feat : number-of-islands

Author: ekgns33

number-of-islands/ekgns33.java

@@ -0,0 +1,74 @@
+/**
+ input : 2d matrix filled with '1', '0'
+ output : number of islands
+
+ 1 indicates land, 0 means water.
+ if lands are surrounded by water we call it island
+
+ example
+
+ 1 1 1.  >> 3 islands
+ 0 0 0
+ 1 0 1
+
+ 1 1 1 1 0
+ 1 1 0 1 0 >> 1 island
+ 0 0 0 0 0
+
+ constraints:
+ 1) m, n range
+ [1, 300]
+ 2) can we change the input grid cell?
+ there is no restriction.
+
+ solution 1)
+
+ ds : array
+ algo : bfs + in-place
+
+ read every cell in grid.
+ if cell is '1'
+ do bfs, change every visiting cell into '2' << if changing input is allowed.
+ increase count
+
+ tc : O(mn), visiting each cell exactly once.
+ sc : O(mn) for queue
+
+ optimize?? maintain visited matrix, set or something else..
+ if we are allowed to modify given input matrix
+ we don't have to maintain visited matrix, but simply change values
+ */
+class Solution {
+  public int numIslands(char[][] grid) {
+    int m = grid.length, n = grid[0].length;
+    int numberOfIslands = 0;
+    for(int i = 0; i < m; i++) {
+      for(int j = 0; j < n; j++) {
+        if(grid[i][j] == '1') {
+          bfsHelper(grid, i, j, m, n);
+          numberOfIslands++;
+        }
+      }
+    }
+    return numberOfIslands;
+  }
+  private static int[][] directions = {
+      {1, 0}, {-1, 0}, {0, 1}, {0, -1}
+  };
+
+  private void bfsHelper(char[][] grid, int x, int y, int m, int n) {
+    Queue<int[]> queue = new LinkedList<>();
+    queue.add(new int[]{x,y});
+    grid[x][y] = '2';
+    while(!queue.isEmpty()) {
+      int[] cur = queue.poll();
+      for(int[] direction : directions) {
+        int nx = direction[0] + cur[0];
+        int ny = direction[1] + cur[1];
+        if(nx < 0 || nx >= m || ny < 0 || ny >= n || grid[nx][ny] != '1') continue;
+        grid[nx][ny] = '2';
+        queue.add(new int[]{nx, ny});
+      }
+    }
+  }
+}