How do you solve Maximum Earnings From Taxi on LeetCode?

Maximum Earnings From Taxi (LeetCode #2008) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the rides by their end points allows for efficient selection of non-overlapping rides.

What is the time complexity of Maximum Earnings From Taxi?

The optimal solution for Maximum Earnings From Taxi runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n), and filling the DP array takes O(n), leading to an overall complexity of O(n log n).

What is the brute force solution for Maximum Earnings From Taxi?

The brute force approach for Maximum Earnings From Taxi is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible combination of rides to see which yields the maximum earnings. This is straightforward but inefficient, as it examines all potential rides without considering overlaps. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Maximum Earnings From Taxi?

Always think about how you can break down the problem into smaller subproblems. Discuss your thought process and potential edge cases with the interviewer. Practice writing code that is clean and easy to read, as clarity is important in interviews.

What are common mistakes when solving Maximum Earnings From Taxi?

Not considering overlapping rides properly when calculating profits. Failing to sort the rides before processing can lead to incorrect results.

#2008

Maximum Earnings From Taxi

Medium

ArrayHash TableBinary SearchDynamic ProgrammingSortingDynamic ProgrammingSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses dynamic programming to keep track of the maximum earnings at each point. By sorting the rides based on their end points, we can efficiently determine the best rides to take without overlap.

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Algorithm

5 steps

1Step 1: Sort the rides based on their end points.
2Step 2: Initialize a DP array where dp[i] represents the maximum earnings up to the i-th ride.
3Step 3: For each ride, calculate the profit and find the last non-overlapping ride using binary search.
4Step 4: Update the DP array with the maximum of taking the current ride or not.
5Step 5: Return the maximum value in the DP array.

solution.py11 lines

1def maxEarnings(n, rides):
2    rides.sort(key=lambda x: x[1])
3    dp = [0] * (len(rides) + 1)
4    for i in range(1, len(rides) + 1):
5        start_i, end_i, tip_i = rides[i - 1]
6        profit = (end_i - start_i + 1) + tip_i
7        j = 0
8        while j < i - 1 and rides[j][1] < start_i:
9            j += 1
10        dp[i] = max(dp[i - 1], profit + dp[j])
11    return dp[len(rides)]

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Complexity note: The sorting step takes O(n log n), and filling the DP array takes O(n), leading to an overall complexity of O(n log n).

1Sorting the rides by their end points allows for efficient selection of non-overlapping rides.
2Dynamic programming helps in building up the maximum earnings incrementally.

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