Number of Islands

Author: Shikha-code36

Graphs/200-Number-of-Islands.py

@@ -0,0 +1,89 @@
+'''
+Leetcode- https://leetcode.com/problems/number-of-islands/
+'''
+
+'''
+200. 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","0","0","0"],
+  ["1","1","0","0","0"],
+  ["0","0","1","0","0"],
+  ["0","0","0","1","1"]
+]
+Output: 3
+'''
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        if not grid:
+            return 0
+        island = 0
+        noRows = len(grid)
+        noCols = len(grid[0])
+        for row in range(noRows):
+            for col in range(noCols):
+                if grid[row][col] == "1":
+                    self.dfs(grid, row, col)
+                    island+=1
+        return island
+
+    def dfs(self,grid,row,col):
+            if row<0 or row>=len(grid) or col<0 or col>=len(grid[0]) or grid[row][col]=="0":
+                return
+
+            grid[row][col] = "0"
+
+            self.dfs(grid,row-1,col)
+            self.dfs(grid,row+1,col)
+            self.dfs(grid,row,col-1)
+            self.dfs(grid,row,col+1)
+
+'''
+Explanation:
+
+1.Initialize island to 0.
+2.Calculate noRows (number of rows) and noCols (number of columns) of the grid.
+3.Start looping through each cell in the grid using nested loops.
+4.For each cell, if the cell contains "1", call the dfs function to explore the island.
+5.Inside the dfs function, check if the current cell is out of bounds or if it's water ("0"). If either condition is met, return from the function.
+6.Mark the current cell as visited by changing its value to "0".
+7.Recursively call dfs on the neighboring cells (up, down, left, right).
+8.Repeat steps 5-7 until all connected land cells are visited and marked as "0". This essentially explores and marks an entire island.
+9.Go back to the main function, increment the island count by 1.
+R10.epeat steps 4-9 for all cells in the grid.
+
+
+Now, let's dry run the code step by step using the provided input grid:
+
+1.Initialize island to 0.
+2.Calculate noRows = 4 and noCols = 5.
+
+Start looping through each cell in the grid:
+
+- Cell at (0, 0) contains "1":
+
+    - Call dfs at (0, 0).
+        -- Mark (0, 0) as "0".
+        -- Call dfs at (0, -1), which returns immediately.
+        -- Call dfs at (0, 1), which recursively calls dfs on (0, 2) and (1, 1), both return immediately.
+        -- Call dfs at (-1, 0), which returns immediately.
+        -- Call dfs at (1, 0), which recursively calls dfs on (2, 0), which calls dfs on (2, 1), (3, 0), and (1, 0), all return immediately.
+        -- Call dfs at (0, 0) again (through recursion), which returns immediately.
+    - Increment island to 1.
+
+- Cell at (0, 1) contains "1":
+
+    - Call dfs at (0, 1), which returns immediately.
+- Continue this process for all cells in the grid.
+
+Final island count: 3
+'''

README.md

@@ -39,3 +39,6 @@ Check the notes for the explaination - [Notes](https://stingy-shallot-4ea.notion
     - [x] [Find the Duplicate Number](Two-Pointers/287-Find-the-Duplicate-Number.py)
     - [x] [Subarray Product Less Than K](Two-Pointers/713-Subarray-Product-Less-Than-K.py)
 
+- [x] [Graphs](Graphs)
+    - [x] [Number of Islands](Graphs/200-Number-of-Islands.py)
+