Subtree Height Evaluator Optimizer — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length N representing system constraints and values, calculate the subtree height evaluator using the Rotated Array Pivot Search methodology. The array is considered rotated if the last element is greater than or equal to the first element. If the array is not rotated, consider the left and right subtrees as the entire array. Handle edge cases like an empty array or an array with a single element.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Subtree Height Evaluator Optimizer"
WHY DOES IT MATTER?
Detecting the rotation pivot in O(log N) is a fundamental pattern for many real‑world problems such as circular buffer management, versioned data stores, and time‑series queries where data wraps around a fixed window.
OPTIMIZATION CHALLENGE
The key insight is that a rotated sorted array retains a monotonic property in at least one half of any interval, allowing a binary search to isolate the minimum element, which transforms a linear‑time scan into logarithmic time.
REAL-WORLD CONNECTION
Imagine a distributed log that continuously appends entries but periodically rolls over to the start of a fixed‑size buffer. Finding the start of the newest segment (the pivot) lets you efficiently compute metrics like the depth of a tree built from log entries without scanning the entire buffer.
During an interview, first verify the rotation condition, then write a clean binary‑search loop that never accesses out‑of‑bounds indices; after finding the pivot, use the mathematical height formula instead of building the tree explicitly.
COMPLEXITY AT A GLANCE
O(log N)O(1)Core Theory — Why This Approach?
The problem blends two classic concepts: computing the height of a binary search tree (BST) when its inorder traversal is stored in a rotated sorted array, and locating the rotation pivot using binary search. In a BST, the height is determined by the depth of the deepest node, which can be derived from the number of levels needed to insert elements in sorted order. When the inorder sequence is rotated, the natural left‑subtree (elements smaller than the root) and right‑subtree (elements larger than the root) are split at the pivot point where the array wraps around. A naive linear scan to find this pivot costs O(N) and defeats the purpose of handling large N (up to 10^6 or more). The optimal paradigm leverages the monotonic property of a rotated sorted array: one half of any sub‑range is always sorted. By repeatedly halving the search space, we locate the smallest element (the pivot) in O(log N) time, then compute subtree heights using the sizes of the left and right partitions, which correspond to the depths of the respective sub‑trees.
Why naive approaches fail: Scanning the entire array to detect the rotation point or to count elements for each subtree incurs linear time per query, leading to O(N) per test case. In interview settings with multiple queries or strict time limits, this quickly becomes a bottleneck. Moreover, recursive height calculation on an unsorted array without recognizing the BST property can cause exponential blow‑up due to repeated sub‑problem recomputation.
Optimal solution: Perform a modified binary search to find the pivot (the index of the minimum element). Once the pivot is known, the array is logically split into two sorted halves: [pivot … N‑1] as the right subtree and [0 … pivot‑1] as the left subtree. The height of a perfectly balanced BST built from a sorted segment of length L is ⌊log₂L⌋ + 1. Using this formula on both halves yields the overall subtree height evaluator in O(log N) time and O(1) extra space.
Interview Questions on This Problem
Q1How would you find the rotation pivot in a sorted array that may have been rotated, and why is binary search applicable?
The pivot is the index of the smallest element. Because at least one half of any sub‑array remains sorted, we can compare mid with the high (or low) element to decide which half contains the unsorted portion where the pivot lies, halving the search space each step, achieving O(log N) time.
Q2Given the pivot, how can you compute the height of the BST that would result from inserting the array elements in order?
The left subtree consists of elements before the pivot (size L) and the right subtree consists of elements from the pivot to the end (size R). The height of a balanced BST built from a sorted segment of size X is ⌊log₂X⌋ + 1. Compute heights for L and R and take max + 1 for the root.
Q3Why does the condition ‘last element >= first element’ imply the array is not rotated, and how does this affect your algorithm?
If the last element is greater than or equal to the first, the array is fully sorted in ascending order, meaning the rotation pivot is at index 0. The algorithm can skip the binary‑search step and directly treat the whole array as the right subtree, simplifying the height calculation.
Examples
Input
[1, 2, 3, 4, 5]
Output
0
Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5], we first check if it's rotated. Since it's not rotated, we consider the left and right subtrees as the entire array. The height of the left subtree is 0 and the height of the right subtree is 1. The maximum of these two heights is 1, so the output is 1.
Input
[1, 1, 1, 1, 1]
Output
0
Explanation: Step-by-step: Given the array [1, 1, 1, 1, 1], we first check if it's rotated. Since it's not rotated, we consider the left and right subtrees as the entire array. The height of the left subtree is 0 and the height of the right subtree is 0. The maximum of these two heights is 0, so the output is 0.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Use a modified binary search to find the pivot in O(log N) and apply the logarithmic height formula on the two sorted partitions, achieving O(log N) time and O(1) extra space.
Brute Force Approach
Scan the entire array to locate the minimum element (pivot) and then compute subtree heights by building the BST explicitly, resulting in O(N) time.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
if (nums.length === 1) return 0;
let max = Math.max(...nums);
let min = Math.min(...nums);
if (nums[nums.length - 1] >= nums[0]) {
let left = nums.slice(0, nums.indexOf(max));
let right = nums.slice(nums.indexOf(max) + 1);
return Math.max(solution(left), solution(right)) + 1;
} else {
let left = nums.slice(0, nums.indexOf(min));
let right = nums.slice(nums.indexOf(min) + 1);
return Math.max(solution(left), solution(right)) + 1;
}
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
if (nums.size() == 1) return 0;
int max = *max_element(nums.begin(), nums.end());
int min = *min_element(nums.begin(), nums.end());
if (nums.back() >= nums[0]) {
vector<int> left(nums.begin(), nums.end());
vector<int> right(nums.begin(), nums.end());
left.erase(find(left.begin(), left.end(), max));
right.erase(find(right.begin(), right.end(), max));
return max(solution(left), solution(right)) + 1;
} else {
vector<int> left(nums.begin(), nums.end());
vector<int> right(nums.begin(), nums.end());
left.erase(find(left.begin(), left.end(), min));
right.erase(find(right.begin(), right.end(), min));
return max(solution(left), solution(right)) + 1;
}
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
if (nums.length == 1) return 0;
int max = Arrays.stream(nums).max().getAsInt();
int min = Arrays.stream(nums).min().getAsInt();
if (nums[nums.length - 1] >= nums[0]) {
int left[] = Arrays.copyOfRange(nums, 0, Arrays.asList(nums).indexOf(max) + 1);
int right[] = Arrays.copyOfRange(nums, Arrays.asList(nums).indexOf(max) + 1, nums.length);
return Math.max(solution(left), solution(right)) + 1;
} else {
int left[] = Arrays.copyOfRange(nums, 0, Arrays.asList(nums).indexOf(min) + 1);
int right[] = Arrays.copyOfRange(nums, Arrays.asList(nums).indexOf(min) + 1, nums.length);
return Math.max(solution(left), solution(right)) + 1;
}
}
}def solution(nums):
if len(nums) == 0:
return 0
if len(nums) == 1:
return 0
max_val = max(nums)
min_val = min(nums)
if nums[-1] >= nums[0]:
left = nums[:nums.index(max_val)]
right = nums[nums.index(max_val) + 1:]
return max(solution(left), solution(right)) + 1
else:
left = nums[:nums.index(min_val)]
right = nums[nums.index(min_val) + 1:]
return max(solution(left), solution(right)) + 1function solution(nums) {
if (nums.length === 0) return 0;
if (nums.length === 1) return 0;
let max = Math.max(...nums);
let min = Math.min(...nums);
if (nums[nums.length - 1] >= nums[0]) {
let left = nums.slice(0, nums.indexOf(max));
let right = nums.slice(nums.indexOf(max) + 1);
return Math.max(solution(left), solution(right)) + 1;
} else {
let left = nums.slice(0, nums.indexOf(min));
let right = nums.slice(nums.indexOf(min) + 1);
return Math.max(solution(left), solution(right)) + 1;
}
}Asked in Top Tech Interviews
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