How do you solve Minimum Sum of Mountain Triplets II on LeetCode?

Minimum Sum of Mountain Triplets II (LeetCode #2909) 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: Preprocessing helps reduce the number of comparisons needed to find valid triplets.

What is the brute force solution for Minimum Sum of Mountain Triplets II?

The brute force approach for Minimum Sum of Mountain Triplets II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible triplets in the array to see if they form a mountain triplet. This 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 Minimum Sum of Mountain Triplets II?

Minimum Sum of Mountain Triplets II uses the following concepts: Array. The recommended patterns to study are: Prefix/Suffix Arrays, Two Pointers.

What are the interview tips for Minimum Sum of Mountain Triplets II?

Always start with a brute force solution to understand the problem. Look for patterns in the problem that can lead to optimizations. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Sum of Mountain Triplets II?

Not considering edge cases where no triplet exists. Overlooking the need to preprocess data for optimization.

#2909

Minimum Sum of Mountain Triplets II

Medium

ArrayPrefix/Suffix ArraysTwo 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 preprocessing the array to find the smallest values to the left and right of each index, we can efficiently find valid mountain triplets without checking all combinations.

⚙️

Algorithm

5 steps

1Step 1: Create a prefix minimum array where prefix_min[i] is the minimum value from nums[0] to nums[i].
2Step 2: Create a suffix minimum array where suffix_min[i] is the minimum value from nums[i] to nums[n-1].
3Step 3: Iterate through each index j from 1 to n-2 and check if nums[j] can form a mountain triplet using prefix_min and suffix_min.
4Step 4: If valid i and k are found, calculate the sum and update the minimum sum.
5Step 5: Return the minimum sum found, or -1 if no triplet exists.

solution.py15 lines

1def minimumMountainTriplet(nums):
2    n = len(nums)
3    prefix_min = [float('inf')] * n
4    suffix_min = [float('inf')] * n
5    prefix_min[0] = nums[0]
6    for i in range(1, n):
7        prefix_min[i] = min(prefix_min[i - 1], nums[i])
8    suffix_min[n - 1] = nums[n - 1]
9    for i in range(n - 2, -1, -1):
10        suffix_min[i] = min(suffix_min[i + 1], nums[i])
11    min_sum = float('inf')
12    for j in range(1, n - 1):
13        if nums[j] > prefix_min[j - 1] and nums[j] > suffix_min[j + 1]:
14            min_sum = min(min_sum, prefix_min[j - 1] + nums[j] + suffix_min[j + 1])
15    return min_sum if min_sum != float('inf') else -1

ℹ

Complexity note: This complexity is linear because we only pass through the array a constant number of times to build the prefix and suffix arrays, and then once more to find the minimum sum.

1Preprocessing helps reduce the number of comparisons needed to find valid triplets.
2Understanding the structure of the problem allows for more efficient algorithms.

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