Number of Islands LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
Share:

Number of Islands 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'.

Number of Islands Solutions

Time: O(mn)
Space: O(min(m,n))

C++

class Solution {
 public:
  int numIslands(vector<vector<char>>& grid) {
    const int m = grid.size();
    const int n = grid[0].size();
    const vector<int> dirs{0, 1, 0, -1, 0};

    auto bfs = [&](int r, int c) {
      queue<pair<int, int>> q{{{r, c}}};
      grid[r][c] = '2';  // mark '2' as visited
      while (!q.empty()) {
        const auto [i, j] = q.front();
        q.pop();
        for (int k = 0; k < 4; ++k) {
          const int x = i + dirs[k];
          const int y = j + dirs[k + 1];
          if (x < 0 || x == m || y < 0 || y == n)
            continue;
          if (grid[x][y] != '1')
            continue;
          q.emplace(x, y);
          grid[x][y] = '2';  // mark '2' as visited
        }
      }
    };

    int ans = 0;

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == '1') {
          bfs(i, j);
          ++ans;
        }

    return ans;
  }
};

Java

 class Solution {
  public int numIslands(char[][] grid) {
    int ans = 0;

    for (int i = 0; i < grid.length; ++i)
      for (int j = 0; j < grid[0].length; ++j)
        if (grid[i][j] == '1') {
          bfs(grid, i, j);
          ++ans;
        }

    return ans;
  }

  private static final int[] dirs = {0, 1, 0, -1, 0};

  private void bfs(char[][] grid, int r, int c) {
    Queue<int[]> q = new ArrayDeque<>();
    q.offer(new int[] {r, c});
    grid[r][c] = '2'; // mark '2' as visited
    while (!q.isEmpty()) {
      final int i = q.peek()[0];
      final int j = q.poll()[1];
      for (int k = 0; k < 4; ++k) {
        final int x = i + dirs[k];
        final int y = j + dirs[k + 1];
        if (x < 0 || x == grid.length || y < 0 || y == grid[0].length)
          continue;
        if (grid[x][y] != '1')
          continue;
        q.offer(new int[] {x, y});
        grid[x][y] = '2'; // mark '2' as visited
      }
    }
  }
}

Python


class Solution:
  def numIslands(self, grid: List[List[str]]) -> int:
    m = len(grid)
    n = len(grid[0])
    dirs = [0, 1, 0, -1, 0]

    def bfs(r, c):
      q = deque([(r, c)])
      grid[r][c] = '2'  # mark '2' as visited
      while q:
        i, j = q.popleft()
        for k in range(4):
          x = i + dirs[k]
          y = j + dirs[k + 1]
          if x < 0 or x == m or y < 0 or y == n:
            continue
          if grid[x][y] != '1':
            continue
          q.append((x, y))
          grid[x][y] = '2'  # mark '2' as visited

    ans = 0

    for i in range(m):
      for j in range(n):
        if grid[i][j] == '1':
          bfs(i, j)
          ans += 1

    return ans

Watch Tutorial

Checkout more Solutions here

Leave a Comment

Your email address will not be published. Required fields are marked *

x