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This approach involves using dynamic programming with a sliding window to maintain the maximum sum at each position. For each position i in the array, calculate the maximum sum that can be achieved till that position, considering the constraint j - i <= k. Use a deque to track the indices which offer the maximum sum within the range of k.
Time Complexity: O(n), where n is the number of elements in the array nums. The space complexity is O(n) due to the storage of the dp array and deque.
1using System;
2using System.Collections.Generic;
3
4class Program {
5 public static int ConstrainedSubsequenceSum(int[] nums, int k) {
6 LinkedList<int> deque = new LinkedList<int>();
7 int[] dp = new int[nums.Length];
8 int maxSum = nums[0];
9 dp[0] = nums[0];
10 deque.AddLast(0);
11
12 for (int i = 1; i < nums.Length; i++) {
13 if (deque.First.Value < i - k) {
14 deque.RemoveFirst();
15 }
16 dp[i] = nums[i] + (deque.Count == 0 ? 0 : dp[deque.First.Value]);
17 maxSum = Math.Max(maxSum, dp[i]);
18 while (deque.Count > 0 && dp[deque.Last.Value] <= dp[i]) {
19 deque.RemoveLast();
20 }
21 deque.AddLast(i);
22 }
23
24 return maxSum;
25 }
26
27 static void Main(string[] args) {
28 int[] nums = { 10, 2, -10, 5, 20 };
29 int k = 2;
30 Console.WriteLine(ConstrainedSubsequenceSum(nums, k));
31 }
32}This C# solution works similarly to other languages shown above, utilizing a double-ended queue for optimal management of the sliding window maximum calculation.
By using a priority queue (or heap), we manage the maximum possible sum within the constraint more efficiently. We employ dynamic programming to calculate the possible maximal sum at each index while maintaining a priority queue to keep track of the relevant maximum sums.
Time Complexity: O(n log k) primarily due to heap operations. Space Complexity: O(n) is utilized by the dp array and the heap.
1#include <vector>
#include <queue>
using namespace std;
int constrainedSubsequenceSum(vector<int>& nums, int k) {
priority_queue<pair<int, int>> pq;
vector<int> dp(nums.size(), 0);
dp[0] = nums[0];
pq.push({nums[0], 0});
int maxSum = nums[0];
for (int i = 1; i < nums.size(); ++i) {
while (!pq.empty() && pq.top().second < i - k) {
pq.pop();
}
int sum = nums[i];
if (!pq.empty()) {
sum += pq.top().first;
}
dp[i] = sum;
pq.push({dp[i], i});
maxSum = max(maxSum, dp[i]);
}
return maxSum;
}
int main() {
vector<int> nums = {10, 2, -10, 5, 20};
int k = 2;
cout << constrainedSubsequenceSum(nums, k) << endl;
return 0;
}C++ uses a max-heap (priority queue) to track potential maxima. It provides a compact way to check out the maximums across a moving window efficiently.