Introduction

In this blog post, we will explore how to solve the LeetCode problem "2530" 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 a 0-indexed integer array nums and an integer k. You have a starting score of 0. In one operation: choose an index i such that 0 <= i < nums.length, increase your score by nums[i], and replace nums[i] with ceil(nums[i] / 3). Return the maximum possible score you can attain after applying exactly k operations. The ceiling function ceil(val) is the least integer greater than or equal to val. Example 1: Input: nums = [10,10,10,10,10], k = 5 Output: 50 Explanation: Apply the operation to each array element exactly once. The final score is 10 + 10 + 10 + 10 + 10 = 50. Example 2: Input: nums = [1,10,3,3,3], k = 3 Output: 17 Explanation: You can do the following operations: Operation 1: Select i = 1, so nums becomes [1,4,3,3,3]. Your score increases by 10. Operation 2: Select i = 1, so nums becomes [1,2,3,3,3]. Your score increases by 4. Operation 3: Select i = 2, so nums becomes [1,2,1,3,3]. Your score increases by 3. The final score is 10 + 4 + 3 = 17. Constraints: 1 <= nums.length, k <= 105 1 <= nums[i] <= 109

Explanation

Here's a breakdown of the approach, complexities, and the Python code:

• Key Idea: Use a max heap (priority queue) to always pick the element that yields the highest score increase in each operation. This greedy approach guarantees the maximum overall score.
• Efficiency: The heap allows us to efficiently find the maximum element and update it. We iterate k times, and each heap operation takes logarithmic time.

• Data Structures: Max Heap

• Runtime Complexity: O(k log n), where n is the length of nums.

• Storage Complexity: O(n) due to the heap.

Code

    import heapq
import math

def max_score(nums, k):
    """
    Calculates the maximum possible score after applying k operations.

    Args:
        nums: A list of integers.
        k: The number of operations.

    Returns:
        The maximum possible score.
    """

    heap = [-num for num in nums]  # Negate for max heap
    heapq.heapify(heap)
    score = 0

    for _ in range(k):
        num = -heapq.heappop(heap)
        score += num
        new_num = math.ceil(num / 3)
        heapq.heappush(heap, -new_num)

    return score