Introduction

In this blog post, we will explore how to solve the LeetCode problem "2830" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. As a salesman, your goal is to maximize your earnings by strategically selecting and selling houses to buyers. Return the maximum amount of gold you can earn. Note that different buyers can't buy the same house, and some houses may remain unsold. Example 1: 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. Example 2: Input: n = 5, offers = [[0,0,1],[0,2,10],[1,3,2]] Output: 10 Explanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers. We sell houses in the range [0,2] to 2nd buyer for 10 golds. It can be proven that 10 is the maximum amount of gold we can achieve. Constraints: 1 <= n <= 105 1 <= offers.length <= 105 offers[i].length == 3 0 <= starti <= endi <= n - 1 1 <= goldi <= 103

Explanation

Here's an efficient solution to the problem:

• Dynamic Programming with Binary Search: We use dynamic programming to store the maximum gold we can obtain up to each house. The key is to iterate through sorted offers and for each offer, decide whether to include it or not. Binary search helps us efficiently find the latest non-overlapping offer.
• Sorting Offers: We sort the offers based on their end indices to facilitate the binary search process and enable efficient dynamic programming.

• Optimal Substructure: The solution leverages the fact that the optimal solution for a range can be built from optimal solutions of smaller ranges by considering if an offer should be taken, or ignored.

• Runtime Complexity: O(N log N + M log M + M log N), where N is the number of houses and M is the number of offers. Sorting takes O(M log M), DP takes O(M) and the binary search in the loop takes O(M log N).

• Storage Complexity: O(N + M), for storing the dp array and the sorted offers.

Code

    def maximize_gold(n: int, offers: list[list[int]]) -> int:
    """
    Calculates the maximum amount of gold that can be earned by strategically selecting and selling houses to buyers.

    Args:
        n: The number of houses on a number line.
        offers: A 2D integer array where offers[i] = [starti, endi, goldi], indicating that the ith buyer wants to buy all the houses from starti to endi for goldi amount of gold.

    Returns:
        The maximum amount of gold that can be earned.
    """

    # Sort offers by end index
    offers.sort(key=lambda x: x[1])

    # dp[i] stores the maximum gold we can obtain considering offers ending at or before index i
    dp = [0] * len(offers)

    for i in range(len(offers)):
        start, end, gold = offers[i]

        # Option 1: Don't include the current offer
        if i > 0:
            dp[i] = dp[i - 1]

        # Option 2: Include the current offer
        current_gold = gold
        # Find the latest non-overlapping offer using binary search
        prev_index = -1
        low = 0
        high = i - 1
        while low <= high:
            mid = (low + high) // 2
            if offers[mid][1] < start:
                prev_index = mid
                low = mid + 1
            else:
                high = mid - 1

        if prev_index != -1:
            current_gold += dp[prev_index]

        dp[i] = max(dp[i], current_gold)

    # Find the last non-overlapping offer
    if len(offers) > 0:
        return dp[-1]
    else:
        return 0