Subtree Height Evaluator Resolver 7 — Problem Statement & Solution Guide
Problem Description
You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the subtree height evaluator using the **Minimum Window Substring** methodology.
Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Subtree Height Evaluator Resolver 7"
WHY DOES IT MATTER?
The sliding‑window pattern converts a combinatorial search over all subarrays into a linear scan, which is essential for any problem that asks for optimal contiguous segments under a coverage constraint. Without it, solutions explode combinatorially and cannot meet real‑world performance SLAs.
OPTIMIZATION CHALLENGE
The breakthrough is maintaining a dynamic count of how many distinct target elements are satisfied (the "formed" variable). This lets the algorithm decide in O(1) whether the current window is valid, avoiding a full scan of the target set at each step.
REAL-WORLD CONNECTION
Think of a streaming log processor that must capture the shortest time interval containing all error codes of interest. The processor slides a time window over the log stream, expanding until all codes appear, then contracts to trim excess, mirroring the MWS algorithm.
When coding, initialize the target frequency map first, then keep a separate window map. Increment "formed" only when a character's count in the window exactly matches its required count; decrement it when it falls below during contraction. This precise bookkeeping prevents off‑by‑one bugs.
COMPLEXITY AT A GLANCE
O(N + |T|)O(|Σ| + |T|)Core Theory — Why This Approach?
The Minimum Window Substring (MWS) problem asks for the smallest contiguous sub‑array of a source sequence that contains all characters (or tokens) of a target multiset. The classic solution relies on a sliding‑window two‑pointer technique that expands the right bound until the window satisfies the requirement, then contracts the left bound to discard unnecessary prefix while still maintaining validity. This dynamic adjustment guarantees each element is visited at most twice, yielding linear time. Naïve brute‑force enumeration of all O(N^2) windows quickly becomes infeasible for N up to 10^5 or higher, because each candidate would need O(M) verification of target coverage, leading to O(N^2·M) worst‑case. The optimal paradigm transforms the verification into constant‑time updates by maintaining frequency counters for the target and the current window, and a "formed" counter that tracks how many distinct target characters meet their required frequency. When "formed" equals the number of distinct target characters, the window is valid and we attempt to shrink it, ensuring the minimal length is captured.
Interview Questions on This Problem
Q1How would you adapt the Minimum Window Substring algorithm to work with an array of integers where the target is a multiset of numbers instead of characters?
Treat each integer as a key in a hash map, store its required count from the target multiset, and apply the same sliding‑window logic: expand the right pointer, update the window frequency map, increment a "formed" counter when a number's count matches the target, then contract from the left while the window remains valid, tracking the smallest window indices.
Q2Explain why the two‑pointer sliding window yields O(N) time for the Minimum Window Substring, even though there are nested loops in the pseudocode.
Each pointer (left and right) only moves forward; the right pointer traverses the string once to include characters, and the left pointer traverses at most once to exclude them. Hence the total number of pointer movements is bounded by 2N, giving linear time despite the appearance of nested loops.
Q3In a distributed system, how could you parallelize the search for a minimum window across shards of data while preserving correctness?
Compute candidate windows locally on each shard using the sliding‑window algorithm, then exchange border information (prefix and suffix character counts) between adjacent shards to potentially merge windows that span shard boundaries. Finally, perform a reduction step to select the globally smallest valid window.
Examples
Input
[1, 2, 3, 4, 5]
Output
6
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we first find all unique elements which are [1, 2, 3, 4, 5]. Then, we find the minimum window that covers all unique elements, which is [1, 2, 3] with a sum of 6.
Input
[1]
Output
1
Explanation: Step-by-step: with input [1], we first find all unique elements which are [1]. Then, we find the minimum window that covers all unique elements, which is [1] with a sum of 1.
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 sliding window with two pointers and hash maps to track frequencies, expanding right until the window is valid then contracting left to minimize length. Each element is processed at most twice, achieving O(N+M) time.
Brute Force Approach
Enumerate every possible subarray, check if it contains all target elements, and keep the smallest valid one. This requires O(N^2) windows and O(M) verification per window, leading to O(N^2·M) time.
Verified Code Solutions
function solution(nums) {
let uniqueElements = new Set(nums);
let minSum = Infinity;
for (let i = 0; i < nums.length; i++) {
let windowSum = 0;
let uniqueInWindow = new Set();
for (let j = i; j < nums.length; j++) {
windowSum += nums[j];
uniqueInWindow.add(nums[j]);
if (uniqueInWindow.size === uniqueElements.size) {
minSum = Math.min(minSum, windowSum);
}
}
}
return minSum;
}class Solution {
public:
int solution(vector<int>& nums) {
set<int> uniqueElements;
for (int num : nums) {
uniqueElements.insert(num);
}
int minSum = INT_MAX;
for (int i = 0; i < nums.size(); i++) {
int windowSum = 0;
set<int> uniqueInWindow;
for (int j = i; j < nums.size(); j++) {
windowSum += nums[j];
uniqueInWindow.insert(nums[j]);
if (uniqueInWindow.size() == uniqueElements.size()) {
minSum = min(minSum, windowSum);
}
}
}
return minSum;
}
};class Solution {
public int solution(int[] nums) {
Set<Integer> uniqueElements = new HashSet<>();
for (int num : nums) {
uniqueElements.add(num);
}
int minSum = Integer.MAX_VALUE;
for (int i = 0; i < nums.length; i++) {
int windowSum = 0;
Set<Integer> uniqueInWindow = new HashSet<>();
for (int j = i; j < nums.length; j++) {
windowSum += nums[j];
uniqueInWindow.add(nums[j]);
if (uniqueInWindow.size() == uniqueElements.size()) {
minSum = Math.min(minSum, windowSum);
}
}
}
return minSum;
}
}def solution(nums):
unique_elements = set(nums)
min_sum = float('inf')
for i in range(len(nums)):
window_sum = 0
unique_in_window = set()
for j in range(i, len(nums)):
window_sum += nums[j]
unique_in_window.add(nums[j])
if len(unique_in_window) == len(unique_elements):
min_sum = min(min_sum, window_sum)
return min_sumfunction solution(nums) {
let uniqueElements = new Set(nums);
let minSum = Infinity;
for (let i = 0; i < nums.length; i++) {
let windowSum = 0;
let uniqueInWindow = new Set();
for (let j = i; j < nums.length; j++) {
windowSum += nums[j];
uniqueInWindow.add(nums[j]);
if (uniqueInWindow.size === uniqueElements.size) {
minSum = Math.min(minSum, windowSum);
}
}
}
return minSum;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.