2012. Sum of Beauty in the Array LeetCode Solution

In this guide, you will get 2012. Sum of Beauty in the Array LeetCode Solution with the best time and space complexity. The solution to Sum of Beauty in the Array problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Sum of Beauty in the Array solution in C++
Sum of Beauty in the Array solution in Java
Sum of Beauty in the Array solution in Python
Additional Resources

2012. Sum of Beauty in the Array LeetCode Solution image

Problem Statement of 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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

2012. Sum of Beauty in the Array LeetCode Solution in C++

class Solution {
 public:
  int sumOfBeauties(vector<int>& nums) {
    const int n = nums.size();
    int ans = 0;
    vector<int> minOfRight(n);
    minOfRight.back() = nums.back();

    for (int i = n - 2; i >= 2; --i)
      minOfRight[i] = min(nums[i], minOfRight[i + 1]);

    int maxOfLeft = nums[0];

    for (int i = 1; i <= n - 2; ++i) {
      if (maxOfLeft < nums[i] && nums[i] < minOfRight[i + 1])
        ans += 2;
      else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1])
        ans += 1;
      maxOfLeft = max(maxOfLeft, nums[i]);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2012. Sum of Beauty in the Array LeetCode Solution in Java

class Solution {
  public int sumOfBeauties(int[] nums) {
    final int n = nums.length;
    int ans = 0;
    int[] minOfRight = new int[n];
    minOfRight[n - 1] = nums[n - 1];

    for (int i = n - 2; i >= 2; --i)
      minOfRight[i] = Math.min(nums[i], minOfRight[i + 1]);

    int maxOfLeft = nums[0];

    for (int i = 1; i <= n - 2; ++i) {
      if (maxOfLeft < nums[i] && nums[i] < minOfRight[i + 1])
        ans += 2;
      else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1])
        ans += 1;
      maxOfLeft = Math.max(maxOfLeft, nums[i]);
    }

    return ans;
  }
}
// code provided by PROGIEZ

2012. Sum of Beauty in the Array LeetCode Solution in Python

class Solution:
  def sumOfBeauties(self, nums: list[int]) -> int:
    n = len(nums)
    ans = 0
    minOfRight = [0] * (n - 1) + [nums[-1]]

    for i in range(n - 2, 1, -1):
      minOfRight[i] = min(nums[i], minOfRight[i + 1])

    maxOfLeft = nums[0]

    for i in range(1, n - 1):
      if maxOfLeft < nums[i] < minOfRight[i + 1]:
        ans += 2
      elif nums[i - 1] < nums[i] < nums[i + 1]:
        ans += 1
      maxOfLeft = max(maxOfLeft, nums[i])

    return ans
 # code by PROGIEZ