#2909

Medium

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Minimum Sum of Mountain Triplets II

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

Problem Statement

You are given a 0-indexed array nums of integers. A triplet of indices (i, j, k) forms a mountain if the following conditions are satisfied: nums[i] < nums[j] > nums[k], and 0 <= i < j < k < nums.length.

Return the minimum possible sum of such a mountain triplet. If no such triplet exists, return -1.

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 <= 105
1 <= nums[i] <= 108

Solution Approach

Fix the middle index and find left and right values

A common approach to solving this problem is to fix the middle index j. For each j, find the smallest value to the left of j (i) and the largest value to the right of j (k). This helps reduce the problem's complexity and ensures the constraints are met.

Efficient traversal using pre-computed arrays

Using pre-computed arrays for left and right minimum/maximum values can speed up the solution. One array will store the smallest value up to index j, and another array will store the largest value from index j onward. This allows quick access while iterating through potential middle indices.

Edge case handling

When the array does not contain any valid mountain triplet, we need to handle this scenario efficiently by returning -1. This can happen when the array is in a strictly increasing or decreasing order.

Complexity Analysis

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

Time complexity depends on the approach used for computing left and right minimum/maximum values. If pre-computed arrays are used, the time complexity is O(n). Space complexity will also be O(n) for storing these arrays.

What Interviewers Usually Probe

Look for a solution that optimizes the left and right value search for each middle index.
Focus on the efficiency of the pre-computation of left and right values.
Evaluate how well the candidate handles edge cases where no valid triplet exists.

Common Pitfalls or Variants

Common pitfalls

Not using efficient array traversal techniques can lead to a solution that does not meet time limits for larger arrays.
Failing to correctly handle the edge case where no valid mountain triplet exists.
Choosing the wrong approach to find the left and right values, leading to higher time complexity.

Follow-up variants

Generalize the solution to handle arrays with varying sizes and edge cases.
Optimize the approach for arrays with large values (up to 10^8) and large lengths (up to 10^5).
Extend the problem to find multiple mountain triplets and return the one with the smallest sum.

FAQ

What is the minimum sum of mountain triplets?

The minimum sum is the sum of the smallest values on the left and right of the middle element of the triplet. This ensures that the conditions of a mountain triplet are met.

How can I efficiently find the left and right values for each index?

You can use pre-computed arrays to store the smallest value up to each index and the largest value from each index onward, allowing for constant-time lookup during the iteration.

How do I handle the case where no mountain triplet exists?

If no valid triplet can be formed, return -1 as per the problem's requirements. This is typically checked after the main traversal.

What is the time complexity of the solution?

The time complexity is O(n), where n is the length of the array. This is achieved by using pre-computed arrays to quickly find the left and right values for each middle index.

How can I apply this problem-solving approach to other similar problems?

This array-driven strategy of pre-computing values for efficient lookups is applicable in many problems involving triplets or other patterns where the left and right values matter, such as in finding local maxima or minima.

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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