README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3221.Maximum%20Array%20Hopping%20Score%20II/README_EN.md	Stack Greedy Array Monotonic Stack

3221. Maximum Array Hopping Score II 🔒

中文文档

Description

Given an array nums, you have to get the maximum score starting from index 0 and hopping until you reach the last element of the array.

In each hop, you can jump from index i to an index j > i, and you get a score of (j - i) * nums[j].

Return the maximum score you can get.

 

Example 1:

Input: nums = [1,5,8]

Output: 16

Explanation:

There are two possible ways to reach the last element:

• 0 -> 1 -> 2 with a score of (1 - 0) * 5 + (2 - 1) * 8 = 13.
• 0 -> 2 with a score of (2 - 0) * 8 = 16.

Example 2:

Input: nums = [4,5,2,8,9,1,3]

Output: 42

Explanation:

We can do the hopping 0 -> 4 -> 6 with a score of (4 - 0) * 9 + (6 - 4) * 3 = 42.

 

Constraints:

• 2 <= nums.length <= 105
• 1 <= nums[i] <= 105

Solutions

Solution 1: Monotonic Stack

We observe that for the current position $i$, we should jump to the next position $j$ with the maximum value to obtain the maximum score.

Therefore, we traverse the array $\textit{nums}$, maintaining a stack $\textit{stk}$ that is monotonically decreasing from the bottom to the top of the stack. For the current position $i$ being traversed, if the value corresponding to the top element of the stack is less than or equal to $\textit{nums}[i]$, we continuously pop the top element of the stack until the stack is empty or the value corresponding to the top element of the stack is greater than $\textit{nums}[i]$, and then push $i$ into the stack.

Next, we initialize the answer $\textit{ans}$ and the current position $i = 0$, traverse the elements in the stack, each time taking out the top element $j$, updating the answer $\textit{ans} += \textit{nums}[j] \times (j - i)$, and then updating $i = j$.

Python3

class Solution:
    def maxScore(self, nums: List[int]) -> int:
        stk = []
        for i, x in enumerate(nums):
            while stk and nums[stk[-1]] <= x:
                stk.pop()
            stk.append(i)
        ans = i = 0
        for j in stk:
            ans += nums[j] * (j - i)
            i = j
        return ans

Java

class Solution {
    public long maxScore(int[] nums) {
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < nums.length; ++i) {
            while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) {
                stk.pop();
            }
            stk.push(i);
        }
        long ans = 0, i = 0;
        while (!stk.isEmpty()) {
            int j = stk.pollLast();
            ans += (j - i) * nums[j];
            i = j;
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long maxScore(vector<int>& nums) {
        vector<int> stk;
        for (int i = 0; i < nums.size(); ++i) {
            while (stk.size() && nums[stk.back()] <= nums[i]) {
                stk.pop_back();
            }
            stk.push_back(i);
        }
        long long ans = 0, i = 0;
        for (int j : stk) {
            ans += (j - i) * nums[j];
            i = j;
        }
        return ans;
    }
};

Go

func maxScore(nums []int) (ans int64) {
	stk := []int{}
	for i, x := range nums {
		for len(stk) > 0 && nums[stk[len(stk)-1]] <= x {
			stk = stk[:len(stk)-1]
		}
		stk = append(stk, i)
	}
	i := 0
	for _, j := range stk {
		ans += int64((j - i) * nums[j])
		i = j
	}
	return
}

TypeScript

function maxScore(nums: number[]): number {
    const stk: number[] = [];
    for (let i = 0; i < nums.length; ++i) {
        while (stk.length && nums[stk.at(-1)!] <= nums[i]) {
            stk.pop();
        }
        stk.push(i);
    }
    let ans = 0;
    let i = 0;
    for (const j of stk) {
        ans += (j - i) * nums[j];
        i = j;
    }
    return ans;
}