Question

1425. Constrained Subsequence Sum

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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

Solution 1

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

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;
  }
};

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
}