Question

1696. Jump Game VI

中文文档

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 $i - k \leq j \leq i - 1$. Therefore, we can use dynamic programming to solve this problem.

The state transition equation is:

$$ f[i] = \max_{j \in [i - k, i - 1]} f[j] + nums[i] $$

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.

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]

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];
  }
}

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];
  }
};

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)]
}

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;
  }
}