Watch 5 video solutions for Sum of Beauty in the Array, a medium level problem involving Array. This walkthrough by CodeClips with Abhishek Ranjan has 1,356 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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 <= 1051 <= nums[i] <= 105Problem Overview: You are given an integer array nums. For every index i (excluding the first and last), you assign a beauty score based on ordering conditions with elements on the left and right. If nums[i] is strictly greater than all elements to its left and strictly smaller than all elements to its right, it contributes 2. Otherwise, if it is strictly between its immediate neighbors (nums[i-1] < nums[i] < nums[i+1]), it contributes 1. The task is to compute the total beauty across the array.
Approach 1: Naive Iterative Approach (O(n^2) time, O(1) space)
Iterate through every valid index i from 1 to n-2. For each position, scan the left portion of the array to compute the maximum value and scan the right portion to compute the minimum value. If max(left) < nums[i] < min(right), add 2 to the score. Otherwise check the local condition nums[i-1] < nums[i] < nums[i+1] and add 1 if it holds. This approach directly implements the definition but repeatedly recomputes left and right ranges, leading to quadratic time complexity.
Approach 2: Optimized Prefix and Suffix Arrays Approach (O(n) time, O(n) space)
Precompute helper arrays so each index can be evaluated in constant time. Build a prefix maximum array where prefixMax[i] stores the maximum value from index 0 to i. Build a suffix minimum array where suffixMin[i] stores the minimum value from index i to the end. When evaluating index i, check whether prefixMax[i-1] < nums[i] < suffixMin[i+1] to assign beauty 2. If that fails, check the local neighbor condition for beauty 1. Because prefix and suffix values are precomputed, each element is processed once, giving linear time.
This pattern appears often in array problems where you need information about elements on both sides of an index. Precomputing prefix maxima and suffix minima is a standard optimization also seen in prefix-based techniques and other dynamic programming style precomputation problems.
Recommended for interviews: The optimized prefix and suffix arrays approach. Interviewers expect you to recognize that repeatedly scanning the left and right sides is inefficient. Demonstrating the brute force first shows understanding of the problem definition, but moving to the O(n) precomputation solution shows algorithmic maturity and familiarity with common array optimization patterns.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Naive Iterative Approach | O(n^2) | O(1) | Useful for understanding the rules or when constraints are very small |
| Optimized Prefix and Suffix Arrays | O(n) | O(n) | Best general solution for large arrays and typical interview expectations |