Introduction

In this blog post, we will explore how to solve the LeetCode problem "2593" 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 array nums consisting of positive integers. Starting with score = 0, apply the following algorithm: Choose the smallest integer of the array that is not marked. If there is a tie, choose the one with the smallest index. Add the value of the chosen integer to score. Mark the chosen element and its two adjacent elements if they exist. Repeat until all the array elements are marked. Return the score you get after applying the above algorithm. Example 1: Input: nums = [2,1,3,4,5,2] Output: 7 Explanation: We mark the elements as follows: - 1 is the smallest unmarked element, so we mark it and its two adjacent elements: [2,1,3,4,5,2]. - 2 is the smallest unmarked element, so we mark it and its left adjacent element: [2,1,3,4,5,2]. - 4 is the only remaining unmarked element, so we mark it: [2,1,3,4,5,2]. Our score is 1 + 2 + 4 = 7. Example 2: Input: nums = [2,3,5,1,3,2] Output: 5 Explanation: We mark the elements as follows: - 1 is the smallest unmarked element, so we mark it and its two adjacent elements: [2,3,5,1,3,2]. - 2 is the smallest unmarked element, since there are two of them, we choose the left-most one, so we mark the one at index 0 and its right adjacent element: [2,3,5,1,3,2]. - 2 is the only remaining unmarked element, so we mark it: [2,3,5,1,3,2]. Our score is 1 + 2 + 2 = 5. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 106

Explanation

Here's the solution:

• High-level approach:

  • Use a boolean array marked to keep track of marked elements.
  • Iterate until all elements are marked. In each iteration, find the smallest unmarked element (and its index if there's a tie).
  • Add the element to the score, mark the element and its adjacent elements.

• Complexity:

  • Runtime: O(n log n) due to sorting (or using a heap)
  • Storage: O(n) for the marked array and potentially for sorting

Code

    def solve():
    nums = [2,1,3,4,5,2]
    print(get_score(nums))

    nums = [2,3,5,1,3,2]
    print(get_score(nums))

def get_score(nums):
    n = len(nums)
    marked = [False] * n
    score = 0

    while not all(marked):
        min_val = float('inf')
        min_index = -1

        for i in range(n):
            if not marked[i] and nums[i] < min_val:
                min_val = nums[i]
                min_index = i
            elif not marked[i] and nums[i] == min_val and i < min_index:
                min_index = i

        score += nums[min_index]
        marked[min_index] = True

        if min_index > 0:
            marked[min_index - 1] = True
        if min_index < n - 1:
            marked[min_index + 1] = True

    return score