How do you solve Maximize Win From Two Segments on LeetCode?

Maximize Win From Two Segments (LeetCode #2555) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using two segments can overlap, maximizing prize collection.

What is the time complexity of Maximize Win From Two Segments?

The optimal solution for Maximize Win From Two Segments runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the list a limited number of times, making it efficient for large inputs.

What is the brute force solution for Maximize Win From Two Segments?

The brute force approach for Maximize Win From Two Segments is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of segments of length k and counting the total prizes collected. This is straightforward but inefficient, as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximize Win From Two Segments?

Maximize Win From Two Segments uses the following concepts: Array, Binary Search, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Maximize Win From Two Segments?

Explain your thought process clearly. Discuss edge cases, such as k=0. Optimize your solution step-by-step.

What are common mistakes when solving Maximize Win From Two Segments?

Not considering overlapping segments. Inefficient counting of prizes for each segment.

#2555

Maximize Win From Two Segments

Medium

ArrayBinary SearchSliding WindowSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window technique to efficiently calculate the number of prizes covered by each segment and combines the results for two segments. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Create an array to store the number of prizes covered by one segment starting at each position.
2Step 2: Use a sliding window to calculate the number of prizes for the first segment of length k.
3Step 3: For each possible starting point of the second segment, calculate the total prizes covered by combining the results from the first segment.

solution.py19 lines

1def maxPrizes(prizePositions, k):
2    n = len(prizePositions)
3    count = [0] * n
4    j = 0
5    for i in range(n):
6        while j < n and prizePositions[j] <= prizePositions[i] + k:
7            j += 1
8        count[i] = j - i
9    max_prizes = 0
10    for i in range(n):
11        second_start = prizePositions[i]
12        second_end = second_start + k
13        total = count[i]
14        j = i
15        while j < n and prizePositions[j] <= second_end:
16            total += count[j]
17            j += 1
18        max_prizes = max(max_prizes, total)
19    return max_prizes

ℹ

Complexity note: The time complexity is O(n) because we only traverse the list a limited number of times, making it efficient for large inputs.

1Using two segments can overlap, maximizing prize collection.
2Sliding window technique is efficient for range queries.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.