How do you solve Minimum Time to Complete Trips on LeetCode?

Minimum Time to Complete Trips (LeetCode #2187) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(1) space complexity. Key insight: Binary search helps in optimizing the search for minimum time.

What is the brute force solution for Minimum Time to Complete Trips?

The brute force approach for Minimum Time to Complete Trips is Brute Force, with O(n * m) time complexity and O(1) space complexity. In the brute force approach, we simulate each time unit and count how many trips each bus can complete until we reach the total number of trips required. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Minimum Time to Complete Trips?

Minimum Time to Complete Trips uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Greedy.

What are the interview tips for Minimum Time to Complete Trips?

Clarify the problem statement and constraints with the interviewer. Discuss your thought process while arriving at the solution. Practice similar problems to get comfortable with binary search.

What are common mistakes when solving Minimum Time to Complete Trips?

Not considering edge cases like minimum time or maximum trips. Confusing the counting of trips with the time taken.

#2187

Minimum Time to Complete Trips

Medium

ArrayBinary SearchBinary SearchGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * m)	O(n log m)
Space	O(1)	O(1)

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Intuition

Time O(n log m)Space O(1)

Using binary search, we can efficiently find the minimum time required to complete the totalTrips by narrowing down the possible time range based on the number of trips completed in a given time.

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Algorithm

5 steps

1Step 1: Set low to 1 and high to the maximum time in the array multiplied by totalTrips.
2Step 2: While low is less than high, calculate mid as the average of low and high.
3Step 3: Count the total trips that can be completed in mid time using the formula sum(mid // t for t in time).
4Step 4: If the total trips are greater than or equal to totalTrips, set high to mid; otherwise, set low to mid + 1.
5Step 5: Return low as the minimum time required.

solution.py12 lines

1# Full working Python code
2
3def minTimeToCompleteTrips(time, totalTrips):
4    low, high = 1, min(time) * totalTrips
5    while low < high:
6        mid = (low + high) // 2
7        trips_completed = sum(mid // t for t in time)
8        if trips_completed >= totalTrips:
9            high = mid
10        else:
11            low = mid + 1
12    return low

ℹ

Complexity note: The time complexity is O(n log m) where n is the number of buses and m is the maximum possible time to complete totalTrips. This is efficient because we reduce the search space logarithmically.

1Binary search helps in optimizing the search for minimum time.
2Understanding how to count trips efficiently is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.