0200 - Number of Islands (Medium)

Problem Link

https://leetcode.com/problems/number-of-islands/

Problem Statement

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

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 1:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

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

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is '0' or '1'.

Approach 1: Flood Fill

We can use 0733 - Flood Fill (Easy) solution in this problem. The idea is to search for $1$ and paint the entire island with different character that does not exist in the grid (says $2$). Every time we start flood fill, we increase our answer by $1$.

Time Complexity: $O(m*n)$ where m is the number of rows, and n is the number of columns in the grid. We must traverse the whole grid in order to determine the number of islands.

Space Complexity: $(m*n)$ where m is the number of rows, and n is the number of columns in the grid. In the worst case, our queue/stack/recursive stack/visited set will grow in direct proportion to the size of the grid.

C++

Written by @wingkwong

class Solution {
public:
    // Check out 0733 - Flood Fill (Easy)
    // we need to find tune the type (int -> char) from that solution
    int dirx[4] = { -1, 0, 0, 1};
    int diry[4] = { 0, 1, -1, 0};
    // the idea is to use bfs to paint the cell with value '1' starting from (sr, sc)
    void floodFill(vector<vector<char>>& image, int sr, int sc, char newColor) {
        // record the original color
        int oriColor = image[sr][sc];
        // if it is same as the one we want to paint, then skip
        if (oriColor == newColor) return;
        int n = image.size(), m = image[0].size();
        // standard bfs
        queue<pair<int, int>> q;
        q.push({sr, sc});
        while(!q.empty()) {
            auto [x, y] = q.front(); q.pop();
            // paint the new color here so that we won't visit it again
            image[x][y] = newColor;
            // after painting the cell at (x, y), we try four different directions
            for(int i = 0; i < 4; i++) {
                int next_x = x + dirx[i];
                int next_y = y + diry[i];
                // we need to make sure that the next cell is valid and the color isn't same as the orginal color
                if(next_x < 0 || next_y < 0 || next_x > n - 1 || next_y > m - 1 || image[next_x][next_y] != oriColor) continue;
                // paint it with the new color
                image[next_x][next_y] = newColor;
                // push the next color to the queue
                q.push({next_x, next_y});
            }
        }
        return;
    }
    
    int numIslands(vector<vector<char>>& grid) {
        int ans = 0;
        // iterate each row
        for(int row = 0; row < grid.size(); row++) {
            // iterate each column
            for(int col = 0; col < grid[0].size(); col++) {
                // if it is land
                if(grid[row][col] == '1'){
                    // perform flood fill and make each cell to 2 or any number except 0 and 1
                    // so that we won't visit it again
                    floodFill(grid, row, col, '2');
                    // after performing flood fill, we color one island 
                    ans++;
                }
            }
       }
       return ans; 
    }
};

Java

Written by @vigneshshiv

// r - # of rows
// c - # of cols
// Time complexity: O(rc)
// Space complexity: O(rc)
class Solution {
    
    public final int NO_DIRS = 4;
    public int[] DIRS = {0, 1, 0, -1, 0};
    
    public int numIslands(char[][] grid) {
        int count = 0;
        // Visited set for string row and col, like "row # col"
        // This helps to avoid to revisiting the same cell again
        Set<String> visited = new HashSet<>();
        //
        for (int r = 0; r < grid.length; r++) {
            for (int c = 0; c < grid[0].length; c++) {
                // Explore area only if it's a land, i.e '1'
                if (grid[r][c] == '1' && exploreGrid(grid, r, c, visited)) {
                    count += 1;
                }
            }
        }
        return count;
    }
    
    public boolean exploreGrid(char[][] grid, int row, int col, Set<String> visited) {
        boolean rowInbounds = 0 <= row && row < grid.length;
        boolean colInbounds = 0 <= col && col < grid[0].length;
        // If row or col out of range, then simply return
        if (!rowInbounds || !colInbounds) {
            return false;
        }
        // Current cell is water, so no need to explore surrounding cells
        if (grid[row][col] == '0') {
            return false;
        }
        String pos = row + "#" + col;
        // Check for already visited cells, if found, then no need to traverse
        if (visited.contains(pos)) {
            return false;
        }
        // Mark as visited
        visited.add(pos);
        for (int dir = 0; dir < NO_DIRS; dir++) {
            // Explore near by cells, by traversing all directions, move UP, DOWN, LEFT, RIGHT
            // Directions can be formed by above mentioned array
            exploreGrid(grid, row + DIRS[dir], col + DIRS[dir + 1], visited);
        }
        return true;
    }
}

Python

Written by @ColeB2

class Solution:
    # Time: O(m*n) as we must traverse each position in the grid.
    # Space: O(m*n) our queue in worst case maybe all land and
    # will grow in proportion to the size of the grid.
    def bfs(self, r: int, c: int, grid: List[List[str]]) -> List[List[str]]:
        # BFS implementation of flood fill, initialize queue with
        # the row and column we passed as first starting point
        q = deque([(r,c)])
        while q:
            # pop off our row and column
            row, col = q.popleft()
            # check we are in bounds, and land.
            if (row < 0 or col < 0 
                or row >= len(grid) or col >= len(grid[0])
                or grid[row][col] != '1'
                ):
                # out of bounds/not land  -> continue
                continue
            # 'colour' our land a different colour. We will set it to -1
            grid[row][col] = -1
            # loop our directions, and add them to the queue
            for (x,y) in ((1,0),(-1,0),(0,1),(0,-1)):
                    q.append((row + x, col + y))
        # return our grid since we changed it.
        return grid


    def numIslands(self, grid: List[List[str]]) -> int:
        # initialize our row, cols and number of islands counter.
        ROWS, COLS = len(grid), len(grid[0])
        num_islands = 0
        # loop through the grid.
        for r in range(ROWS):
            for c in range(COLS):
                # if we find land
                if grid[r][c] == '1':
                    # flood fill it, making sure to overwrite our 
                    # grid as we will change it during our algorithm.
                    grid = self.bfs(r, c, grid)
                    num_islands += 1
        
        return num_islands