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Reach End of Array With Max Score

#1812 Most AskedMedium

Reach End of Array With Max Score #1812 Most Asked

Medium

Google

11 views

Topics:

ArraysDynamic Programming

You are given an integer array nums of length n.

Your goal is to start at index 0 and reach index n - 1. You can only jump to indices greater than your current index.

The score for a jump from index i to index j is calculated as (j - i) * nums[i].

Return the maximum possible total score by the time you reach the last index.

Example 1:

Input: nums = [1,3,1,5]

Output: 7

Explanation:

First, jump to index 1 and then jump to the last index. The final score is 1 * 1 + 2 * 3 = 7.

Example 2:

Input: nums = [4,3,1,3,2]

Output: 16

Explanation:

Jump directly to the last index. The final score is 4 * 4 = 16.

Constraints:

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

Try coding it in LeetCode

Reach End of Array With Max Score

#1812 Most AskedMedium

Reach End of Array With Max Score #1812 Most Asked

Medium

Google

11 views

Topics:

ArraysDynamic Programming

You are given an integer array nums of length n.

Your goal is to start at index 0 and reach index n - 1. You can only jump to indices greater than your current index.

The score for a jump from index i to index j is calculated as (j - i) * nums[i].

Return the maximum possible total score by the time you reach the last index.

Example 1:

Input: nums = [1,3,1,5]

Output: 7

Explanation:

First, jump to index 1 and then jump to the last index. The final score is 1 * 1 + 2 * 3 = 7.

Example 2:

Input: nums = [4,3,1,3,2]

Output: 16

Explanation:

Jump directly to the last index. The final score is 4 * 4 = 16.

Constraints:

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

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 values within the array? Can they be negative, zero, or have a specific range?
What should be returned if it is impossible to reach the end of the array? Should I return a specific value like -1, or throw an exception?
Are there any constraints on the starting index? Is it always index 0, or could it be a different index?
Could the array be empty or null? If so, what is the expected behavior?
By 'max score', do you mean the maximum sum of the values of the indices we land on as we traverse the array?

Brute Force Solution

Approach

The brute force method is like trying every single path through a maze. We explore every possible jump we can make and see if it gets us to the end, remembering the best score we've found so far.

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

Start at the beginning of the array.
From your current spot, consider all the possible jump lengths you are allowed to make.
For each possible jump, calculate what your score would be if you made that jump.
Imagine making that jump, and from your new spot, again consider all the possible jumps you can make.
Keep doing this until you either reach the end of the array or you can't make any more jumps.
If you reach the end of the array, remember the total score you got for that particular path.
Go back and try a different initial jump at the beginning, and repeat the process to the end, also saving the new score if the end is reachable.
Do this for every possible combination of jumps until you've explored every single path from the beginning.
Finally, compare all the total scores you saved from the paths that reached the end of the array, and choose the path with the highest score.

Code Implementation

def reach_end_with_max_score_brute_force(numbers, max_jump):
    max_score_found = float('-inf')

    def explore_paths(current_index, current_score):
        nonlocal max_score_found

        # If we've reached the end, update the max score.
        if current_index == len(numbers) - 1:
            max_score_found = max(max_score_found, current_score + numbers[current_index])
            return

        # If we've gone past the end, this path is invalid.
        if current_index >= len(numbers):
            return

        # Iterate through all possible jump lengths.
        for jump_length in range(1, max_jump + 1):
            next_index = current_index + jump_length

            # Explore the path with the new index and score.
            explore_paths(next_index, current_score + numbers[current_index])

    # Start exploring paths from the beginning of the array.
    explore_paths(0, 0)

    # Return the maximum score found, handling the case where the end is unreachable.
    if max_score_found == float('-inf'):
        return -1
    else:
        return max_score_found

Big(O) Analysis

Time Complexity

O(k^n) – The brute force approach explores every possible jump sequence. In the worst case, from each position, we can make up to k jumps (where k is the maximum jump allowed). If the array has n elements, the number of possible paths is roughly k * k * ... * k (n times), which equals k^n. Therefore, the time complexity is O(k^n), where k is the maximum jump length, and n is the length of the array.

Space Complexity

O(N) – The brute force approach uses recursion to explore all possible paths. In the worst-case scenario, the recursion depth can go up to N, where N is the length of the array, as we might make a jump of size 1 at each step. Each recursive call consumes memory on the call stack. Therefore, the auxiliary space used by the recursion stack is proportional to N, resulting in O(N) space complexity.

Optimal Solution

Approach

The best way to solve this is to work backward from the end, figuring out the best score you can get at each position. Instead of exploring every path, we focus on remembering the best result for each step along the way to avoid repeating work.

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

Start at the very end of the list of numbers. The best score you can get at the very end is just that last number itself.
Now, move one spot to the left. What's the best score you can get from here? It's the number you're on, plus the best score you can get from jumping ahead a certain number of spots (as allowed by the problem).
Keep track of the best score you can achieve from each spot as you move leftward. If you've already calculated the best score for a spot, don't calculate it again; just use the number you saved earlier.
When you get to the beginning of the list, the best score you've calculated for that starting spot is the overall best score you can achieve to reach the end.
Essentially, you are only doing the work once for each position, and looking up values when needed, to arrive at the end result.

Code Implementation

def reach_end_with_max_score(numbers):
    array_length = len(numbers)
    best_scores = [float('-inf')] * array_length

    # The last element's best score is itself.
    best_scores[array_length - 1] = numbers[array_length - 1]

    for i in range(array_length - 2, -1, -1):
        # Find max reachable score from current pos.
        for jump_length in range(1, min(array_length - i, 7)):
            next_index = i + jump_length
            best_scores[i] = max(best_scores[i],
                                numbers[i] + best_scores[next_index])

        # Handles situation where you cannot reach end
        if best_scores[i] == float('-inf'):
            best_scores[i] = float('-inf')

    # Return the best score starting from index 0
    return best_scores[0]

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates backward through the array of size n. For each index, it calculates the best score by considering possible jumps, as defined in the problem (e.g. based on the value at that index). Crucially, the best score for each reachable position is already calculated and stored, so it's a constant-time lookup. Thus, the algorithm performs a constant amount of work for each of the n positions, leading to O(n) time complexity.

Space Complexity

O(N) – The solution described uses a data structure to store the best score achievable from each position in the input list. This implies creating an array or list to store these intermediate best scores. Since we calculate and store a best score for each of the N positions in the input list, the auxiliary space required grows linearly with the input size N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 or throw an IllegalArgumentException, depending on problem specifics.
Array with a single element	If the problem states we need to traverse at least one step return -infinity, or, if able to start and end at the same spot, return the value of the starting point in the array.
Array with all elements being 0	Handle to prevent infinite loops if 0 valued cells allow free movement, potentially returning the maximum reachable index.
Array with all negative numbers	The problem may not allow for going back from negative numbers, so handle accordingly to return the most optimum or indicate no possible solution.
Array with extremely large positive or negative numbers, potentially causing integer overflow	Use long data types to store intermediate calculations and the maximum score to prevent overflow.
No path exists to the end of the array	Return a default value representing failure (e.g., -infinity or null), or throw an exception.
Array contains cycles or loops, where a path leads back to a previously visited index	Use dynamic programming or memoization to avoid recomputing scores for the same index and detect cycles.
The problem is ambiguous on whether to maximize score or the number of steps to get to the end of array	Clarify the problem with the interviewer and modify the logic accordingly.

Null or empty input array

How to Handle:

Return 0 or throw an IllegalArgumentException, depending on problem specifics.

Array with a single element

How to Handle:

If the problem states we need to traverse at least one step return -infinity, or, if able to start and end at the same spot, return the value of the starting point in the array.

Array with all elements being 0

How to Handle:

Handle to prevent infinite loops if 0 valued cells allow free movement, potentially returning the maximum reachable index.

Array with all negative numbers

How to Handle:

The problem may not allow for going back from negative numbers, so handle accordingly to return the most optimum or indicate no possible solution.

Array with extremely large positive or negative numbers, potentially causing integer overflow

How to Handle:

Use long data types to store intermediate calculations and the maximum score to prevent overflow.

No path exists to the end of the array

How to Handle:

Return a default value representing failure (e.g., -infinity or null), or throw an exception.

Array contains cycles or loops, where a path leads back to a previously visited index

How to Handle:

Use dynamic programming or memoization to avoid recomputing scores for the same index and detect cycles.

The problem is ambiguous on whether to maximize score or the number of steps to get to the end of array

How to Handle:

Clarify the problem with the interviewer and modify the logic accordingly.

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 values within the array? Can they be negative, zero, or have a specific range?
What should be returned if it is impossible to reach the end of the array? Should I return a specific value like -1, or throw an exception?
Are there any constraints on the starting index? Is it always index 0, or could it be a different index?
Could the array be empty or null? If so, what is the expected behavior?
By 'max score', do you mean the maximum sum of the values of the indices we land on as we traverse the array?

Brute Force Solution

Approach

The brute force method is like trying every single path through a maze. We explore every possible jump we can make and see if it gets us to the end, remembering the best score we've found so far.

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

Start at the beginning of the array.
From your current spot, consider all the possible jump lengths you are allowed to make.
For each possible jump, calculate what your score would be if you made that jump.
Imagine making that jump, and from your new spot, again consider all the possible jumps you can make.
Keep doing this until you either reach the end of the array or you can't make any more jumps.
If you reach the end of the array, remember the total score you got for that particular path.
Go back and try a different initial jump at the beginning, and repeat the process to the end, also saving the new score if the end is reachable.
Do this for every possible combination of jumps until you've explored every single path from the beginning.
Finally, compare all the total scores you saved from the paths that reached the end of the array, and choose the path with the highest score.

Code Implementation

def reach_end_with_max_score_brute_force(numbers, max_jump):
    max_score_found = float('-inf')

    def explore_paths(current_index, current_score):
        nonlocal max_score_found

        # If we've reached the end, update the max score.
        if current_index == len(numbers) - 1:
            max_score_found = max(max_score_found, current_score + numbers[current_index])
            return

        # If we've gone past the end, this path is invalid.
        if current_index >= len(numbers):
            return

        # Iterate through all possible jump lengths.
        for jump_length in range(1, max_jump + 1):
            next_index = current_index + jump_length

            # Explore the path with the new index and score.
            explore_paths(next_index, current_score + numbers[current_index])

    # Start exploring paths from the beginning of the array.
    explore_paths(0, 0)

    # Return the maximum score found, handling the case where the end is unreachable.
    if max_score_found == float('-inf'):
        return -1
    else:
        return max_score_found

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start at the very end of the list of numbers. The best score you can get at the very end is just that last number itself.
Now, move one spot to the left. What's the best score you can get from here? It's the number you're on, plus the best score you can get from jumping ahead a certain number of spots (as allowed by the problem).
Keep track of the best score you can achieve from each spot as you move leftward. If you've already calculated the best score for a spot, don't calculate it again; just use the number you saved earlier.
When you get to the beginning of the list, the best score you've calculated for that starting spot is the overall best score you can achieve to reach the end.
Essentially, you are only doing the work once for each position, and looking up values when needed, to arrive at the end result.

Code Implementation

def reach_end_with_max_score(numbers):
    array_length = len(numbers)
    best_scores = [float('-inf')] * array_length

    # The last element's best score is itself.
    best_scores[array_length - 1] = numbers[array_length - 1]

    for i in range(array_length - 2, -1, -1):
        # Find max reachable score from current pos.
        for jump_length in range(1, min(array_length - i, 7)):
            next_index = i + jump_length
            best_scores[i] = max(best_scores[i],
                                numbers[i] + best_scores[next_index])

        # Handles situation where you cannot reach end
        if best_scores[i] == float('-inf'):
            best_scores[i] = float('-inf')

    # Return the best score starting from index 0
    return best_scores[0]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 or throw an IllegalArgumentException, depending on problem specifics.
Array with a single element	If the problem states we need to traverse at least one step return -infinity, or, if able to start and end at the same spot, return the value of the starting point in the array.
Array with all elements being 0	Handle to prevent infinite loops if 0 valued cells allow free movement, potentially returning the maximum reachable index.
Array with all negative numbers	The problem may not allow for going back from negative numbers, so handle accordingly to return the most optimum or indicate no possible solution.
Array with extremely large positive or negative numbers, potentially causing integer overflow	Use long data types to store intermediate calculations and the maximum score to prevent overflow.
No path exists to the end of the array	Return a default value representing failure (e.g., -infinity or null), or throw an exception.
Array contains cycles or loops, where a path leads back to a previously visited index	Use dynamic programming or memoization to avoid recomputing scores for the same index and detect cycles.
The problem is ambiguous on whether to maximize score or the number of steps to get to the end of array	Clarify the problem with the interviewer and modify the logic accordingly.

Null or empty input array

How to Handle:

Return 0 or throw an IllegalArgumentException, depending on problem specifics.

Array with a single element

How to Handle:

If the problem states we need to traverse at least one step return -infinity, or, if able to start and end at the same spot, return the value of the starting point in the array.

Array with all elements being 0

How to Handle:

Handle to prevent infinite loops if 0 valued cells allow free movement, potentially returning the maximum reachable index.

Array with all negative numbers

How to Handle:

The problem may not allow for going back from negative numbers, so handle accordingly to return the most optimum or indicate no possible solution.

Array with extremely large positive or negative numbers, potentially causing integer overflow

How to Handle:

Use long data types to store intermediate calculations and the maximum score to prevent overflow.

No path exists to the end of the array

How to Handle:

Return a default value representing failure (e.g., -infinity or null), or throw an exception.

Array contains cycles or loops, where a path leads back to a previously visited index

How to Handle:

Use dynamic programming or memoization to avoid recomputing scores for the same index and detect cycles.

The problem is ambiguous on whether to maximize score or the number of steps to get to the end of array

How to Handle:

Clarify the problem with the interviewer and modify the logic accordingly.

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