Introduction

In this blog post, we will explore how to solve the LeetCode problem "2195" 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 array nums and an integer k. Append k unique positive integers that do not appear in nums to nums such that the resulting total sum is minimum. Return the sum of the k integers appended to nums. Example 1: Input: nums = [1,4,25,10,25], k = 2 Output: 5 Explanation: The two unique positive integers that do not appear in nums which we append are 2 and 3. The resulting sum of nums is 1 + 4 + 25 + 10 + 25 + 2 + 3 = 70, which is the minimum. The sum of the two integers appended is 2 + 3 = 5, so we return 5. Example 2: Input: nums = [5,6], k = 6 Output: 25 Explanation: The six unique positive integers that do not appear in nums which we append are 1, 2, 3, 4, 7, and 8. The resulting sum of nums is 5 + 6 + 1 + 2 + 3 + 4 + 7 + 8 = 36, which is the minimum. The sum of the six integers appended is 1 + 2 + 3 + 4 + 7 + 8 = 25, so we return 25. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109 1 <= k <= 108

Explanation

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

• Key Idea: The most efficient way to minimize the sum is to append the smallest possible unique positive integers that are not already present in nums. We iterate, keeping track of the smallest missing integer and adding it until we've appended k integers.
• Leveraging Sorted Input: Sorting the input array nums allows us to quickly identify the missing positive integers in a systematic manner. We can skip over duplicates efficiently too.

• Sum Calculation: Once we identify the sequence of integers to be appended, we can compute their sum using the arithmetic series formula for efficient summation.

• Complexity: O(n log n) time for sorting, O(1) space (excluding the space used to store the sorted array in place).

Code

    def minimalKSum(nums, k):
    nums = sorted(list(set(nums)))  # Remove duplicates and sort
    appended_sum = 0
    last_num = 0  # Initialize the last number seen (before the start)

    for num in nums:
        if num > last_num + 1:
            diff = num - last_num - 1
            if k >= diff:
                appended_sum += (last_num + 1 + num - 1) * diff // 2
                k -= diff
                last_num = num
            else:
                appended_sum += (last_num + 1 + last_num + k) * k // 2
                k = 0
                break
        last_num = num

    if k > 0:
        appended_sum += (last_num + 1 + last_num + k) * k // 2

    return appended_sum