Leetcode

Number of Islands

May 26, 2022 by aAshish

Approach 1: BFS

  • 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};
    int ans = 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
        }
      }
    };

    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

Approach 2: DFS

  • Time:O(mn)
  • Space:O(mn)

C++

 
class Solution {
 public:
  int numIslands(vector<vector<char>>& grid) {
    int ans = 0;

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

    return ans;
  }

 private:
  void dfs(vector<vector<char>>& grid, int i, int j) {
    if (i < 0 || i == grid.size() || j < 0 || j == grid[0].size())
      return;
    if (grid[i][j] != '1')
      return;

    grid[i][j] = '2';  // mark '2' as visited
    dfs(grid, i + 1, j);
    dfs(grid, i - 1, j);
    dfs(grid, i, j + 1);
    dfs(grid, i, j - 1);
  }
};

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') {
          dfs(grid, i, j);
          ++ans;
        }

    return ans;
  }

  private void dfs(char[][] grid, int i, int j) {
    if (i < 0 || i == grid.length || j < 0 || j == grid[0].length)
      return;
    if (grid[i][j] != '1')
      return;

    grid[i][j] = '2'; // mark '2' as visited
    dfs(grid, i + 1, j);
    dfs(grid, i - 1, j);
    dfs(grid, i, j + 1);
    dfs(grid, i, j - 1);
  }
}

Python

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

    def dfs(i: int, j: int) -> None:
      if i < 0 or i == m or j < 0 or j == n:
        return
      if grid[i][j] != '1':
        return

      grid[i][j] = '2'  # mark '2' as visited
      dfs(i + 1, j)
      dfs(i - 1, j)
      dfs(i, j + 1)
      dfs(i, j - 1)

    ans = 0

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

    return ans