Leetcode #2532: Time to Cross a Bridge
In this guide, we solve Leetcode #2532 Time to Cross a Bridge in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There are k workers who want to move n boxes from the right (old) warehouse to the left (new) warehouse. You are given the two integers n and k, and a 2D integer array time of size k x 4 where time[i] = [righti, picki, lefti, puti].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Simulation, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
From 0 to 1 minutes: worker 2 crosses the bridge to the right.
From 1 to 2 minutes: worker 2 picks up a box from the right warehouse.
From 2 to 6 minutes: worker 2 crosses the bridge to the left.
From 6 to 7 minutes: worker 2 puts a box at the left warehouse.
The whole process ends after 7 minutes. We return 6 because the problem asks for the instance of time at which the last worker reaches the left side of the bridge.
Python Solution
class Solution:
def findCrossingTime(self, n: int, k: int, time: List[List[int]]) -> int:
time.sort(key=lambda x: x[0] + x[2])
cur = 0
wait_in_left, wait_in_right = [], []
work_in_left, work_in_right = [], []
for i in range(k):
heappush(wait_in_left, -i)
while 1:
while work_in_left:
t, i = work_in_left[0]
if t > cur:
break
heappop(work_in_left)
heappush(wait_in_left, -i)
while work_in_right:
t, i = work_in_right[0]
if t > cur:
break
heappop(work_in_right)
heappush(wait_in_right, -i)
left_to_go = n > 0 and wait_in_left
right_to_go = bool(wait_in_right)
if not left_to_go and not right_to_go:
nxt = inf
if work_in_left:
nxt = min(nxt, work_in_left[0][0])
if work_in_right:
nxt = min(nxt, work_in_right[0][0])
cur = nxt
continue
if right_to_go:
i = -heappop(wait_in_right)
cur += time[i][2]
if n == 0 and not wait_in_right and not work_in_right:
return cur
heappush(work_in_left, (cur + time[i][3], i))
else:
i = -heappop(wait_in_left)
cur += time[i][0]
n -= 1
heappush(work_in_right, (cur + time[i][1], i))
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.