Leetcode #2830: Maximize the Profit as the Salesman
In this guide, we solve Leetcode #2830 Maximize the Profit as the Salesman 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
You are given an integer n representing the number of houses on a number line, numbered from 0 to n - 1. Additionally, you are given a 2D integer array offers where offers[i] = [starti, endi, goldi], indicating that ith buyer wants to buy all the houses from starti to endi for goldi amount of gold.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Binary Search, Dynamic Programming, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: n = 5, offers = [[0,0,1],[0,2,2],[1,3,2]]
Output: 3
Explanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.
We sell houses in the range [0,0] to 1st buyer for 1 gold and houses in the range [1,3] to 3rd buyer for 2 golds.
It can be proven that 3 is the maximum amount of gold we can achieve.
Python Solution
class Solution:
def maximizeTheProfit(self, n: int, offers: List[List[int]]) -> int:
offers.sort(key=lambda x: x[1])
f = [0] * (len(offers) + 1)
g = [x[1] for x in offers]
for i, (s, _, v) in enumerate(offers, 1):
j = bisect_left(g, s)
f[i] = max(f[i - 1], f[j] + v)
return f[-1]
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.