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1425. Constrained Subsequence Sum 👍

  • Time: $O(n)$
  • Space: $O(k)$
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class Solution {
 public:
  int constrainedSubsetSum(vector<int>& nums, int k) {
    // dp[i] := max sum of non-empty subsequence in nums[0..i]
    vector<int> dp(nums.size());
    // Q stores dp[i - k], dp[i - k + 1], ..., dp[i - 1] whose values are > 0 in
    // Decreasing order
    deque<int> q;

    for (int i = 0; i < nums.size(); ++i) {
      if (q.empty())
        dp[i] = nums[i];
      else
        dp[i] = max(q.front(), 0) + nums[i];
      while (!q.empty() && q.back() < dp[i])
        q.pop_back();
      q.push_back(dp[i]);
      if (i >= k && dp[i - k] == q.front())
        q.pop_front();
    }

    return *max_element(begin(dp), end(dp));
  }
};

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class Solution {
  public int constrainedSubsetSum(int[] nums, int k) {
    // dp[i] := max sum of non-empty subsequence in nums[0..i]
    int[] dp = new int[nums.length];
    // Q stores dp[i - k], dp[i - k + 1], ..., dp[i - 1] whose values are > 0 in decreasing order
    Deque<Integer> q = new ArrayDeque<>();

    for (int i = 0; i < nums.length; ++i) {
      if (q.isEmpty())
        dp[i] = nums[i];
      else
        dp[i] = Math.max(q.peekFirst(), 0) + nums[i];
      while (!q.isEmpty() && q.peekLast() < dp[i])
        q.pollLast();
      q.offerLast(dp[i]);
      if (i >= k && dp[i - k] == q.peekFirst())
        q.pollFirst();
    }

    return Arrays.stream(dp).max().getAsInt();
  }
}

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class Solution:
  def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
    # dp[i] := max sum of non-empty subsequence in nums[0..i]
    dp = [0] * len(nums)
    # Q stores dp[i - k], dp[i - k + 1], ..., dp[i - 1] whose values are > 0 in decreasing order
    q = collections.deque()

    for i, num in enumerate(nums):
      if q:
        dp[i] = max(q[0], 0) + num
      else:
        dp[i] = num
      while q and q[-1] < dp[i]:
        q.pop()
      q.append(dp[i])
      if i >= k and dp[i - k] == q[0]:
        q.popleft()

    return max(dp)