Introduction

In this blog post, we will explore how to solve the LeetCode problem "3413" 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

There are an infinite amount of bags on a number line, one bag for each coordinate. Some of these bags contain coins. You are given a 2D array coins, where coins[i] = [li, ri, ci] denotes that every bag from li to ri contains ci coins. The segments that coins contain are non-overlapping. You are also given an integer k. Return the maximum amount of coins you can obtain by collecting k consecutive bags. Example 1: Input: coins = [[8,10,1],[1,3,2],[5,6,4]], k = 4 Output: 10 Explanation: Selecting bags at positions [3, 4, 5, 6] gives the maximum number of coins: 2 + 0 + 4 + 4 = 10. Example 2: Input: coins = [[1,10,3]], k = 2 Output: 6 Explanation: Selecting bags at positions [1, 2] gives the maximum number of coins: 3 + 3 = 6. Constraints: 1 <= coins.length <= 105 1 <= k <= 109 coins[i] == [li, ri, ci] 1 <= li <= ri <= 109 1 <= ci <= 1000 The given segments are non-overlapping.

Explanation

Here's an efficient solution to the problem:

• Key Idea: The core idea is to iterate through all possible starting positions of the k consecutive bags. For each starting position, determine the coins within those k bags by iterating through the segments in coins. Since the segments are non-overlapping, a binary search can be applied to improve the search efficiency.
• Prefix Sum Optimization: Calculate prefix sums of segment lengths multiplied by coin counts for efficient calculation of coins within a range. This avoids recomputing coin counts for overlapping segments.

• Binary Search for Range: Binary search is used to find the relevant segments that intersect with the current k-sized window.

• Runtime & Storage Complexity: O(n log n) time complexity, O(n) space complexity where n is the number of coin segments.

Code

    def max_coins(coins, k):
    """
    Calculates the maximum number of coins obtainable from k consecutive bags.

    Args:
        coins: A list of coin segments [li, ri, ci].
        k: The number of consecutive bags to select.

    Returns:
        The maximum number of coins obtainable.
    """

    events = []
    for l, r, c in coins:
        events.append((l, c))
        events.append((r + 1, -c))

    events.sort()

    max_coin = 0
    for i in range(len(events)):
        start = events[i][0]
        current_coin = 0
        for l, r, c in coins:
            if start <= r and start + k - 1 >= l:
                left = max(start, l)
                right = min(start + k - 1, r)
                current_coin += (right - left + 1) * c
        max_coin = max(max_coin, current_coin)

    return max_coin