1695-maximum-erasure-value

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Description

You are given an array of positive integers nums and want to erase a subarray containing unique elements. The score you get by erasing the subarray is equal to the sum of its elements.

Return the maximum score you can get by erasing exactly one subarray.

An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is, if it is equal to a[l],a[l+1],...,a[r] for some (l,r).

 

Example 1:

Input: nums = [4,2,4,5,6]
Output: 17
Explanation: The optimal subarray here is [2,4,5,6].

Example 2:

Input: nums = [5,2,1,2,5,2,1,2,5]
Output: 8
Explanation: The optimal subarray here is [5,2,1] or [1,2,5].

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 104

Solution(Python)

class Solution:
    def maximumUniqueSubarray(self, nums: List[int]) -> int:
        return self.optimal(nums)

    # time complexity: O(n^2)
    # Space complexity: O(n)
    def bruteforce(self, nums: List[int]) -> int:
        n = len(nums)
        max_sum = 0
        seen = set()
        for i in range(n):
            seen.clear()
            cur_sum = 0

            for j in range(i, n):
                if nums[j] in seen:
                    break
                cur_sum += nums[j]
                seen.add(nums[j])
            max_sum = max(max_sum, cur_sum)
        return max_sum

    # time complexity: O(n)
    # Space complexity: O(n)
    def optimal(self, nums: List[int]) -> int:
        n = len(nums)
        max_sum = 0
        cur_sum = 0
        seen = set()
        l = 0
        for num in nums:
            while num in seen:
                seen.remove(nums[l])
                cur_sum -= nums[l]
                l += 1
            cur_sum += num
            seen.add(num)
            max_sum = max(max_sum, cur_sum)
        return max_sum