1696. Jump Game VI

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.

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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]

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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];
    }
}

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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[q.back()] <= f[i]) q.pop_back();
            q.push_back(i);
        }
        return f[n - 1];
    }
};

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func maxResult(nums []int, k int) int {
	n := len(nums)
	f := make([]int, n)
	q := []int{0}
	for i, v := range nums {
		if i-q[0] > k {
			q = q[1:]
		}
		f[i] = v + f[q[0]]
		for len(q) > 0 && f[q[len(q)-1]] <= f[i] {
			q = q[:len(q)-1]
		}
		q = append(q, i)
	}
	return f[n-1]
}