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feat: implement BFS solution for Jump Game III (#1306) #2168

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hackdartstorm merged 1 commit into from
Feb 19, 2026
Merged

feat: implement BFS solution for Jump Game III (#1306) #2168

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36 changes: 36 additions & 0 deletions exercises/1000_programs/medium/1306_jump_game_iii.py
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def canReach(arr: list[int], start: int) -> bool:
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Copilot AI Feb 19, 2026

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The PR description says "Closes #1306", but the referenced issue #1306 is about screenshot automation (exercises/automation/098_screenshot_auto.py), while this change adds a Jump Game III solution. Please update the PR description/linked issue so the PR closes the correct issue (or remove the close directive).

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Function name canReach doesn't follow the repo naming convention for functions/variables (lowercase_with_underscores). Consider renaming to can_reach (and updating the call sites in this file).

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# 1. Initialize the queue with our starting position
queue = [start]
# 2. Keep track of visited indices so we don't get stuck in a loop
visited = set()
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This function is missing a docstring. CONTRIBUTING.md specifies adding docstrings for functions/classes; please add a short docstring describing parameters, return value, and the BFS approach.

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while queue:
# Get the current position
curr = queue.pop(0)
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queue.pop(0) makes the BFS dequeue operation O(n) per pop, which can degrade to O(n^2). Use collections.deque and popleft() (and ideally initialize the queue as a deque) to keep the BFS truly O(n).

Suggested change
def canReach(arr: list[int], start: int) -> bool:
# 1. Initialize the queue with our starting position
queue = [start]
# 2. Keep track of visited indices so we don't get stuck in a loop
visited = set()
while queue:
# Get the current position
curr = queue.pop(0)
from collections import deque
def canReach(arr: list[int], start: int) -> bool:
# 1. Initialize the queue with our starting position
queue = deque([start])
# 2. Keep track of visited indices so we don't get stuck in a loop
visited = set()
while queue:
# Get the current position
curr = queue.popleft()

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# SUCCESS: We found a zero!
if arr[curr] == 0:
return True
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There is no input validation for arr being empty or start being out of bounds; arr[curr] will raise IndexError in those cases. Add an early guard (e.g., return False or raise a clear ValueError) before accessing arr[start].

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# Mark current as visited
visited.add(curr)

# Calculate possible jumps
for next_pos in [curr + arr[curr], curr - arr[curr]]:
# Check if the jump is within the array bounds and not visited
if 0 <= next_pos < len(arr) and next_pos not in visited:
queue.append(next_pos)
Comment on lines +7 to +22
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Visited bookkeeping happens only after dequeuing, which can allow the same index to be enqueued multiple times before it's processed. Mark nodes visited when enqueuing (or skip processing if curr is already visited right after dequeue) to avoid duplicate work.

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# If the queue is empty and we never hit 0, it's impossible
return False

# --- Test Cases ---
if __name__ == "__main__":
# Test 1: Should be True (can reach index 5 or 6)
print(f"Test 1: {canReach([4,2,3,0,3,1,2], 5)}")

# Test 2: Should be True (start is already 0)
print(f"Test 2: {canReach([4,2,3,0,3,1,2], 0)}")

# Test 3: Should be False (cannot reach 0)
print(f"Test 3: {canReach([3,0,2,1,2], 2)}")
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