Description
You are given a 0-indexed integer array nums and an integer k.
You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive.
You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array.
Return the maximum score you can get.
Example 1:
Input: nums = [1,-1,-2,4,-7,3], k = 2
Output: 7
Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
Example 2:
Input: nums = [10,-5,-2,4,0,3], k = 3
Output: 17
Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
Example 3:
Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
Output: 0
Constraints:
1 <= nums.length, k <= 105-104 <= nums[i] <= 104
Solutions
Solution 1: Dynamic Programming + Monotonic Queue Optimization
We define f[i] as the maximum score when reaching index i. The value of f[i] can be transferred from f[j], where j satisfies i - k ≤ j ≤ i - 1. Therefore, we can use dynamic programming to solve this problem.
The state transition equation is:
f[i] = max_{j ∈ [i - k, i - 1]} f[j] + nums[i]
We can use a monotonic queue to optimize the state transition equation. Specifically, we maintain a monotonically decreasing queue, which stores the index j, and the f[j] values corresponding to the indices in the queue are monotonically decreasing. When performing state transition, we only need to take out the index j at the front of the queue to get the maximum value of f[j], and then update the value of f[i] to f[j] + nums[i].
The time complexity is O(n), and the space complexity is O(n). Here, n is the length of the array.
PythonJavaC++GoTypeScript
class Solution:
def maxResult(self, nums: List[int], k: int) -> int:
n = len(nums)
f = [0] * n
q = deque([0])
for i in range(n):
if i - q[0] > k:
q.popleft()
f[i] = nums[i] + f[q[0]]
while q and f[q[-1]] <= f[i]:
q.pop()
q.append(i)
return f[-1](code-box)
class Solution {
public int maxResult(int[] nums, int k) {
int n = nums.length;
int[] f = new int[n];
Deque<Integer> q = new ArrayDeque<>();
q.offer(0);
for (int i = 0; i < n; ++i) {
if (i - q.peekFirst() > k) {
q.pollFirst();
}
f[i] = nums[i] + f[q.peekFirst()];
while (!q.isEmpty() && f[q.peekLast()] <= f[i]) {
q.pollLast();
}
q.offerLast(i);
}
return f[n - 1];
}
}(code-box)
class Solution {
public:
int maxResult(vector<int>& nums, int k) {
int n = nums.size();
int f[n];
f[0] = 0;
deque<int> q = {0};
for (int i = 0; i < n; ++i) {
if (i - q.front() > k) {
q.pop_front();
}
f[i] = nums[i] + f[q.front()];
while (!q.empty() && f[i] >= f[q.back()]) {
q.pop_back();
}
q.push_back(i);
}
return f[n - 1];
}
};(code-box)
func maxResult(nums []int, k int) int {
n := len(nums)
f := make([]int, n)
q := Deque{}
q.PushBack(0)
for i := 0; i < n; i++ {
if i-q.Front() > k {
q.PopFront()
}
f[i] = nums[i] + f[q.Front()]
for !q.Empty() && f[i] >= f[q.Back()] {
q.PopBack()
}
q.PushBack(i)
}
return f[n-1]
}
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 maxResult(nums: number[], k: number): number {
const n = nums.length;
const f: number[] = Array(n).fill(0);
const q = new Deque<number>();
q.pushBack(0);
for (let i = 0; i < n; ++i) {
if (i - q.frontValue()! > k) {
q.popFront();
}
f[i] = nums[i] + f[q.frontValue()!];
while (!q.isEmpty() && f[i] >= f[q.backValue()!]) {
q.popBack();
}
q.pushBack(i);
}
return f[n - 1];
}
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)
