README_EN.md

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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