2012. Sum of Beauty in the Array

Problem

You are given a 0-indexed integer array nums. For each index i (1 <= i <= nums.length - 2) the beauty of nums[i] equals:

• 2, if nums[j] < nums[i] < nums[k], for all 0 <= j < i and for all i < k <= nums.length - 1.

• 1, if nums[i - 1] < nums[i] < nums[i + 1], and the previous condition is not satisfied.

• 0, if none of the previous conditions holds.

Return** the sum of beauty of all nums[i] where **1 <= i <= nums.length - 2.

  Example 1:

Input: nums = [1,2,3]
Output: 2
Explanation: For each index i in the range 1 <= i <= 1:
- The beauty of nums[1] equals 2.

Example 2:

Input: nums = [2,4,6,4]
Output: 1
Explanation: For each index i in the range 1 <= i <= 2:
- The beauty of nums[1] equals 1.
- The beauty of nums[2] equals 0.

Example 3:

Input: nums = [3,2,1]
Output: 0
Explanation: For each index i in the range 1 <= i <= 1:
- The beauty of nums[1] equals 0.

  Constraints:

• 3 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^5

Solution (Java)

class Solution {
    public int sumOfBeauties(int[] nums) {
        int[] maxArr = new int[nums.length];
        maxArr[0] = nums[0];
        for (int i = 1; i < nums.length - 1; i++) {
            maxArr[i] = Math.max(maxArr[i - 1], nums[i]);
        }
        int[] minArr = new int[nums.length];
        minArr[nums.length - 1] = nums[nums.length - 1];
        for (int i = nums.length - 2; i >= 0; i--) {
            minArr[i] = Math.min(minArr[i + 1], nums[i]);
        }

        int sum = 0;
        for (int i = 1; i < nums.length - 1; i++) {
            if (nums[i] > maxArr[i - 1] && nums[i] < minArr[i + 1]) {
                sum += 2;
            } else if (nums[i] > nums[i - 1] && nums[i] < nums[i + 1]) {
                sum += 1;
            }
        }

        return sum;
    }
}

Explain:

nope.

Complexity:

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