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

Minimum Sum of Mountain Triplets I (LeetCode #2908) 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: A mountain triplet requires a peak with valid left and right values.

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

The brute force approach for Minimum Sum of Mountain Triplets I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible triplet in the array to see if it forms a mountain triplet. This is straightforward but can be slow for larger arrays since it involves nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Sum of Mountain Triplets I?

Minimum Sum of Mountain Triplets I uses the following concepts: Array. The recommended patterns to study are: Two Pointers, Sliding Window, Dynamic Programming.

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

Clearly explain your thought process while coding. Consider edge cases and constraints upfront. Discuss time and space complexity as you develop your solution.

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

Not checking bounds for indices i, j, k. Overlooking the condition for mountain triplet formation.

#2908

Minimum Sum of Mountain Triplets I

Easy

ArrayTwo PointersSliding WindowDynamic Programming

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 single pass to find the minimum values for potential mountain triplets. This reduces the number of checks needed and ensures we only look at valid indices.

⚙️

Algorithm

4 steps

1Step 1: Create two arrays, leftMin and rightMin, to store the minimum values to the left and right of each index.
2Step 2: Fill leftMin by iterating from left to right, keeping track of the minimum value seen so far.
3Step 3: Fill rightMin by iterating from right to left, keeping track of the minimum value seen so far.
4Step 4: Loop through each possible middle index j and check if leftMin[j] < nums[j] and rightMin[j] < nums[j]. If true, calculate the sum and update the minimum sum.

solution.py13 lines

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

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times. The space complexity is O(n) due to the additional arrays used to store the minimum values.

1A mountain triplet requires a peak with valid left and right values.
2Using auxiliary arrays can help optimize the search for valid triplets.

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