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Implement numIslands function to count islands in a grid #93

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26 changes: 24 additions & 2 deletions src/graph/number_of_islands.py
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Original file line number Diff line number Diff line change
Expand Up @@ -36,8 +36,30 @@ def numIslands(self, grid):
Time Complexity: O(m * n)
Space Complexity: O(m * n) for recursion stack
"""
# TODO: Implement solution
pass
row_num = len(grid)
col_num = len(grid[0])
visit_record = []
num_island = 0
Comment on lines 36 to +42
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Copilot AI Feb 19, 2026

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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).

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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)
Comment on lines +41 to +54
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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)

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Comment on lines +44 to +54
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Copilot AI Feb 19, 2026

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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).

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for i in range(0, len(grid)):
for j in range(0, len(grid[0])):
Comment on lines +56 to +57
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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):

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if (i,j) not in visit_record and grid[i][j] == "1":
num_island += 1
dfs(i, j)

return num_island


# Example usage (for testing locally)
Expand Down