Implement numIslands function to count islands in a grid

[cd155]

This pull request implements the solution for counting the number of islands in a grid using depth-first search (DFS) in the numIslands function. The previously unimplemented method now tracks visited cells and recursively explores connected land cells.

Implementation of DFS-based island counting:

• Added a DFS helper function to recursively explore connected land cells, marking them as visited in visit_record.
• Updated the main logic to iterate through the grid, incrementing the island count for each unvisited land cell and invoking DFS to mark all connected cells.
• Replaced the placeholder pass statement with the complete solution and returned the final count of islands.

[Copilot]

Pull request overview

Implements the previously placeholder numIslands method to count connected land components (“islands”) in a 2D grid using DFS traversal.

Changes:

• Added a DFS helper to traverse and mark connected land cells as visited.
• Added grid iteration logic to count islands by starting DFS from each unvisited land cell.
• Returned the computed island count instead of the prior pass placeholder.

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[Copilot]

The docstring's space complexity note says it's for the recursion stack, but the implementation also stores all visited cells in visit_record, which is O(m*n) additional space. Please update the complexity comment to match the actual approach (and if you move to iterative traversal, adjust accordingly).

[Copilot]

row_num/col_num are computed but the loops still use len(grid) / len(grid[0]). Using the already-computed variables avoids repeated length lookups and keeps the code consistent (also helps if you later add an early return for empty input).

Suggested change

	for i in range(0, len(grid)):
	for j in range(0, len(grid[0])):
	for i in range(row_num):
	for j in range(col_num):

[Copilot]

visit_record is a list, so (i, j) not in visit_record is an O(k) linear scan. Since this check is done for every cell (and again within DFS), the worst-case runtime can degrade to ~O((mn)^2) and contradicts the stated O(mn) complexity. Use a set for visited (or a boolean matrix / in-place marking) so membership tests are O(1).

Suggested change

	visit_record = []
	num_island = 0
	def dfs(i, j):
	if (i,j) not in visit_record and grid[i][j] == "1":
	visit_record.append((i, j))
	if 0<=i-1<row_num:
	dfs(i-1, j)
	if 0<=i+1<row_num:
	dfs(i+1, j)
	if 0<=j-1<col_num:
	dfs(i, j-1)
	if 0<=j+1<col_num:
	dfs(i, j+1)
	visit_record = set()
	num_island = 0
	def dfs(i, j):
	if (i, j) not in visit_record and grid[i][j] == "1":
	visit_record.add((i, j))
	if 0 <= i - 1 < row_num:
	dfs(i - 1, j)
	if 0 <= i + 1 < row_num:
	dfs(i + 1, j)
	if 0 <= j - 1 < col_num:
	dfs(i, j - 1)
	if 0 <= j + 1 < col_num:
	dfs(i, j + 1)

[Copilot]

This recursive DFS can hit Python's recursion limit on worst-case inputs (constraints allow up to 300x300; a single large island can require tens of thousands of recursive calls), causing RecursionError and failing on valid inputs. Consider switching to an iterative DFS/BFS using an explicit stack/queue (or otherwise avoid deep recursion).