Problem Description

You have prizes positioned along the X-axis. You're given an integer array prizePositions that is sorted in non-decreasing order, where prizePositions[i] represents the position of the i-th prize. Multiple prizes can exist at the same position.

You're also given an integer k which represents the length of segments you can select.

Your task is to select two segments, each of length k, to collect prizes. A prize is collected if its position falls within at least one of your selected segments (including the segment endpoints). The two segments you choose can overlap.

For example, if k = 2, you could choose segments [1, 3] and [2, 4]. This would collect any prize at positions between 1 and 3 (inclusive) OR between 2 and 4 (inclusive).

The goal is to find the maximum number of prizes you can collect by optimally choosing your two segments.

Key Points:

• Each segment must have length exactly k
• Segments can have any integer endpoints
• Segments can overlap
• A prize is collected if it falls within at least one segment
• You want to maximize the total number of prizes collected

Intuition

The key insight is that we need to choose two segments optimally. Let's think about this problem step by step.

First, since the prizes are already sorted by position, any segment of length k starting at position p will collect all prizes between positions p and p + k (inclusive). This means we can think of each segment as a contiguous subarray of prizes.

Now, for choosing two segments optimally, we can consider each possible position as the end of the second segment. For each ending position, we need to:

1. Determine how many prizes the second segment collects
2. Find the best possible first segment that doesn't interfere with maximizing the second segment

This leads us to a dynamic programming approach. We can precompute f[i] which represents the maximum number of prizes we can collect using one segment from the first i prizes. This helps us avoid recalculating the best first segment repeatedly.

When we're at position i (considering it as part of the second segment), we can:

• Use binary search to find the leftmost position j where prizePositions[j] >= prizePositions[i] - k. This tells us the starting point of a segment of length k that ends at or after position i.
• The second segment collects i - j prizes (from index j to i-1)
• The first segment should be chosen optimally from prizes before index j, which is exactly f[j]
• Total prizes = f[j] + (i - j)

By iterating through all positions and tracking the maximum total, we find the optimal solution. We also update f[i] at each step to maintain our precomputed values for future positions.

The beauty of this approach is that it avoids overlap issues naturally - by choosing the first segment from prizes before index j, we ensure it doesn't interfere with counting prizes in the second segment.

Learn more about Binary Search and Sliding Window patterns.

Solution Approach

We implement the solution using dynamic programming combined with binary search.

Step 1: Initialize Data Structures

• Create an array f of size n + 1 where f[i] stores the maximum number of prizes that can be collected using one segment from the first i prizes
• Initialize f[0] = 0 since no prizes can be collected from zero prizes
• Initialize ans = 0 to track the maximum total prizes we can collect with two segments

Step 2: Iterate Through Each Prize Position We enumerate through each prize position using 1-based indexing (for cleaner DP transitions):

for i, x in enumerate(prizePositions, 1):

Step 3: Find the Left Boundary Using Binary Search For each prize at position x, we need to find the leftmost prize that would be included in a segment of length k ending at or after position x:

j = bisect_left(prizePositions, x - k)

This finds the smallest index j where prizePositions[j] >= x - k. The segment [x - k, x] would collect all prizes from index j to i - 1.

Step 4: Calculate Maximum Prizes

• The second segment (ending at current position) collects i - j prizes
• The first segment should be chosen optimally from the first j prizes, which gives us f[j] prizes
• Update the answer: ans = max(ans, f[j] + i - j)

Step 5: Update DP Array We update f[i] for future iterations:

f[i] = max(f[i - 1], i - j)

This means the maximum prizes we can get from the first i prizes is either:

• The maximum from the first i - 1 prizes (not including current prize in our segment)
• Or i - j prizes (using a segment that includes the current prize)

Step 6: Return Result After processing all positions, ans contains the maximum number of prizes we can collect with two segments.

Time Complexity: O(n log n) where n is the number of prizes. We iterate through each prize once and perform a binary search for each.

Space Complexity: O(n) for the DP array f.

Example Walkthrough

Let's walk through the solution with a concrete example:

• prizePositions = [1, 1, 2, 2, 3, 3, 5]
• k = 2

We want to find two segments of length 2 that collect the maximum number of prizes.

Initialization:

• n = 7 (number of prizes)
• f = [0, 0, 0, 0, 0, 0, 0, 0] (size n+1)
• ans = 0

Iteration 1: i=1, x=1

• Binary search: Find leftmost position where prize ≥ (1-2) = -1
• j = 0 (all prizes are ≥ -1)
• Second segment [−1, 1] collects prizes at indices 0 to 0: i - j = 1 - 0 = 1 prize
• First segment from first j=0 prizes: f[0] = 0 prizes
• Total: 0 + 1 = 1
• Update: ans = max(0, 1) = 1
• Update: f[1] = max(f[0], 1 - 0) = max(0, 1) = 1

Iteration 2: i=2, x=1

• Binary search: Find leftmost position where prize ≥ (1-2) = -1
• j = 0
• Second segment [−1, 1] collects prizes at indices 0 to 1: i - j = 2 - 0 = 2 prizes
• First segment: f[0] = 0 prizes
• Total: 0 + 2 = 2
• Update: ans = max(1, 2) = 2
• Update: f[2] = max(f[1], 2 - 0) = max(1, 2) = 2

Iteration 3: i=3, x=2

• Binary search: Find leftmost position where prize ≥ (2-2) = 0
• j = 0 (prize at index 0 is 1 ≥ 0)
• Second segment [0, 2] collects prizes at indices 0 to 2: i - j = 3 - 0 = 3 prizes
• First segment: f[0] = 0 prizes
• Total: 0 + 3 = 3
• Update: ans = max(2, 3) = 3
• Update: f[3] = max(f[2], 3 - 0) = max(2, 3) = 3

Iteration 4: i=4, x=2

• Binary search: Find leftmost position where prize ≥ (2-2) = 0
• j = 0
• Second segment [0, 2] collects prizes at indices 0 to 3: i - j = 4 - 0 = 4 prizes
• First segment: f[0] = 0 prizes
• Total: 0 + 4 = 4
• Update: ans = max(3, 4) = 4
• Update: f[4] = max(f[3], 4 - 0) = max(3, 4) = 4

Iteration 5: i=5, x=3

• Binary search: Find leftmost position where prize ≥ (3-2) = 1
• j = 0 (prize at index 0 is 1 ≥ 1)
• Second segment [1, 3] collects prizes at indices 0 to 4: i - j = 5 - 0 = 5 prizes
• First segment: f[0] = 0 prizes
• Total: 0 + 5 = 5
• Update: ans = max(4, 5) = 5
• Update: f[5] = max(f[4], 5 - 0) = max(4, 5) = 5

Iteration 6: i=6, x=3

• Binary search: Find leftmost position where prize ≥ (3-2) = 1
• j = 0
• Second segment [1, 3] collects prizes at indices 0 to 5: i - j = 6 - 0 = 6 prizes
• First segment: f[0] = 0 prizes
• Total: 0 + 6 = 6
• Update: ans = max(5, 6) = 6
• Update: f[6] = max(f[5], 6 - 0) = max(5, 6) = 6

Iteration 7: i=7, x=5

• Binary search: Find leftmost position where prize ≥ (5-2) = 3
• j = 4 (prize at index 4 is 3 ≥ 3)
• Second segment [3, 5] collects prizes at indices 4 to 6: i - j = 7 - 4 = 3 prizes
• First segment from first j=4 prizes: f[4] = 4 prizes
• Total: 4 + 3 = 7
• Update: ans = max(6, 7) = 7
• Update: f[7] = max(f[6], 7 - 4) = max(6, 3) = 6

Final Answer: 7

The optimal solution uses:

• First segment: [0, 2] collecting prizes at positions 1, 1, 2, 2 (4 prizes)
• Second segment: [3, 5] collecting prizes at positions 3, 3, 5 (3 prizes)
• Total: 7 prizes

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The algorithm iterates through all prize positions once with a for loop, which takes O(n) time. Within each iteration, it performs a binary search using bisect_left(prizePositions, x - k) to find the leftmost position where we can start a segment that ends at position x. Since prizePositions is sorted and has length n, each binary search operation takes O(log n) time. Therefore, the overall time complexity is O(n) × O(log n) = O(n × log n).

Space Complexity: O(n)

The algorithm uses an array f of size n + 1 to store the maximum number of prizes that can be collected up to each position. This requires O(n + 1) = O(n) additional space. The other variables (ans, i, j, x) use constant space O(1). Therefore, the total space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Segment Length

A critical pitfall is confusing the segment length k with the number of positions it covers. The segment has length k, meaning it spans from position x to position x + k inclusive. This is a continuous range of length k, not k discrete positions.

Incorrect interpretation: Thinking we need to select exactly k prizes. Correct interpretation: We select all prizes within a continuous interval of length k.

Example: If k = 2 and we have prizes at positions [1, 2, 3, 4]:

• Segment [1, 3] has length 2 and covers prizes at positions 1, 2, and 3
• This collects 3 prizes, not 2

2. Off-by-One Error in Binary Search

When finding the leftmost prize position for a segment ending at position x, developers often make mistakes with the boundary calculation:

Common mistake:

# Wrong: This finds prizes strictly less than x - k
j = bisect_left(prizePositions, x - k - 1)

Correct approach:

# Correct: Find the first position >= x - k
j = bisect_left(prizePositions, x - k)

The segment [x - k, x] includes both endpoints, so any prize at position x - k should be included.

3. Incorrect DP State Update

A subtle error occurs when updating the DP array. The value f[i] should represent the maximum prizes collectible from the first i prizes using ONE segment, not the cumulative sum.

Incorrect:

# Wrong: Accumulating all possible prizes
f[i] = f[i - 1] + (i - j)

Correct:

# Right: Taking maximum between previous best and current segment
f[i] = max(f[i - 1], i - j)

4. Not Handling Edge Cases

Failing to consider edge cases can lead to incorrect results:

• Empty array: When prizePositions is empty
• Single prize: When there's only one prize
• Large k value: When k is larger than the entire range of positions
• All prizes at same position: Multiple prizes at identical positions

Solution: The provided code handles these naturally through the DP approach, but explicit checks can improve clarity:

def maximizeWin(self, prizePositions: List[int], k: int) -> int:
    if not prizePositions:
        return 0
  
    n = len(prizePositions)
    if n <= 2:
        return n
  
    # If k covers the entire range, we can collect all prizes with one segment
    if k >= prizePositions[-1] - prizePositions[0]:
        return n
  
    # Continue with main algorithm...

5. Misunderstanding Segment Overlap

Some developers mistakenly try to ensure segments don't overlap, but the problem explicitly allows overlapping segments. This leads to unnecessarily complex solutions.

Key insight: Overlapping segments are beneficial when prizes are densely packed. The algorithm naturally handles this by considering all possible segment placements without explicitly checking for overlap.