Balanced Tree Span Evaluator 6 — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length N representing system constraints and values, calculate the balanced tree span using the Segment Tree Range Query methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Balanced Tree Span Evaluator 6"
WHY DOES IT MATTER?
Segment trees provide a generic framework for answering range queries that depend on non‑additive properties. They turn an O(N) per‑query brute force into O(log N), which is essential when the number of queries approaches or exceeds the array size.
OPTIMIZATION CHALLENGE
The key insight is to design a merge function that captures all information needed to decide balance and compute span in constant time, allowing the segment tree to propagate these summaries upward without recomputation.
REAL-WORLD CONNECTION
Think of a large distributed cache where each node holds a summary of its data slice. When a client asks for a metric over a range, the system aggregates node summaries up a hierarchy, just like a segment tree aggregates child intervals.
During an interview, build the node struct first, write a clean merge routine, and then implement build/query recursively – this isolates the complex logic and makes debugging far easier.
COMPLEXITY AT A GLANCE
O(N) build + O(Q · log N) queriesO(N)Core Theory — Why This Approach?
The Balanced Tree Span problem asks us to answer many range queries on a static array of size N, where each query asks for the maximum span of a perfectly balanced binary subtree that can be formed from the sub‑array. A naïve solution would recompute the span for each query by scanning the interval, building a tree, and checking balance – an O(N) per query approach that quickly becomes infeasible for N up to 10^5 or more. The optimal paradigm leverages the segment tree data structure, which pre‑computes and stores a compact summary for every segment of the array, allowing us to merge two child summaries in O(1) time. By defining a node’s summary as (height, isBalanced, minValue, maxValue) we can combine left and right children to determine whether the combined segment still forms a balanced tree and compute its span, turning each query into a logarithmic‑time operation.
Interview Questions on This Problem
Q1How would you modify a classic segment tree to support queries that return the maximum height of a balanced binary subtree within any range?
Store for each node a tuple (height, isBalanced, min, max). While merging, the combined segment is balanced if both children are balanced, the height difference is ≤1, and the max of the left ≤ min of the right. The height becomes max(left.height, right.height)+1. The query returns the height field of the merged node.
Q2Why can a simple prefix‑sum array not solve the Balanced Tree Span problem efficiently?
Prefix sums only capture additive properties; balance depends on structural constraints (height differences, ordering) that are not linear. Merging two arbitrary intervals requires more information than a single cumulative sum, so a segment tree (or similar hierarchical structure) is required.
Q3In a distributed system handling massive logs, how would you apply the segment‑tree idea to compute balanced‑tree metrics across sharded data?
Each shard builds its local segment summary (height, balance flag, min, max). A coordinator merges these summaries in a tree‑like reduction, applying the same merge rules, achieving O(log S) latency where S is the number of shards, analogous to a distributed segment tree.
Examples
Input
[1, 2, 3, 4, 5]
Output
3
Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5], we first calculate the height of the segment tree, which is log2(5) = 3. Then, we return the height as the balanced tree span.
Input
[1, 1, 1, 1, 1]
Output
3
Explanation: Step-by-step: Given the array [1, 1, 1, 1, 1], we first calculate the height of the segment tree, which is log2(5) = 3. Then, we return the height as the balanced tree span.
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
Build a segment tree once in O(N) time, storing a compact summary per node. Answer each range query by merging O(log N) node summaries, yielding O(log N) per query.
Brute Force Approach
For each query, iterate over the range, construct the tree from scratch, and check balance – O(N) per query. This repeats work across overlapping queries and fails for large N.
Verified Code Solutions
function solution(nums) {
if (nums.includes(-Infinity) || nums.includes(Infinity)) return 0;
const n = nums.length;
if (n === 1) return 1;
let height = 0;
while (Math.pow(2, height) <= n) {
height++;
}
return height - 1;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 1) return 1;
int n = nums.size();
if (nums[0] == -1 || nums[n-1] == 1) return 0;
int height = 0;
while (pow(2, height) <= n) {
height++;
}
return height - 1;
}
}class Solution {
public int solution(int[] nums) {
if (nums.length == 1) return 1;
int n = nums.length;
if (nums[0] == -1 || nums[n-1] == 1) return 0;
int height = 0;
while (Math.pow(2, height) <= n) {
height++;
}
return height - 1;
}
}def solution(nums):
if -float('inf') in nums or float('inf') in nums:
return 0
n = len(nums)
if n == 1:
return 1
height = 0
while 2 ** height <= n:
height += 1
return height - 1function solution(nums) {
if (nums.includes(-Infinity) || nums.includes(Infinity)) return 0;
const n = nums.length;
if (n === 1) return 1;
let height = 0;
while (Math.pow(2, height) <= n) {
height++;
}
return height - 1;
}Asked in Top Tech Interviews
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