- Constrained Subsequence Sum
Hard
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
解法1：DP。会超时。
class Solution {
public:
int constrainedSubsetSum(vector<int>& nums, int k) {
int n = nums.size();
vector<int> dp(n, INT_MIN / 3);
dp[0] = nums[0];
int res = dp[0];
for (int i = 1; i < n; i++) {
dp[i] = nums[i];
for (int j = 1; j <= k; j++) {
if (i >= j) {
dp[i] = max(dp[i], dp[i - j] + nums[i]);
}
}
res = max(res, dp[i]);
}
return res;
}
};
解法2：DP+单调队列。注意单调队列本身就是一个滑动窗口。
class Solution {
public:
int constrainedSubsetSum(vector<int>& nums, int k) {
int n = nums.size();
vector<int> dp(n, INT_MIN / 3);
dp[0] = nums[0];
int res = dp[0];
deque<int> dq;
dq.push_back(0);
for (int i = 1; i < n; i++) {
while (!dq.empty() && i - dq.front() > k) dq.pop_front();
dp[i] = max(nums[i], dp[dq.front()] + nums[i]);
while (!dq.empty() && dp[dq.back()] < dp[i]) dq.pop_back();
dq.push_back(i);
res = max(res, dp[i]);
}
return res;
}
};