# 2908. Minimum Sum of Mountain Triplets I

## Description

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.

Example 1:

Input: nums = [8,6,1,5,3]
Output: 9
Explanation: 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
Explanation: 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
Explanation: It can be shown that there are no mountain triplets in nums.


Constraints:

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

## Solutions

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 $nums[i+1..n-1]$.

Next, we enumerate the middle element $nums[i]$ of the mountain triplet from left to right, and use a variable $left$ to represent the minimum value in $ums[0..i-1]$, and a variable $ans$ to represent the current minimum element sum found. For each $i$, we need to find the element $nums[i]$ that satisfies $left < nums[i]$ and $right[i+1] < nums[i]$, and update $ans$.

Finally, if $ans$ is still the initial value, it means that there is no mountain triplet, and we return $-1$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

• class Solution {
public int minimumSum(int[] nums) {
int n = nums.length;
int[] right = new int[n + 1];
final int inf = 1 << 30;
right[n] = inf;
for (int i = n - 1; i >= 0; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
int ans = inf, left = inf;
for (int i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = Math.min(ans, left + nums[i] + right[i + 1]);
}
left = Math.min(left, nums[i]);
}
return ans == inf ? -1 : ans;
}
}

• class Solution {
public:
int minimumSum(vector<int>& nums) {
int n = nums.size();
const int inf = 1 << 30;
int right[n + 1];
right[n] = inf;
for (int i = n - 1; ~i; --i) {
right[i] = min(right[i + 1], nums[i]);
}
int ans = inf, left = inf;
for (int i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = min(ans, left + nums[i] + right[i + 1]);
}
left = min(left, nums[i]);
}
return ans == inf ? -1 : ans;
}
};

• 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


• func minimumSum(nums []int) int {
n := len(nums)
const inf = 1 << 30
right := make([]int, n+1)
right[n] = inf
for i := n - 1; i >= 0; i-- {
right[i] = min(right[i+1], nums[i])
}
ans, left := inf, inf
for i, x := range nums {
if left < x && right[i+1] < x {
ans = min(ans, left+x+right[i+1])
}
left = min(left, x)
}
if ans == inf {
return -1
}
return ans
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

• function minimumSum(nums: number[]): number {
const n = nums.length;
const right: number[] = Array(n + 1).fill(Infinity);
for (let i = n - 1; ~i; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
let [ans, left] = [Infinity, Infinity];
for (let i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = Math.min(ans, left + nums[i] + right[i + 1]);
}
left = Math.min(left, nums[i]);
}
return ans === Infinity ? -1 : ans;
}