How do you solve Sum of Beauty in the Array on LeetCode?

Sum of Beauty in the Array (LeetCode #2012) 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 prefix and suffix arrays can significantly reduce the time complexity.

What is the brute force solution for Sum of Beauty in the Array?

The brute force approach for Sum of Beauty in the Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each element in the array and compares it with all previous and subsequent elements to determine its beauty. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Beauty in the Array?

Sum of Beauty in the Array uses the following concepts: Array. The recommended patterns to study are: Prefix Sum, Suffix Array, Two Pointers.

What are the interview tips for Sum of Beauty in the Array?

Always clarify the problem requirements before jumping into coding. Discuss your thought process and potential optimizations with the interviewer. Practice explaining your code as you write it to simulate explaining it in an interview.

What are common mistakes when solving Sum of Beauty in the Array?

Not considering edge cases where the array length is minimal. Overlooking the need for prefix/suffix arrays for optimization.

#2012

Sum of Beauty in the Array

Medium

ArrayPrefix SumSuffix ArrayTwo 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)

By using prefix and suffix arrays, we can efficiently determine the maximum values before and after each index, allowing us to compute the beauty in linear time.

⚙️

Algorithm

6 steps

1Step 1: Create a prefix array to store the maximum value from the start to each index.
2Step 2: Create a suffix array to store the maximum value from each index to the end.
3Step 3: Loop through each index 'i' from 1 to nums.length - 2.
4Step 4: For each 'i', check if nums[i] is greater than the prefix[i-1] and less than the suffix[i+1] for beauty 2.
5Step 5: If the above condition is not satisfied, check if nums[i-1] < nums[i] < nums[i+1] for beauty 1.
6Step 6: Sum the beauties and return the total.

solution.py19 lines

1def sumOfBeauty(nums):
2    n = len(nums)
3    prefix = [0] * n
4    suffix = [0] * n
5    prefix[0] = nums[0]
6    for i in range(1, n):
7        prefix[i] = max(prefix[i - 1], nums[i])
8    suffix[n - 1] = nums[n - 1]
9    for i in range(n - 2, -1, -1):
10        suffix[i] = max(suffix[i + 1], nums[i])
11    total_beauty = 0
12    for i in range(1, n - 1):
13        beauty = 0
14        if prefix[i - 1] < nums[i] < suffix[i + 1]:
15            beauty = 2
16        elif nums[i - 1] < nums[i] < nums[i + 1]:
17            beauty = 1
18        total_beauty += beauty
19    return total_beauty

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times (once for prefix, once for suffix, and once for beauty calculation). The space complexity is O(n) due to the additional prefix and suffix arrays.

1Using prefix and suffix arrays can significantly reduce the time complexity.
2Understanding the conditions for beauty helps in optimizing checks.

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