2909. Minimum Sum of Mountain Triplets II LeetCode Solution
In this guide, you will get 2909. Minimum Sum of Mountain Triplets II LeetCode Solution with the best time and space complexity. The solution to Minimum Sum of Mountain Triplets II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Sum of Mountain Triplets II
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.
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
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(n)
2909. Minimum Sum of Mountain Triplets II LeetCode Solution in C++
class Solution {
public:
// Same as 2908. Minimum Sum of Mountain Triplets I
int minimumSum(vector<int>& nums) {
const int n = nums.size();
int ans = INT_MAX;
vector<int> minPrefix(n);
vector<int> minSuffix(n);
partial_sum(nums.begin(), nums.end(), minPrefix.begin(),
[](int x, int y) { return min(x, y); });
partial_sum(nums.rbegin(), nums.rend(), minSuffix.begin(),
[](int x, int y) { return min(x, y); });
ranges::reverse(minSuffix);
for (int i = 0; i < n; ++i)
if (nums[i] > minPrefix[i] && nums[i] > minSuffix[i])
ans = min(ans, nums[i] + minPrefix[i] + minSuffix[i]);
return ans == INT_MAX ? -1 : ans;
}
}; /* code provided by PROGIEZ */
2909. Minimum Sum of Mountain Triplets II LeetCode Solution in Java
class Solution {
// Same as 2908. Minimum Sum of Mountain Triplets I
public int minimumSum(int[] nums) {
final int n = nums.length;
int ans = Integer.MAX_VALUE;
int[] minPrefix = new int[n];
int[] minSuffix = new int[n];
minPrefix[0] = nums[0];
minSuffix[n - 1] = nums[n - 1];
for (int i = 1; i < n; ++i)
minPrefix[i] = Math.min(minPrefix[i - 1], nums[i]);
for (int i = n - 2; i >= 0; --i)
minSuffix[i] = Math.min(minSuffix[i + 1], nums[i]);
for (int i = 0; i < n; ++i)
if (nums[i] > minPrefix[i] && nums[i] > minSuffix[i])
ans = Math.min(ans, nums[i] + minPrefix[i] + minSuffix[i]);
return ans == Integer.MAX_VALUE ? -1 : ans;
}
} // code provided by PROGIEZ
2909. Minimum Sum of Mountain Triplets II LeetCode Solution in Python
class Solution:
# Same as 2908. Minimum Sum of Mountain Triplets I
def minimumSum(self, nums: list[int]) -> int:
ans = math.inf
minPrefix = list(itertools.accumulate(nums, min))
minSuffix = list(itertools.accumulate(reversed(nums), min))[::-1]
for i, num in enumerate(nums):
if num > minPrefix[i] and num > minSuffix[i]:
ans = min(ans, num + minPrefix[i] + minSuffix[i])
return -1 if ans == math.inf else ans # code by PROGIEZ
