1695. Maximum Erasure Value

1695. Maximum Erasure Value (Medium)

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

Companies:
Cashfree

Related Topics:
Array, Hash Table, Sliding Window

Similar Questions:

• Longest Substring Without Repeating Characters (Medium)

Solution 1. Shrinkable Sliding Window + Hash Map

// OJ: https://leetcode.com/problems/maximum-erasure-value/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int maximumUniqueSubarray(vector<int>& A) {
        int i = 0, ans = 0, N = A.size(); // window [i, j] is a window which only contains unique elements.
        unordered_map<int, int> m; // number -> index of last occurrence.
        vector<int> sum(N + 1);
        partial_sum(begin(A), end(A), begin(sum) + 1);
        for (int j = 0; j < N; ++j) {
            if (m.count(A[j])) i = max(i, m[A[j]] + 1);
            m[A[j]] = j;
            ans = max(ans, sum[j + 1] - sum[i]);
        }
        return ans;
    }
};

Solution 2. Shrinkable Sliding Window + Hash Set

// OJ: https://leetcode.com/problems/maximum-erasure-value/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(U) where U is the number of unique elements in `A`
class Solution {
public:
    int maximumUniqueSubarray(vector<int>& A) {
        int ans = 0, N = A.size(), sum = 0;
        unordered_set<int> s;
        for (int i = 0, j = 0; j < N; ++j) {
            while (s.count(A[j])) {
                s.erase(A[i]);
                sum -= A[i++];
            }
            s.insert(A[j]);
            sum += A[j];
            ans = max(ans, sum);
        }
        return ans;
    }
};

Solution 3. Shrinkable Sliding Window + Prefix State Map

Check out "C++ Maximum Sliding Window Cheatsheet Template!"

// OJ: https://leetcode.com/problems/maximum-erasure-value/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(U) where U is the number of unique elements in `A`
class Solution {
public:
    int maximumUniqueSubarray(vector<int>& A) {
        int i = 0, j = 0, N = A.size(), ans = 0, dup = 0, sum = 0;
        unordered_map<int, int> cnt;
        while (j < N) {
            dup += ++cnt[A[j]] == 2;
            sum += A[j++];
            while (dup) {
                dup -= --cnt[A[i]] == 1;
                sum -= A[i++];
            }
            ans = max(ans, sum);
        }
        return ans;
    }
};

Discuss

https://leetcode.com/problems/maximum-erasure-value/discuss/1504271