Introduction

In this blog post, we will explore how to solve the LeetCode problem "3487" 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. You are allowed to delete any number of elements from nums without making it empty. After performing the deletions, select a subarray of nums such that: All elements in the subarray are unique. The sum of the elements in the subarray is maximized. Return the maximum sum of such a subarray. Example 1: Input: nums = [1,2,3,4,5] Output: 15 Explanation: Select the entire array without deleting any element to obtain the maximum sum. Example 2: Input: nums = [1,1,0,1,1] Output: 1 Explanation: Delete the element nums[0] == 1, nums[1] == 1, nums[2] == 0, and nums[3] == 1. Select the entire array [1] to obtain the maximum sum. Example 3: Input: nums = [1,2,-1,-2,1,0,-1] Output: 3 Explanation: Delete the elements nums[2] == -1 and nums[3] == -2, and select the subarray [2, 1] from [1, 2, 1, 0, -1] to obtain the maximum sum. Constraints: 1 <= nums.length <= 100 -100 <= nums[i] <= 100

Explanation

Here's the solution to the problem:

• Sliding Window with a Set: Use a sliding window approach to maintain a subarray with unique elements. A set is used to efficiently track the elements currently within the window.

• Expanding and Contracting the Window: Expand the window to the right as long as the new element is not already present in the set. If a duplicate is encountered, contract the window from the left until the duplicate is removed from the set.

• Maximize the Sum: Keep track of the current sum of the elements in the window and update the maximum sum encountered so far.

• Runtime Complexity: O(n), where n is the length of the input array. Storage Complexity: O(k), where k is the number of unique elements in the input array (in the worst case O(n)).

Code

    def maximumUniqueSubarray(nums):
    """
    Finds the maximum sum of a subarray with unique elements.

    Args:
        nums: A list of integers.

    Returns:
        The maximum sum of a subarray with unique elements.
    """

    max_sum = 0
    current_sum = 0
    seen = set()
    left = 0

    for right in range(len(nums)):
        while nums[right] in seen:
            current_sum -= nums[left]
            seen.remove(nums[left])
            left += 1

        current_sum += nums[right]
        seen.add(nums[right])
        max_sum = max(max_sum, current_sum)

    return max_sum