#2908

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Minimum Sum of Mountain Triplets I

Find the minimum sum of a mountain triplet in an array of integers, or return -1 if no valid triplet exists.

array

Problem Statement

You are given a 0-indexed array nums of integers. A triplet of indices (i, j, k) is a mountain if:

i < j < k
nums[i] < nums[j] and nums[k] < nums[j]. Return the minimum possible sum of a mountain triplet of nums. If no such triplet exists, return -1.

The task involves finding a valid triplet where nums[j] is the peak, and the two other values are smaller than nums[j]. If multiple valid triplets exist, the one with the smallest sum should be returned. Otherwise, return -1 if no valid mountain triplet is found.

Examples

Example 1

Input: nums = [8,6,1,5,3]

Output: 9

Triplet (2, 3, 4) is a mountain triplet of sum 9 since:

2 < 3 < 4
nums[2] < nums[3] and nums[4] < nums[3] And the sum of this triplet is nums[2] + nums[3] + nums[4] = 9. It can be shown that there are no mountain triplets with a sum of less than 9.

Example 2

Input: nums = [5,4,8,7,10,2]

Output: 13

Triplet (1, 3, 5) is a mountain triplet of sum 13 since:

1 < 3 < 5
nums[1] < nums[3] and nums[5] < nums[3] And the sum of this triplet is nums[1] + nums[3] + nums[5] = 13. It can be shown that there are no mountain triplets with a sum of less than 13.

Example 3

Input: nums = [6,5,4,3,4,5]

Output: -1

It can be shown that there are no mountain triplets in nums.

Constraints

3 <= nums.length <= 50
1 <= nums[i] <= 50

Solution Approach

Brute Force Solution

A simple approach would involve iterating over all possible triplets in the array and checking if they form a valid mountain triplet. This can be done by checking for conditions nums[i] < nums[j] and nums[k] < nums[j] for each triplet (i, j, k). The time complexity for this approach would be O(n^3), where n is the length of the input array.

Optimized Approach Using Two Loops

Instead of iterating over all triplets, the problem can be optimized by iterating over each element and using two loops to find the smallest element to the left of nums[j] and the smallest element to the right of nums[j]. This reduces the complexity of finding the triplets to O(n^2).

Final Optimized Approach

The final approach involves improving upon the previous method by utilizing dynamic programming or binary search techniques to reduce redundant computations. The solution can be implemented in O(n^2) time with O(n) space complexity.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The brute force solution requires checking each triplet, leading to O(n^3) time complexity. By optimizing to two nested loops for left and right comparisons, the time complexity is reduced to O(n^2). Using dynamic programming or binary search techniques could reduce the complexity further, potentially improving both time and space requirements.

What Interviewers Usually Probe

Candidate uses brute force but recognizes its inefficiency.
Candidate attempts to optimize the brute force solution.
Candidate demonstrates an understanding of reducing complexity by leveraging dynamic programming or binary search.

Common Pitfalls or Variants

Common pitfalls

Failing to properly check the triplet conditions may lead to incorrect results.
Overlooking the case where no valid triplet exists, resulting in an incorrect return value.
Choosing the wrong triplet without considering the minimum sum might lead to incorrect answers.

Follow-up variants

What if the array is in descending order?
What if the array contains only equal elements?
What if the array contains duplicate elements?

FAQ

What is a mountain triplet?

A mountain triplet consists of three indices (i, j, k) where i < j < k, nums[i] < nums[j], and nums[k] < nums[j].

What is the time complexity of the brute force solution?

The brute force solution checks all possible triplets, leading to a time complexity of O(n^3), where n is the length of the array.

How can I optimize the brute force solution?

You can optimize the brute force solution by using two loops to find the smallest elements to the left and right of nums[j], reducing the time complexity to O(n^2).

What should I do if no valid mountain triplet exists?

If no valid mountain triplet exists, you should return -1 as specified in the problem statement.

How does GhostInterview help with the problem-solving process?

GhostInterview provides a structured approach to solving the problem, highlights common mistakes, and suggests optimizations to improve both time and space complexity.

terminal

Solution

Solution 1: Preprocessing + Enumeration

We can preprocess the minimum value on the right side of each position and record it in the array $right[i]$, where $right[i]$ represents the minimum value in $nums[i+1..n-1]$.

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class Solution:
    def minimumSum(self, nums: List[int]) -> int:
        n = len(nums)
        right = [inf] * (n + 1)
        for i in range(n - 1, -1, -1):
            right[i] = min(right[i + 1], nums[i])
        ans = left = inf
        for i, x in enumerate(nums):
            if left < x and right[i + 1] < x:
                ans = min(ans, left + x + right[i + 1])
            left = min(left, x)
        return -1 if ans == inf else ans

Continue Practicing

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