of islands (sets)

4. I-mplement

Implement the code to solve the algorithm.

# Java Solution
    public int numIslands(char[][] grid) {
            if(grid == null || grid.length < 1 || grid[0].length < 1){
                return 0;
            }
            int row = grid.length;
            int col = grid[0].length;
            int[] root = new int[row * col];
            Arrays.fill(root, -1);
            int n = 0;
            for(int i = 0; i < row; i++){
                for(int j = 0; j < col; j++){
                    root[i*col + j] = i*col + j;
                    if(grid[i][j] == '1'){
                        n++;
                    }
                }
            }
            for(int i = 0; i < row; i++){
                for(int j = 0; j < col; j++){
                    if(grid[i][j] == '1'){
                        int p = i * col + j;
                        int q;
                        if((i+1) < row && grid[i+1][j] == '1'){
                            q = p + col;
                            n = union(root, p, q, n);
                        }
                        if((j+1) < col && grid[i][j+1] == '1'){
                            q = p + 1;
                            n = union(root, p, q, n);
                        }
                    }
                }
            }
            return n;
        }
        private int union(int[] root, int p, int q, int count){
            int proot = find(root, p);
            int qroot = find(root, q);
            if(proot!= qroot){
                root[proot] = qroot;
                count--;
            }
            return count;
        }
        private int find(int[] root, int p){
            while(p != root[p]){
                root[p] = root[root[p]];
                p = root[p];
            }
            return p;
        }

# Python Solution
    class Solution(object):
        def numIslands(self, grid):
            """
            :type grid: List[List[str]]
            :rtype: int
            """
            if len(grid) == 0: return 0
            row = len(grid); col = len(grid[0])
            self.count = sum(grid[i][j]=='1' for i in range(row) for j in range(col))
            parent = [i for i in range(row*col)]
            
            def find(x):
                if parent[x]!= x:
                    return find(parent[x])
                return parent[x]
            
            def union(x,y):
                xroot, yroot = find(x),find(y)
                if xroot == yroot: return 
                parent[xroot] = yroot
                self.count -= 1
            
            
            
            for i in range(row):
                for j in range(col):
                    if grid[i][j] == '0':
                        continue
                    index = i*col + j
                    if j < col-1 and grid[i][j+1] == '1':
                        union(index, index+1)
                    if i < row-1 and grid[i+1][j] == '1':
                        union(index, index+col)
            return self.count

5. R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errorS and verify the code works for the happy and edge cases you created in the “Understand” section

6. E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Calculate complexity in terms of:

- the number of grid elements, `V`
- the size of the maximum island, `I`

Time complexity:

- `O(V)` to build array
- `O(V)` iterations
    - `O(log I)` for union operation
- Total: **O(V log I)**
- Can ensure `O(V log V)` by tuning union function

Space complexity: O(V)