2908. Minimum Sum of Mountain Triplets I 

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class Solution {
    private static final int MAX = Integer.MAX_VALUE / 2;

    public int minimumSum(int[] nums) {
        int n = nums.length;
        int minSum = MAX;

        int[] rightMin = new int[n]; // 后缀最小值，包含 nums[i]
        rightMin[n - 1] = nums[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            rightMin[i] = Math.min(rightMin[i + 1], nums[i]);
        }

        int leftMin = nums[0];
        for (int i = 1; i < n - 1; i++) {
            if (leftMin < nums[i] && nums[i] > rightMin[i + 1]) {
                int sum = leftMin + nums[i] + rightMin[i];
                minSum = Math.min(minSum, sum);
            }
            leftMin = Math.min(leftMin, nums[i]);
        }

        return minSum == MAX ? -1 : minSum;
    }
}

References

2908. Minimum Sum of Mountain Triplets I