How do you solve Maximize the Profit as the Salesman on LeetCode?

Maximize the Profit as the Salesman (LeetCode #2830) 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: Understanding the overlap of offers is crucial to maximizing profit.

What is the time complexity of Maximize the Profit as the Salesman?

The optimal solution for Maximize the Profit as the Salesman runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n) time, while filling the DP array takes O(n) time, leading to an overall complexity of O(n log n).

What is the brute force solution for Maximize the Profit as the Salesman?

The brute force approach for Maximize the Profit as the Salesman is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we consider every possible combination of offers to see which houses can be sold for maximum profit. This involves checking all buyers and their ranges, which can be quite slow for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximize the Profit as the Salesman?

Maximize the Profit as the Salesman uses the following concepts: Array, Hash Table, Binary Search, Dynamic Programming, Sorting. The recommended patterns to study are: Dynamic Programming, Sorting, Interval Scheduling.

What are the interview tips for Maximize the Profit as the Salesman?

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can break down the problem into smaller parts, especially when dealing with overlapping intervals. Practice explaining your thought process clearly while coding, as communication is key in interviews.

What are common mistakes when solving Maximize the Profit as the Salesman?

Failing to consider overlapping offers properly. Not sorting the offers before processing them, which can lead to incorrect results.

#2830

Maximize the Profit as the Salesman

Medium

ArrayHash TableBinary SearchDynamic ProgrammingSortingDynamic ProgrammingSortingInterval Scheduling

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 efficiently calculate the maximum gold by considering each offer and the best possible outcome for the houses sold up to that point. By sorting the offers and using a DP array, we can avoid redundant calculations.

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Algorithm

3 steps

1Step 1: Sort the offers based on the start index.
2Step 2: Initialize a DP array where dp[i] represents the maximum gold earned by selling houses up to index i.
3Step 3: For each offer, update the DP array considering the current offer and the best previous non-overlapping offer.

solution.py11 lines

1def maxGold(n, offers):
2    offers.sort()
3    dp = [0] * n
4    for start, end, gold in offers:
5        dp[end] = max(dp[end], (dp[start - 1] if start > 0 else 0) + gold)
6    for i in range(1, n):
7        dp[i] = max(dp[i], dp[i - 1])
8    return dp[-1]
9
10# Example usage
11print(maxGold(5, [[0,0,1],[0,2,2],[1,3,2]]))

ℹ

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

1Understanding the overlap of offers is crucial to maximizing profit.
2Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.

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