LeetCode 3487: Maximum Unique Subarray Sum After Deletion Solution

Master LeetCode problem 3487 (Maximum Unique Subarray Sum After Deletion), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3487. Maximum Unique Subarray Sum After Deletion

Easy

Array Hash Table Greedy

Problem Explanation

Explanation

To solve this problem, we can iterate through the input array and maintain a set to keep track of unique elements encountered so far. We also keep track of the maximum sum achieved by considering elements up to the current index. We use two pointers to maintain a sliding window containing only unique elements.

Initialize a set to store unique elements and variables start and maxSum to keep track of the current window and maximum sum.
Iterate through the input array:
- For each element, check if it's already in the set:
  - If it is, remove elements from the set and update the start pointer until the current element is not in the set.
- Add the current element to the set.
- Update the maximum sum if needed.
Return the maximum sum.

Time Complexity: O(n) where n is the length of the input array nums. Space Complexity: O(n) for the set storing unique elements.

Solution Code

class Solution {
    public int maximumUniqueSubarray(int[] nums) {
        Set<Integer> uniqueSet = new HashSet<>();
        int start = 0, maxSum = 0, currSum = 0;

        for (int i = 0; i < nums.length; i++) {
            while (uniqueSet.contains(nums[i])) {
                currSum -= nums[start];
                uniqueSet.remove(nums[start]);
                start++;
            }

            currSum += nums[i];
            uniqueSet.add(nums[i]);
            maxSum = Math.max(maxSum, currSum);
        }

        return maxSum;
    }
}

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