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A simple C++ solution to Leetcode 1695. Maximum Erasure Value using a sliding window approach and prefix sums.
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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).
Input: nums = [4,2,4,5,6]
Output: 17
Explanation: The optimal subarray here is [2,4,5,6].
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].
1 <= nums.length <= 10^5.
1 <= nums[i] <= 10^4.
You can use a map to store the position of the elements of nums. Then when iterating nums you can identify if an element has been visited before. That helps you to decide if a subarray contains unique elements.
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
int maximumUniqueSubarray(vector<int>& nums) {
vector<int> sum(nums.size(), 0);
sum[0] = nums[0];
int maxSum = sum[0];
unordered_map<int, int> position;
position[nums[0]] = 0;
int start = -1;
for (int i = 1; i < nums.size(); i++) {
sum[i] = sum[i - 1] + nums[i];
if (position.find(nums[i]) != position.end()) {
start = max(start, position[nums[i]]);
}
position[nums[i]] = i;
maxSum = (start == -1) ? sum[i] : max(maxSum, sum[i] - sum[start]);
}
return maxSum;
}
int main() {
vector<int> nums{4,2,4,5,6};
cout << maximumUniqueSubarray(nums) << endl;
nums = {5,2,1,2,5,2,1,2,5};
cout << maximumUniqueSubarray(nums) << endl;
}
Output:
17
8
This C++ code defines a function maximumUniqueSubarray that finds the maximum sum of a subarray in the given vector nums with the constraint that all elements must be unique. It uses two key techniques: prefix sums and a sliding window approach.
Runtime: O(N), where N = nums.length.
Extra space: O(N).
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