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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 $nums$.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
  }
  
  

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