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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 1:
This can be solved in multiple ways, the easiest is to maintain a queue and verify the value read from the given nums array is unique. If not remove the element and the count from the queue and the current count respectively.

class Solution {
    public int maximumUniqueSubarray(int[] nums) {
        Queue<Integer> q = new LinkedList<>();
        int maxCount = 0;
        int count = 0;
        for(int i =0;i< nums.length;i++){
            if(!q.contains(nums[i])){
                count += nums[i];
                q.add(nums[i]);
            }else {
                
                maxCount = count>maxCount ? count: maxCount;
                while(!q.isEmpty() && q.peek() != nums[i]){
                    int del = q.remove();
                    count -= del;
                }
                q.remove();
                q.add(nums[i]);
            }
            
        }
        maxCount = count>maxCount ? count: maxCount;
        return maxCount;
    }
}

Solution 2:
The optimal way to solve this would be to use two pointer approach, which can be solved using O(n) time and O(m) space where m is the number of unique elements.

public int maximumUniqueSubarray(int[] nums) {        
 int maxScore = 0, currScore = 0;
 Set<Integer> set = new HashSet<>();

 for (int l=0, r=0; r<nums.length; r++) {
  while (!set.add(nums[r])) {
   currScore -= nums[l];
   set.remove(nums[l++]);
  }
  currScore += nums[r];
  maxScore = Math.max(maxScore, currScore);
 }

 return maxScore;
}