# 2012. Sum of Beauty in the Array

## Description

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 <= 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 $nums$.

• class Solution {
public int sumOfBeauties(int[] nums) {
int n = nums.length;
int[] right = new int[n];
right[n - 1] = nums[n - 1];
for (int i = n - 2; i > 0; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
int ans = 0;
int l = nums[0];
for (int i = 1; i < n - 1; ++i) {
int r = right[i + 1];
if (l < nums[i] && nums[i] < r) {
ans += 2;
} else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
ans += 1;
}
l = Math.max(l, nums[i]);
}
return ans;
}
}

• class Solution {
public:
int sumOfBeauties(vector<int>& nums) {
int n = nums.size();
vector<int> right(n, nums[n - 1]);
for (int i = n - 2; i; --i) {
right[i] = min(right[i + 1], nums[i]);
}
int ans = 0;
for (int i = 1, l = nums[0]; i < n - 1; ++i) {
int r = right[i + 1];
if (l < nums[i] && nums[i] < r) {
ans += 2;
} else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
ans += 1;
}
l = max(l, nums[i]);
}
return ans;
}
};

• class Solution:
def sumOfBeauties(self, nums: List[int]) -> int:
n = len(nums)
right = [nums[-1]] * n
for i in range(n - 2, -1, -1):
right[i] = min(right[i + 1], nums[i])
ans = 0
l = nums[0]
for i in range(1, n - 1):
r = right[i + 1]
if l < nums[i] < r:
ans += 2
elif nums[i - 1] < nums[i] < nums[i + 1]:
ans += 1
l = max(l, nums[i])
return ans


• func sumOfBeauties(nums []int) (ans int) {
n := len(nums)
right := make([]int, n)
right[n-1] = nums[n-1]
for i := n - 2; i > 0; i-- {
right[i] = min(right[i+1], nums[i])
}
for i, l := 1, nums[0]; i < n-1; i++ {
r := right[i+1]
if l < nums[i] && nums[i] < r {
ans += 2
} else if nums[i-1] < nums[i] && nums[i] < nums[i+1] {
ans++
}
l = max(l, nums[i])
}
return
}

• function sumOfBeauties(nums: number[]): number {
const n = nums.length;
const right: number[] = Array(n).fill(nums[n - 1]);
for (let i = n - 2; i; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
let ans = 0;
for (let i = 1, l = nums[0]; i < n - 1; ++i) {
const r = right[i + 1];
if (l < nums[i] && nums[i] < r) {
ans += 2;
} else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
ans += 1;
}
l = Math.max(l, nums[i]);
}
return ans;
}