Maximize Win From Two Segments

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ArraysBinary SearchDynamic ProgrammingGreedy Algorithms

There are some prizes on the X-axis. You are given an integer array prizePositions that is sorted in non-decreasing order, where prizePositions[i] is the position of the i^th prize. There could be different prizes at the same position on the line. You are also given an integer k.

You are allowed to select two segments with integer endpoints. The length of each segment must be k. You will collect all prizes whose position falls within at least one of the two selected segments (including the endpoints of the segments). The two selected segments may intersect.

For example if k = 2, you can choose segments [1, 3] and [2, 4], and you will win any prize i that satisfies 1 <= prizePositions[i] <= 3 or 2 <= prizePositions[i] <= 4.

Return the maximum number of prizes you can win if you choose the two segments optimally.

Example 1:

Input: prizePositions = [1,1,2,2,3,3,5], k = 2
Output: 7
Explanation: In this example, you can win all 7 prizes by selecting two segments [1, 3] and [3, 5].

Example 2:

Input: prizePositions = [1,2,3,4], k = 0
Output: 2
Explanation: For this example, one choice for the segments is [3, 3] and [4, 4], and you will be able to get 2 prizes.

Constraints:

1 <= prizePositions.length <= 10⁵
1 <= prizePositions[i] <= 10⁹
0 <= k <= 10⁹
prizePositions is sorted in non-decreasing order.

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 allowed ranges and data types for the input array elements? Can they be negative or zero?
What should I return if it's impossible to select two segments that cover all unique numbers in the input array?
Are the segments required to be non-overlapping, adjacent, or can they overlap freely?
What constitutes a 'win'? Is it maximizing the sum of the values within the two segments or something else?
What are the constraints on the segment sizes? Is there a minimum or maximum length for each segment?

Brute Force Solution

Approach

The brute force approach for this problem involves examining all possible segment combinations. We essentially try out every possible way to pick our two segments from the given data. Then, from all of those segment combinations, we keep the segments that give us the highest total win.

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

Imagine drawing a line to mark where the first segment begins and ends; try every single possibility for where this line could be drawn.
For each of those possibilities for the first segment, draw another line to mark where the second segment begins and ends, again trying every single possible location for the second line.
Calculate the 'win' you would get from each of these pairs of segments.
Compare all the 'win' values you calculated, and keep track of the largest one you've seen so far.
After trying every possible pair of segments, the largest 'win' you kept track of is the answer.

Code Implementation

def maximize_win_from_two_segments_brute_force(
    input_array
):    max_win = 0
    array_length = len(input_array)

    # Iterate to define the first segment
    for first_segment_start in range(array_length):        for first_segment_end in range(
            first_segment_start, array_length
        ):
            first_segment_win = sum(
                input_array[
                    first_segment_start : first_segment_end + 1
                ]
            )

            # Iterate to define the second segment
            for second_segment_start in range(array_length):
                for second_segment_end in range(
                    second_segment_start, array_length
                ):

                    # Calculate the win for the second segment
                    second_segment_win = sum(
                        input_array[
                            second_segment_start : second_segment_end
                            + 1
                        ]
                    )

                    # Calculate the total win and update max if needed
                    total_win = (
                        first_segment_win + second_segment_win
                    )
                    max_win = max(max_win, total_win)

    return max_win

Big(O) Analysis

Time Complexity

O(n^4) – The described brute force approach involves iterating through all possible segment combinations. The outer loops define the start and end points of the first segment, leading to approximately n^2 possibilities. For each of these, we iterate through all possible start and end points for the second segment, resulting in another n^2 possibilities. Thus, the total number of operations is proportional to n^2 * n^2, which simplifies to O(n^4).

Space Complexity

O(1) – The provided brute force approach only uses a few variables to keep track of the start and end indices of the two segments, as well as the maximum win found so far. The number of these variables does not depend on the input size N (where N represents the size of the input data from which segments are selected). Therefore, the auxiliary space used is constant, regardless of the input size.

Optimal Solution

Approach

The problem involves selecting two segments from a larger range to maximize the combined win value. The clever approach involves efficiently identifying the best segments without examining every possible pair. This involves pre-calculating information to make smarter choices when choosing the segments.

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

First, calculate the win value for every possible segment of the specified length for each segment.
Store these values so you can quickly look them up later. This is like creating a lookup table.
Then, iterate through the range and consider each segment as the first segment you'll use.
For each first segment, determine where the second segment can start without overlapping. Remember the second segment also has a specified length.
Look up the win value of the best possible second segment that can be formed, utilizing the pre-calculated values from the lookup table. Choose the second segment such that it has the maximum win value possible.
Calculate the combined win value of the first segment and the best non-overlapping second segment.
Keep track of the maximum combined win value you've found so far, and update it whenever you find a better combination.
Finally, after checking all possible first segments, the maximum combined win value you've recorded is the answer.

Code Implementation

def maximize_win_from_two_segments(win_values, first_segment_length, second_segment_length):
    number_of_days = len(win_values)
    segment_wins = {}

    # Calculate and store win values for all possible segments of both lengths.
    for i in range(number_of_days - first_segment_length + 1):
        segment_wins[(i, first_segment_length)] = sum(win_values[i:i + first_segment_length])
    for i in range(number_of_days - second_segment_length + 1):
        segment_wins[(i, second_segment_length)] = sum(win_values[i:i + second_segment_length])

    maximum_combined_win = 0
    for first_segment_start in range(number_of_days - first_segment_length + 1):

        # Consider each possible first segment.
        first_segment_win = segment_wins[(first_segment_start, first_segment_length)]
        best_second_segment_win = 0

        # Find the best non-overlapping second segment.
        for second_segment_start in range(number_of_days - second_segment_length + 1):
            if second_segment_start >= first_segment_start + first_segment_length or \
               first_segment_start >= second_segment_start + second_segment_length:

                second_segment_win = segment_wins[(second_segment_start, second_segment_length)]
                best_second_segment_win = max(best_second_segment_win, second_segment_win)

        # Update the overall maximum win.
        maximum_combined_win = max(maximum_combined_win, first_segment_win + best_second_segment_win)

    return maximum_combined_win

Big(O) Analysis

Time Complexity

O(n) – Calculating the win value for every possible segment of length `firstLen` and `secondLen` involves iterating through the array of size `n` once for each length, which takes O(n) time. Subsequently, iterating through the range to consider each segment as the first segment also takes O(n) time. Finding the best non-overlapping second segment involves a single pass to find the maximum. Thus, the overall time complexity is dominated by the single loop that iterates through the input array, making the solution O(n).

Space Complexity

O(N) – The algorithm pre-calculates and stores the win value for every possible segment. Since there can be up to N possible segments (where N is the size of the range), this creates a lookup table that could store up to N win values. Therefore, the auxiliary space required to store these pre-calculated win values is proportional to N. This dominates the space complexity.

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as no segments are possible.
Array with a single element	Return 0 immediately as two segments are not possible.
All elements are zero	The maximum win will be 0, which the algorithm should naturally return.
Array contains negative numbers	The algorithm should correctly handle negative sums as it seeks to maximize the sum of two segments.
Array contains very large positive numbers (potential overflow)	Use long long or appropriate data type to store segment sums to avoid integer overflow.
Two segments overlap	Ensure the implementation explicitly prohibits overlapping segments by enforcing `j < i` in segment sum calculation or similar logic.
No combination of segments yields a positive result	Return 0, indicating that the maximum possible win is zero.
Array is sorted in descending order	The algorithm should still correctly find the optimal non-overlapping segments, potentially picking the first few elements.

Empty input array

How to Handle:

Return 0 immediately as no segments are possible.

Array with a single element

How to Handle:

Return 0 immediately as two segments are not possible.

All elements are zero

How to Handle:

The maximum win will be 0, which the algorithm should naturally return.

Array contains negative numbers

How to Handle:

The algorithm should correctly handle negative sums as it seeks to maximize the sum of two segments.

Array contains very large positive numbers (potential overflow)

How to Handle:

Use long long or appropriate data type to store segment sums to avoid integer overflow.

Two segments overlap

How to Handle:

Ensure the implementation explicitly prohibits overlapping segments by enforcing `j < i` in segment sum calculation or similar logic.

No combination of segments yields a positive result

How to Handle:

Return 0, indicating that the maximum possible win is zero.

Array is sorted in descending order

How to Handle:

The algorithm should still correctly find the optimal non-overlapping segments, potentially picking the first few elements.