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 <= 105
-104 <= nums[i] <= 104
Solution:
class Solution {
public int constrainedSubsetSum(int[] nums, int k) {
int n = nums.length;
int[] dp = new int[n];
Deque<Integer> decreasing = new LinkedList();
int ans = Integer.MIN_VALUE;
for (int i = 0; i < n; i ++) {
int prevMaxSum = Math.max(0, decreasing.isEmpty() ? 0 : dp[decreasing.peekFirst()]);
dp[i] = nums[i] + prevMaxSum;
ans = Math.max(ans, dp[i]);
while (!decreasing.isEmpty() && dp[i] >= dp[decreasing.peekLast()]) {
decreasing.pollLast();
}
decreasing.offer(i);
if (i - decreasing.peekFirst() >= k) {
decreasing.pollFirst();
}
}
return ans;
}
}