README.md

2012. Sum of Beauty in the Array

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 (Rust)

1. Solution

impl Solution {
    pub fn sum_of_beauties(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut max = vec![i32::MIN; nums.len()];
        let mut min = vec![i32::MAX; nums.len()];
        let mut ret = 0;

        for i in 1..nums.len() {
            max[i] = max[i - 1].max(nums[i - 1]);
            min[n - 1 - i] = min[n - i].min(nums[n - i]);
        }

        for i in 1..nums.len() - 1 {
            if max[i] < nums[i] && nums[i] < min[i] {
                ret += 2;
            } else if nums[i - 1] < nums[i] && nums[i] < nums[i + 1] {
                ret += 1;
            }
        }

        ret
    }
}