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: Dynamic Programming + Monotonic Queue
We define f[i] to represent the maximum sum of the subsequence ending at nums[i] that meets the conditions. Initially, f[i] = 0, and the answer is max_{0 ≤ i \lt n} f(i).
We notice that the problem requires us to maintain the maximum value of a sliding window, which is a typical application scenario for a monotonic queue. We can use a monotonic queue to optimize the dynamic programming transition.
We maintain a monotonic queue q that is decreasing from the front to the back, storing the indices i. Initially, we add a sentinel 0 to the queue.
We traverse i from 0 to n - 1. For each i, we perform the following operations:
- If the front element q[0] satisfies i - q[0] > k, it means the front element is no longer within the sliding window, and we need to remove the front element from the queue;
- Then, we calculate f[i] = max(0, f[q[0]]) + nums[i], which means we add nums[i] to the sliding window to get the maximum subsequence sum;
- Next, we update the answer ans = max(ans, f[i]);
- Finally, we add i to the back of the queue and maintain the monotonicity of the queue. If f[q[back]] ≤ f[i], we need to remove the back element until the queue is empty or f[q[back]] > f[i].
The final answer is ans.
The time complexity is O(n), and the space complexity is O(n). Here, n is the length of the array nums.
PythonJavaC++GoTypeScript
class Solution:
def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
q = deque([0])
n = len(nums)
f = [0] * n
ans = -inf
for i, x in enumerate(nums):
while i - q[0] > k:
q.popleft()
f[i] = max(0, f[q[0]]) + x
ans = max(ans, f[i])
while q and f[q[-1]] <= f[i]:
q.pop()
q.append(i)
return ans(code-box)
class Solution {
public int constrainedSubsetSum(int[] nums, int k) {
Deque<Integer> q = new ArrayDeque<>();
q.offer(0);
int n = nums.length;
int[] f = new int[n];
int ans = -(1 << 30);
for (int i = 0; i < n; ++i) {
while (i - q.peekFirst() > k) {
q.pollFirst();
}
f[i] = Math.max(0, f[q.peekFirst()]) + nums[i];
ans = Math.max(ans, f[i]);
while (!q.isEmpty() && f[q.peekLast()] <= f[i]) {
q.pollLast();
}
q.offerLast(i);
}
return ans;
}
}(code-box)
class Solution {
public:
int constrainedSubsetSum(vector<int>& nums, int k) {
deque<int> q = {0};
int n = nums.size();
int f[n];
f[0] = 0;
int ans = INT_MIN;
for (int i = 0; i < n; ++i) {
while (i - q.front() > k) {
q.pop_front();
}
f[i] = max(0, f[q.front()]) + nums[i];
ans = max(ans, f[i]);
while (!q.empty() && f[q.back()] <= f[i]) {
q.pop_back();
}
q.push_back(i);
}
return ans;
}
};(code-box)
func constrainedSubsetSum(nums []int, k int) int {
q := Deque{}
q.PushFront(0)
n := len(nums)
f := make([]int, n)
ans := nums[0]
for i, x := range nums {
for i-q.Front() > k {
q.PopFront()
}
f[i] = max(0, f[q.Front()]) + x
ans = max(ans, f[i])
for !q.Empty() && f[q.Back()] <= f[i] {
q.PopBack()
}
q.PushBack(i)
}
return ans
}
// template
type Deque struct{ l, r []int }
func (q Deque) Empty() bool {
return len(q.l) == 0 && len(q.r) == 0
}
func (q Deque) Size() int {
return len(q.l) + len(q.r)
}
func (q *Deque) PushFront(v int) {
q.l = append(q.l, v)
}
func (q *Deque) PushBack(v int) {
q.r = append(q.r, v)
}
func (q *Deque) PopFront() (v int) {
if len(q.l) > 0 {
q.l, v = q.l[:len(q.l)-1], q.l[len(q.l)-1]
} else {
v, q.r = q.r[0], q.r[1:]
}
return
}
func (q *Deque) PopBack() (v int) {
if len(q.r) > 0 {
q.r, v = q.r[:len(q.r)-1], q.r[len(q.r)-1]
} else {
v, q.l = q.l[0], q.l[1:]
}
return
}
func (q Deque) Front() int {
if len(q.l) > 0 {
return q.l[len(q.l)-1]
}
return q.r[0]
}
func (q Deque) Back() int {
if len(q.r) > 0 {
return q.r[len(q.r)-1]
}
return q.l[0]
}
func (q Deque) Get(i int) int {
if i < len(q.l) {
return q.l[len(q.l)-1-i]
}
return q.r[i-len(q.l)]
}(code-box)
function constrainedSubsetSum(nums: number[], k: number): number {
const q = new Deque<number>();
const n = nums.length;
q.pushBack(0);
let ans = nums[0];
const f: number[] = Array(n).fill(0);
for (let i = 0; i < n; ++i) {
while (i - q.frontValue()! > k) {
q.popFront();
}
f[i] = Math.max(0, f[q.frontValue()!]!) + nums[i];
ans = Math.max(ans, f[i]);
while (!q.isEmpty() && f[q.backValue()!]! <= f[i]) {
q.popBack();
}
q.pushBack(i);
}
return ans;
}
class Node<T> {
value: T;
next: Node<T> | null;
prev: Node<T> | null;
constructor(value: T) {
this.value = value;
this.next = null;
this.prev = null;
}
}
class Deque<T> {
private front: Node<T> | null;
private back: Node<T> | null;
private size: number;
constructor() {
this.front = null;
this.back = null;
this.size = 0;
}
pushFront(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.next = this.front;
this.front!.prev = newNode;
this.front = newNode;
}
this.size++;
}
pushBack(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.prev = this.back;
this.back!.next = newNode;
this.back = newNode;
}
this.size++;
}
popFront(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.front!.value;
this.front = this.front!.next;
if (this.front !== null) {
this.front.prev = null;
} else {
this.back = null;
}
this.size--;
return value;
}
popBack(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.back!.value;
this.back = this.back!.prev;
if (this.back !== null) {
this.back.next = null;
} else {
this.front = null;
}
this.size--;
return value;
}
frontValue(): T | undefined {
return this.front?.value;
}
backValue(): T | undefined {
return this.back?.value;
}
getSize(): number {
return this.size;
}
isEmpty(): boolean {
return this.size === 0;
}
}(code-box)
