Problem
Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i <= k is satisfied.
A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.
Example 1:
Input: nums = [10,2,-10,5,20], k = 2
Output: 37
Explanation: The subsequence is [10, 2, 5, 20].
Example 2:
Input: nums = [-1,-2,-3], k = 1
Output: -1
Explanation: The subsequence must be non-empty, so we choose the largest number.
Example 3:
Input: nums = [10,-2,-10,-5,20], k = 2
Output: 23
Explanation: The subsequence is [10, -2, -5, 20].
Constraints:
1 <= k <= nums.length <= 10^5-10^4 <= nums[i] <= 10^4
Solution
class Solution {
public int constrainedSubsetSum(int[] nums, int k) {
int n = nums.length;
int res = Integer.MIN_VALUE;
LinkedList<int[]> mono = new LinkedList<>();
for (int i = 0; i < n; i++) {
int take = nums[i];
while (!mono.isEmpty() && i - mono.getFirst()[0] > k) {
mono.removeFirst();
}
if (!mono.isEmpty()) {
int mx = Math.max(0, mono.getFirst()[1]);
take += mx;
}
while (!mono.isEmpty() && take > mono.getLast()[1]) {
mono.removeLast();
}
mono.add(new int[] {i, take});
res = Math.max(res, take);
}
return res;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).