Question

2012. Sum of Beauty in the Array

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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:
  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

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;
  }
};

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;
}