What is the time complexity of Maximum Earnings From Taxi?

The optimal solution has a time complexity of O(n + m).

What is the space complexity of Maximum Earnings From Taxi?

The space complexity is O(n + m).

What data structures are used to solve Maximum Earnings From Taxi?

This problem uses Array, Hash Table, Binary Search, Dynamic Programming, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Earnings From Taxi?

Insight 1: Dynamic programming is suitable because the problem has optimal substructure; the maximum earnings up to a point 'i' can be derived from the maximum earnings at previous points. Insight 2: Sorting rides directly is not necessary, but grouping rides ending at the same point simplifies the dynamic programming step.

Which companies ask LeetCode 2008. Maximum Earnings From Taxi in interviews?

This problem is frequently asked at Myntra.

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Maximum Earnings From Taxi - LeetCode 2008 Solution

Maximum Earnings From Taxi - Complete Solution Guide

Maximum Earnings From Taxi is LeetCode problem 2008, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n points on a road you are driving your taxi on. The n points on the road are labeled from 1 to n in the direction you are going, and you want to drive from point 1 to point n to make money by picking up passengers. You cannot change the direction of the taxi. The passengers are represented by a 0-indexed 2D integer array rides , where rides[i] = [start i , end i , tip i ] denotes the i th passenger requesting a ride from point start i to point end i who is willing to give a tip i doll

Detailed Explanation

The problem asks us to find the maximum earnings a taxi driver can make by picking up passengers on a road with 'n' points, numbered from 1 to n. Each passenger ride is defined by a start point, an end point, and a tip. The driver can only carry one passenger at a time. The earnings for a ride are calculated as (end - start + tip). The objective is to maximize the total earnings by optimally selecting which passengers to pick up.

Solution Approach

The solution utilizes dynamic programming to determine the maximum earnings at each point 'i' on the road. A DP array 'dp' of size (n+1) is used, where dp[i] represents the maximum earnings possible up to point 'i'. The algorithm iterates from 1 to n. At each point 'i', it considers two possibilities: either not picking up any new ride and simply carrying over the maximum earnings from the previous point (dp[i-1]), or picking up a ride that ends at point 'i'. If a ride ends at point 'i', we calculate the potential earnings from that ride (end - start + tip) and add it to the maximum earnings achievable at the ride's starting point (dp[start]). We then update dp[i] to be the maximum of these two possibilities.

Step-by-Step Algorithm

Step 1: Initialize a dynamic programming array `dp` of size `n+1` with all values set to 0. `dp[i]` represents the maximum earnings up to point `i`.
Step 2: Create a list of lists `rides_by_end` to group rides by their end points. This allows us to efficiently access all rides ending at a particular point.
Step 3: Iterate through the `rides` array and add each ride to the appropriate list in `rides_by_end` based on its end point.
Step 4: Iterate from `i = 1` to `n`. For each point `i`, calculate the maximum earnings:
Step 5: Option 1: `dp[i] = dp[i-1]` (carry over the maximum earnings from the previous point).
Step 6: Option 2: Iterate through all rides ending at point `i`. For each such ride, calculate the profit (`i - start + tip`). Update `dp[i]` to be the maximum of its current value and `dp[start] + profit`.
Step 7: After the loop finishes, `dp[n]` will contain the maximum earnings achievable up to point `n`, which is the final result.

Key Insights

Insight 1: Dynamic programming is suitable because the problem has optimal substructure; the maximum earnings up to a point 'i' can be derived from the maximum earnings at previous points.
Insight 2: Sorting rides directly is not necessary, but grouping rides ending at the same point simplifies the dynamic programming step.
Insight 3: The core idea is to consider each point 'i' on the road and decide whether to extend the earnings from the previous point or to pick up a ride that ends at point 'i'. Choosing the maximum of these two options leads to the optimal solution.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n + m)

Topics

This problem involves: Array, Hash Table, Binary Search, Dynamic Programming, Sorting.

Companies

Asked at: Myntra.

Maximum Earnings From Taxi | LeetCode 2008

Medium

Array Hash Table Binary Search Dynamic Programming Sorting

Time: O(n + m)Space: O(n + m)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public long maxTaxiEarnings(int n, int[][] rides) {
        long[] dp = new long[n + 1];

        List<List<int[]>> rides_by_end = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            rides_by_end.add(new ArrayList<>());
        }

        for (int[] ride : rides) {
            int start = ride[0];
            int end = ride[1];
            int tip = ride[2];
            rides_by_end.get(end).add(new int[]{start, tip});
        }

        for (int i = 1; i <= n; i++) {
            // Option 1: Drive from i-1 to i without a passenger.
            dp[i] = dp[i - 1];

            // Option 2: A ride ends at point i.
            for (int[] ride : rides_by_end.get(i)) {
                int start = ride[0];
                int tip = ride[1];
                long profit = (long) (i - start + tip);
                dp[i] = Math.max(dp[i], dp[start] + profit);
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 2008: Maximum Earnings From Taxi - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Dynamic Programming Sorting

Companies

Myntra

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2008

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a dynamic programming array `dp` of size `n+1` with all values set to 0. `dp[i]` represents the maximum earnings up to point `i`.
2
Step 2: Create a list of lists `rides_by_end` to group rides by their end points. This allows us to efficiently access all rides ending at a particular point.
3
Step 3: Iterate through the `rides` array and add each ride to the appropriate list in `rides_by_end` based on its end point.
4
Step 4: Iterate from `i = 1` to `n`. For each point `i`, calculate the maximum earnings:
5
Step 5: Option 1: `dp[i] = dp[i-1]` (carry over the maximum earnings from the previous point).
6
Step 6: Option 2: Iterate through all rides ending at point `i`. For each such ride, calculate the profit (`i - start + tip`). Update `dp[i]` to be the maximum of its current value and `dp[start] + profit`.
7
Step 7: After the loop finishes, `dp[n]` will contain the maximum earnings achievable up to point `n`, which is the final result.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n + m)

💾

Space Complexity

O(n + m)

📝

Why?

The time complexity is derived from the iteration through the road points (O(n)) and the iteration through rides (O(m)). The space complexity is determined by the DP array (O(n)) and the storage of rides organized by end points (O(m)).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting Rides by Start Time

Sort the rides by their start times and use a similar dynamic programming approach. For each ride, we have two options: either take the ride or skip it. If we take the ride, we add its profit to the maximum earnings achievable up to its start time. If we skip it, we simply carry over the maximum earnings from the previous ride.

Time: O(m log m + m), where 'm' is the number of rides. O(m log m) for sorting and O(m) for the DP iteration.Space: O(m), primarily for the DP array.

Trade-offs: This approach is slightly less efficient than the main solution due to the sorting step (O(m log m)). It can be useful if sorting is needed for other reasons or if the number of rides is significantly smaller than the number of points on the road.

Binary Search

Sort the rides by their end times. For each ride i, find the latest non-overlapping ride j using binary search. DP[i] = max(DP[i-1], profit[i] + DP[j]).

Time: O(m log m) for sorting + O(m log m) for binary search in the DP loop. Overall, O(m log m).Space: O(m) for the DP array.

Trade-offs: Another approach that is also based on sorting and then finding the non-overlapping ride using binary search. The overhead of sorting and binary search usually makes it slightly slower than the bucket approach. Good if rides are sparse, and n is very large so bucket is not efficient.

✨

Key Insights

1Insight 1: Dynamic programming is suitable because the problem has optimal substructure; the maximum earnings up to a point 'i' can be derived from the maximum earnings at previous points.
2Insight 2: Sorting rides directly is not necessary, but grouping rides ending at the same point simplifies the dynamic programming step.
3Insight 3: The core idea is to consider each point 'i' on the road and decide whether to extend the earnings from the previous point or to pick up a ride that ends at point 'i'. Choosing the maximum of these two options leads to the optimal solution.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: No rides available. In this case, the DP array will remain initialized with 0s, and the function will correctly return 0.
Edge case 2: A ride starts and ends at the same point (start == end). The problem statement says start < end, so this isn't possible. But if hypothetically start=end, the profit will be tip, and the algorithm still correctly handles this because it correctly calculates profit = end - start + tip.
Edge case 3: All rides end at the same point. The algorithm still processes all rides and picks the optimal combination to maximize profit at that point.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using dynamic programming efficiently. A naive recursive approach will lead to exponential time complexity due to overlapping subproblems. Dynamic programming avoids recalculating these subproblems.

✗Common mistake 2: Incorrectly calculating the profit for a ride. It's important to include the tip in the profit calculation.

✗Common mistake 3: Not considering the option of skipping a ride. The algorithm must choose between taking a ride and not taking it at each point to ensure optimality.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing scheduling for delivery services. Given a set of delivery requests with different locations and rewards, the algorithm can be used to maximize the total earnings by optimally selecting which deliveries to fulfill.

🏦Real-world scenario 2: Resource allocation in a project. The algorithm can be applied to select the most profitable tasks to complete within a limited timeframe, given that each task requires certain resources and yields a specific profit.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Earnings From Taxi"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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