200. Number of Islands BFS

Author: mahesh-3pillar

README.md

@@ -41,7 +41,7 @@ Approach #1: DFS
 
 //modifying grid, pass as a pointer
 //when try to change or get (len or value) dereference `(*grid)[i][j]==1`
-//Pass to recursive function, pass as it is
+//Pass grid to recursive function, pass as it is
 //byte use '' instead of ""
 
 Approach #2: BFS
@@ -50,7 +50,7 @@ Approach #2: BFS
 3. add i and j pair into queue
 4. Mark visited by setting `grid[i][j]=0`
 5. while queue is empty
-   1. Pop from queue
+   1. dqueue pair i, j from queue
    2. change i and j to travel in all four direction 
    3. check if they are in boundary of grid and value is 1
       1. add into the queue

leetcode/200.number-of-islands/200.number-of-islands-bfs.go

@@ -0,0 +1,35 @@
+//BFS
+
+func numIslands(grid [][]byte) int {
+	if len(grid) == 0 || len(grid[0]) == 0 {
+		return 0
+	}
+
+	landCount := 0
+	queue := [][2]int{}
+	for i := 0; i < len(grid); i++ {
+		for j := 0; j < len(grid[i]); j++ {
+			if grid[i][j] == '1' {
+				landCount = landCount + 1
+				grid[i][j] = '0'
+				queue = append(queue, [2]int{i, j})
+				offsets := [4][2]int{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}
+				for len(queue) > 0 {
+					currPair := queue[0]
+					queue = queue[1:]
+					row := len(grid)
+					col := len(grid[0])
+					for _, offset := range offsets {
+						m := currPair[0] + offset[0]
+						n := currPair[1] + offset[1]
+						if m >= 0 && m < row && n >= 0 && n < col && grid[m][n] == '1' {
+							grid[m][n] = '0'
+							queue = append(queue, [2]int{m, n})
+						}
+					}
+				}
+			}
+		}
+	}
+	return landCount
+}