Constrained Subsequence Sum

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

class Solution {
    public int constrainedSubsetSum(int[] nums, int k) {
        // This deque is going to be a strictly increasing deque(right -> left)
        Deque<Integer> deque = new LinkedList<>();
        int n = nums.length;
        int[] dp = new int[n];
        // dp[i] is the maximum sum we can get from nums[:i] and nums[i] is guaranteed
        // to be included.
        int res = Integer.MIN_VALUE;

        for (int i = 0; i < n; i++) {
            // Kicking out the values out of index window -> k
            while (!deque.isEmpty() && i - deque.peekFirst() > k)
                deque.pollFirst();
            // The max value is on top of deque
            dp[i] = Math.max(nums[i], (deque.isEmpty() ? 0 : dp[deque.peekFirst()]) + nums[i]);
            // Maintaining increasing deque
            while (!deque.isEmpty() && dp[i] > dp[deque.peekLast()])
                deque.pollLast();
            // Adding to the last
            deque.addLast(i);
            res = Math.max(res, dp[i]);
        }
        return res;
    }
}