LeetCode 2909: Minimum Sum of Mountain Triplets II Solution

Master LeetCode problem 2909 (Minimum Sum of Mountain Triplets II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2909. Minimum Sum of Mountain Triplets II

Medium

Array

Problem Explanation

Explanation

To find the minimum sum of a mountain triplet in the given array, we need to iterate through the array and for each element nums[j], we need to find the minimum elements nums[i] and nums[k] such that i < j and j < k and nums[i] < nums[j] and nums[k] < nums[j]. We can maintain two arrays, left and right, to store the minimum elements on the left and right side of each element respectively. Then we can iterate through the array and for each element nums[j], we can calculate the sum of the triplet and keep track of the minimum sum.

Initialize two arrays, left and right, to store the minimum element on the left and right side of each element.
Iterate through the array to fill the left array with minimum values on the left side.
Iterate through the array in reverse to fill the right array with minimum values on the right side.
Iterate through the array to find the minimum sum of mountain triplets.

Time complexity: O(n) Space complexity: O(n)

Solution Code

class Solution {
    public int minMountainTriples(int[] nums) {
        int n = nums.length;
        int[] left = new int[n];
        int[] right = new int[n];
        
        left[0] = Integer.MAX_VALUE;
        for (int i = 1; i < n; i++) {
            left[i] = Math.min(left[i - 1], nums[i - 1]);
        }
        
        right[n - 1] = Integer.MAX_VALUE;
        for (int i = n - 2; i >= 0; i--) {
            right[i] = Math.min(right[i + 1], nums[i + 1]);
        }
        
        int minSum = Integer.MAX_VALUE;
        for (int j = 1; j < n - 1; j++) {
            if (nums[j] > left[j] && nums[j] > right[j]) {
                minSum = Math.min(minSum, nums[j] + left[j] + right[j]);
            }
        }
        
        return minSum == Integer.MAX_VALUE ? -1 : minSum;
    }
}

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