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Jump Game VI

Medium

Asked by:

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Topics:

ArraysDynamic ProgrammingGreedy Algorithms

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

Try coding it in LeetCode

Jump Game VI

Medium

Asked by:

27 views

Topics:

ArraysDynamic ProgrammingGreedy Algorithms

You are given a 0-indexed integer array nums and an integer k.

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums` and the value of `k`?
Can the values within the `nums` array be negative, zero, or only positive?
If it's not possible to reach the end of the array (index n-1), what value should I return?
If there are multiple ways to reach the end with the same maximum score, is any one of them acceptable, or is there a specific path I should aim for?
Is `k` always guaranteed to be less than the length of the `nums` array?

Brute Force Solution

Approach

The goal is to reach the end, making the best score possible. The brute force method explores every path, jumping different lengths to see where we end up.

Here's how the algorithm would work step-by-step:

Start at the beginning.
From your current position, consider all possible jump lengths that are allowed.
For each possible jump, pretend you took that jump and calculate your new score.
Keep doing this from each new position until you reach the end.
Remember the score for each possible path to the end.
After exploring all paths, choose the path that gave you the highest score.

Code Implementation

def jump_game_vi_brute_force(numbers, max_jump):
    def find_max_score(current_index):
        # If we've reached the end, return the score.
        if current_index == len(numbers) - 1:
            return numbers[current_index]

        max_score = float('-inf')

        # Iterate through all possible jump lengths.
        for jump_length in range(1, min(max_jump + 1, len(numbers) - current_index)):

            next_index = current_index + jump_length

            # Recursively call find_max_score from the next position.
            score = numbers[current_index] + find_max_score(next_index)
            max_score = max(max_score, score)

        return max_score

    # Start the recursion from the first index
    return find_max_score(0)

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach described involves exploring all possible jump combinations. From each position, we consider up to k possible jumps. Since we can make jumps from n positions in the array, and for each of those n positions we are branching up to k times, and we are making up to n jumps to reach the end, this can lead to k multiplied by itself up to n times. Hence, the time complexity is O(n) multiplied by O(k^n), or O(n * k^n). Due to the exponential portion dominating the complexity in brute force, we can approximate this to O(k^n). Since k is a constant and not dependent on n, we can say the time complexity is O(n^k) because the depth of the recursion can be at most n.

Space Complexity

O(N^K) – The brute force approach explores all possible paths to the end. In the worst case, from each position, we consider up to K jump lengths. This leads to a branching factor of K at each level of the recursion. Since the maximum depth of the recursion is N (the length of the input array), the total number of paths explored can be approximated as K^N. Storing the score for each path to the end requires an auxiliary data structure of size K^N, where K is the maximum jump length, and N is the number of elements in the input array. Furthermore, the recursion depth can reach N, contributing O(N) to the space complexity, but the path storage dominates. Therefore, the space complexity is primarily determined by the storage of scores for each path, which is O(N^K) in the absolute worst-case scenario where K approaches N and we store scores for every possible path.

Optimal Solution

Approach

The best approach isn't about trying every possible jump. It's about smartly keeping track of the best possible score we can reach at each position, and using that to efficiently find the best final score. We maintain a 'window' of the best previous scores, allowing us to quickly decide the next best jump.

Here's how the algorithm would work step-by-step:

Imagine you're moving along a game board and want to find the biggest score you can get. Start at the beginning.
As you move, always remember the biggest score you could possibly have at each spot you visited.
Keep a sliding 'window' that shows the positions within your jumping range that gave you the best scores recently.
For each position, check what the best score was within that window, and add that to the current position's value to determine your best score at the current spot.
Slide the window along. Remove old, irrelevant scores (positions outside of your jumping range) and add new, promising scores (positions you can reach).
Always ensure the window helps you quickly find the best score from the immediately reachable positions.
By the time you reach the end, you will know the highest score achievable if you always made optimal moves from the beginning.

Code Implementation

from collections import deque

def jump_game_vi(numbers, max_jump):
    score = [0] * len(numbers)
    score[0] = numbers[0]
    window = deque([0])

    for i in range(1, len(numbers)):
        # Keep the queue within the max jump range
        while window and window[0] < i - max_jump:
            window.popleft()

        # The best score at the current position.
        score[i] = score[window[0]] + numbers[i]

        # Maintain the window's relevance by removing
        # worse scores that are further behind.
        while window and score[window[-1]] <= score[i]:
            window.pop()
        window.append(i)

    return score[-1]

Big(O) Analysis

Time Complexity

O(n) – We iterate through the array of n elements once. Within the loop, we use a deque (double-ended queue) to maintain a sliding window of potential best previous scores. The operations on the deque (adding and removing elements) take constant time on average. Each element is added and removed from the deque at most once, ensuring that the total number of deque operations is proportional to n. Therefore, the dominant operation is the single pass through the array, resulting in O(n) time complexity.

Space Complexity

O(k) – The algorithm maintains a sliding window of best previous scores using a data structure like a deque or priority queue. In the worst case, this window could contain up to k elements, where k is the maximum jump distance. Therefore, the auxiliary space required is proportional to k. The space used for the window does not depend on the number of elements in the input array beyond the maximum jump distance. This leads to an auxiliary space complexity of O(k).

Edge Cases

Case	How to Handle
Empty input array	Return 0, as no jumps are possible and the problem statement implies the array will always be non-empty.
Array with a single element	Return the value of the single element, as no jumps are needed.
k = 0	If k is 0, we can only jump to the current position; therefore, the result is simply the first element if the array's length is 1.
All numbers are negative	The algorithm should correctly select the maximum score within the allowed jump range, even if all values are negative.
The first element is very large, and other elements are small or negative	Ensure integer overflow does not occur when calculating the cumulative score.
k is greater than or equal to the length of the array	Handle this case by treating k as array.length - 1, as we can jump to the end in a single step.
Large input array with large k	The solution needs to scale efficiently, avoiding naive O(n*k) approaches that would lead to timeouts; prioritize a heap or deque-based solution.
Array with only zeros after the first element.	The algorithm should still find the path to the end from the first element even if subsequent jumps yield no gain in score.

Empty input array

How to Handle:

Return 0, as no jumps are possible and the problem statement implies the array will always be non-empty.

Array with a single element

How to Handle:

Return the value of the single element, as no jumps are needed.

k = 0

How to Handle:

If k is 0, we can only jump to the current position; therefore, the result is simply the first element if the array's length is 1.

All numbers are negative

How to Handle:

The algorithm should correctly select the maximum score within the allowed jump range, even if all values are negative.

The first element is very large, and other elements are small or negative

How to Handle:

Ensure integer overflow does not occur when calculating the cumulative score.

k is greater than or equal to the length of the array

How to Handle:

Handle this case by treating k as array.length - 1, as we can jump to the end in a single step.

Large input array with large k

How to Handle:

The solution needs to scale efficiently, avoiding naive O(n*k) approaches that would lead to timeouts; prioritize a heap or deque-based solution.

Array with only zeros after the first element.

How to Handle:

The algorithm should still find the path to the end from the first element even if subsequent jumps yield no gain in score.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums` and the value of `k`?
Can the values within the `nums` array be negative, zero, or only positive?
If it's not possible to reach the end of the array (index n-1), what value should I return?
If there are multiple ways to reach the end with the same maximum score, is any one of them acceptable, or is there a specific path I should aim for?
Is `k` always guaranteed to be less than the length of the `nums` array?

Brute Force Solution

Approach

The goal is to reach the end, making the best score possible. The brute force method explores every path, jumping different lengths to see where we end up.

Here's how the algorithm would work step-by-step:

Start at the beginning.
From your current position, consider all possible jump lengths that are allowed.
For each possible jump, pretend you took that jump and calculate your new score.
Keep doing this from each new position until you reach the end.
Remember the score for each possible path to the end.
After exploring all paths, choose the path that gave you the highest score.

Code Implementation

def jump_game_vi_brute_force(numbers, max_jump):
    def find_max_score(current_index):
        # If we've reached the end, return the score.
        if current_index == len(numbers) - 1:
            return numbers[current_index]

        max_score = float('-inf')

        # Iterate through all possible jump lengths.
        for jump_length in range(1, min(max_jump + 1, len(numbers) - current_index)):

            next_index = current_index + jump_length

            # Recursively call find_max_score from the next position.
            score = numbers[current_index] + find_max_score(next_index)
            max_score = max(max_score, score)

        return max_score

    # Start the recursion from the first index
    return find_max_score(0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine you're moving along a game board and want to find the biggest score you can get. Start at the beginning.
As you move, always remember the biggest score you could possibly have at each spot you visited.
Keep a sliding 'window' that shows the positions within your jumping range that gave you the best scores recently.
For each position, check what the best score was within that window, and add that to the current position's value to determine your best score at the current spot.
Slide the window along. Remove old, irrelevant scores (positions outside of your jumping range) and add new, promising scores (positions you can reach).
Always ensure the window helps you quickly find the best score from the immediately reachable positions.
By the time you reach the end, you will know the highest score achievable if you always made optimal moves from the beginning.

Code Implementation

from collections import deque

def jump_game_vi(numbers, max_jump):
    score = [0] * len(numbers)
    score[0] = numbers[0]
    window = deque([0])

    for i in range(1, len(numbers)):
        # Keep the queue within the max jump range
        while window and window[0] < i - max_jump:
            window.popleft()

        # The best score at the current position.
        score[i] = score[window[0]] + numbers[i]

        # Maintain the window's relevance by removing
        # worse scores that are further behind.
        while window and score[window[-1]] <= score[i]:
            window.pop()
        window.append(i)

    return score[-1]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0, as no jumps are possible and the problem statement implies the array will always be non-empty.
Array with a single element	Return the value of the single element, as no jumps are needed.
k = 0	If k is 0, we can only jump to the current position; therefore, the result is simply the first element if the array's length is 1.
All numbers are negative	The algorithm should correctly select the maximum score within the allowed jump range, even if all values are negative.
The first element is very large, and other elements are small or negative	Ensure integer overflow does not occur when calculating the cumulative score.
k is greater than or equal to the length of the array	Handle this case by treating k as array.length - 1, as we can jump to the end in a single step.
Large input array with large k	The solution needs to scale efficiently, avoiding naive O(n*k) approaches that would lead to timeouts; prioritize a heap or deque-based solution.
Array with only zeros after the first element.	The algorithm should still find the path to the end from the first element even if subsequent jumps yield no gain in score.

Empty input array

How to Handle:

Return 0, as no jumps are possible and the problem statement implies the array will always be non-empty.

Array with a single element

How to Handle:

Return the value of the single element, as no jumps are needed.

k = 0

How to Handle:

If k is 0, we can only jump to the current position; therefore, the result is simply the first element if the array's length is 1.

All numbers are negative

How to Handle:

The algorithm should correctly select the maximum score within the allowed jump range, even if all values are negative.

The first element is very large, and other elements are small or negative

How to Handle:

Ensure integer overflow does not occur when calculating the cumulative score.

k is greater than or equal to the length of the array

How to Handle:

Handle this case by treating k as array.length - 1, as we can jump to the end in a single step.

Large input array with large k

How to Handle:

The solution needs to scale efficiently, avoiding naive O(n*k) approaches that would lead to timeouts; prioritize a heap or deque-based solution.

Array with only zeros after the first element.

How to Handle:

The algorithm should still find the path to the end from the first element even if subsequent jumps yield no gain in score.