In this guide, you will get 1695. Maximum Erasure Value LeetCode Solution with the best time and space complexity. The solution to Maximum Erasure Value problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(n)
1695. Maximum Erasure Value LeetCode Solution in C++
class Solution {
public:
int maximumUniqueSubarray(vector<int>& nums) {
int ans = 0;
int score = 0;
unordered_set<int> seen;
for (int l = 0, r = 0; r < nums.size(); ++r) {
while (!seen.insert(nums[r]).second) {
score -= nums[l];
seen.erase(nums[l++]);
}
score += nums[r];
ans = max(ans, score);
}
return ans;
}
}; /* code provided by PROGIEZ */
1695. Maximum Erasure Value LeetCode Solution in Java
class Solution {
public int maximumUniqueSubarray(int[] nums) {
int ans = 0;
int score = 0;
Set<Integer> seen = new HashSet<>();
for (int l = 0, r = 0; r < nums.length; ++r) {
while (!seen.add(nums[r])) {
score -= nums[l];
seen.remove(nums[l++]);
}
score += nums[r];
ans = Math.max(ans, score);
}
return ans;
}
} // code provided by PROGIEZ
1695. Maximum Erasure Value LeetCode Solution in Python
class Solution:
def maximumUniqueSubarray(self, nums: list[int]) -> int:
ans = 0
score = 0
seen = set()
l = 0
for r, num in enumerate(nums):
while num in seen:
score -= nums[l]
seen.remove(nums[l])
l += 1
seen.add(nums[r])
score += nums[r]
ans = max(ans, score)
return ans # code by PROGIEZ
