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 all0 <= j < i and for alli < 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 <= 105
1 <= nums[i] <= 105
Solutions
Solution 1: Preprocessing Right Minimum + Traversing to Maintain Left Maximum
We can preprocess the right minimum array $right$, where $right[i]$ represents the minimum value in $nums[i..n-1]$.
Then we traverse the array $nums$ from left to right, while maintaining the maximum value $l$ on the left. For each position $i$, we judge whether $l < nums[i] < right[i + 1]$ holds. If it does, we add $2$ to the answer. Otherwise, we judge whether $nums[i - 1] < nums[i] < nums[i + 1]$ holds. If it does, we add $1$ to the answer.
After the traversal, we can get the answer.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.