How do you solve Maximum Score of a Good Subarray on LeetCode?

Maximum Score of a Good Subarray (LeetCode #1793) 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: The minimum value in a subarray heavily influences the score.

What is the brute force solution for Maximum Score of a Good Subarray?

The brute force approach for Maximum Score of a Good Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray that includes the index k. For each subarray, we calculate the minimum value and the score, keeping track of the maximum score found. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Score of a Good Subarray?

Maximum Score of a Good Subarray uses the following concepts: Array, Two Pointers, Binary Search, Stack, Monotonic Stack. The recommended patterns to study are: Monotonic Stack, Two Pointers, Sliding Window.

What are the interview tips for Maximum Score of a Good Subarray?

Always clarify the problem constraints and edge cases with the interviewer. Explain your thought process clearly while coding. Consider both time and space complexity when discussing your solution.

What are common mistakes when solving Maximum Score of a Good Subarray?

Not considering edge cases where k is at the start or end of the array. Failing to correctly maintain the minimum value when expanding the subarray.

#1793

Maximum Score of a Good Subarray

Hard

ArrayTwo PointersBinary SearchStackMonotonic StackMonotonic StackTwo PointersSliding Window

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 monotonic stack to efficiently find the maximum score by determining the range of subarrays that can include the index k. This approach reduces the time complexity significantly by avoiding the need to check every subarray explicitly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a stack to keep track of indices and a variable for the maximum score.
2Step 2: Iterate through the array to find the left and right bounds for each element using the stack.
3Step 3: For each element, calculate the score using its minimum value and the length of the valid subarray that includes k.
4Step 4: Update the maximum score if the current score is higher.

solution.py34 lines

1# Full working Python code
2
3def maxScore(nums, k):
4    n = len(nums)
5    left = [0] * n
6    right = [0] * n
7    stack = []
8
9    # Calculate left bounds
10    for i in range(n):
11        while stack and nums[stack[-1]] >= nums[i]:
12            stack.pop()
13        left[i] = stack[-1] if stack else -1
14        stack.append(i)
15
16    stack.clear()
17
18    # Calculate right bounds
19    for i in range(n - 1, -1, -1):
20        while stack and nums[stack[-1]] > nums[i]:
21            stack.pop()
22        right[i] = stack[-1] if stack else n
23        stack.append(i)
24
25    max_score = 0
26
27    # Calculate scores
28    for i in range(n):
29        if left[i] < k < right[i]:
30            length = right[i] - left[i] - 1
31            score = nums[i] * length
32            max_score = max(max_score, score)
33
34    return max_score

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times (once for left bounds and once for right bounds). The space complexity is O(n) due to the additional arrays for left and right bounds.

1The minimum value in a subarray heavily influences the score.
2Using a monotonic stack allows us to efficiently find valid subarray bounds.

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