# 1425. Constrained Subsequence Sum

## Description

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

## Solutions

• class Solution {
public int constrainedSubsetSum(int[] nums, int k) {
int n = nums.length;
int[] dp = new int[n];
int ans = Integer.MIN_VALUE;
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0; i < n; ++i) {
if (!q.isEmpty() && i - q.peek() > k) {
q.poll();
}
dp[i] = Math.max(0, q.isEmpty() ? 0 : dp[q.peek()]) + nums[i];
while (!q.isEmpty() && dp[q.peekLast()] <= dp[i]) {
q.pollLast();
}
q.offer(i);
ans = Math.max(ans, dp[i]);
}
return ans;
}
}

• class Solution {
public:
int constrainedSubsetSum(vector<int>& nums, int k) {
int n = nums.size();
vector<int> dp(n);
int ans = INT_MIN;
deque<int> q;
for (int i = 0; i < n; ++i) {
if (!q.empty() && i - q.front() > k) q.pop_front();
dp[i] = max(0, q.empty() ? 0 : dp[q.front()]) + nums[i];
ans = max(ans, dp[i]);
while (!q.empty() && dp[q.back()] <= dp[i]) q.pop_back();
q.push_back(i);
}
return ans;
}
};

• class Solution:
def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
n = len(nums)
dp = [0] * n
ans = -inf
q = deque()
for i, v in enumerate(nums):
if q and i - q[0] > k:
q.popleft()
dp[i] = max(0, 0 if not q else dp[q[0]]) + v
while q and dp[q[-1]] <= dp[i]:
q.pop()
q.append(i)
ans = max(ans, dp[i])
return ans


• func constrainedSubsetSum(nums []int, k int) int {
n := len(nums)
dp := make([]int, n)
ans := math.MinInt32
q := []int{}
for i, v := range nums {
if len(q) > 0 && i-q[0] > k {
q = q[1:]
}
dp[i] = v
if len(q) > 0 && dp[q[0]] > 0 {
dp[i] += dp[q[0]]
}
for len(q) > 0 && dp[q[len(q)-1]] < dp[i] {
q = q[:len(q)-1]
}
q = append(q, i)
ans = max(ans, dp[i])
}
return ans
}