Leetcode #2012: Sum of Beauty in the Array
In this guide, we solve Leetcode #2012 Sum of Beauty in the Array in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.