Question

2909. Minimum Sum of Mountain Triplets II

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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 <= 105
• 1 <= nums[i] <= 108

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:
  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

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;
  }
};

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
}

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;
}