Subtree Height Evaluator Optimizer 5 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing a distributed sensor network where data packets are transmitted as a continuous stream of characters. The network requires a specific set of critical sensor identifiers (represented by a target string) to be present in any valid transmission window to ensure data integrity. Your goal is to identify the shortest contiguous segment of the input stream that contains all the required identifiers, accounting for their exact frequencies. If the input stream does not contain all required identifiers, the system must return an empty string. This problem models the Minimum Window Substring pattern, where you must efficiently slide a window over the input to minimize its length while satisfying the coverage constraint of the target set.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Subtree Height Evaluator Optimizer 5"
WHY DOES IT MATTER?
The sliding‑window pattern efficiently solves problems that require finding optimal subarrays or substrings under a coverage constraint, turning exponential brute‑force checks into linear scans.
OPTIMIZATION CHALLENGE
The key insight is that the window only needs to move forward; by tracking how many required characters are still missing, we can decide when to shrink the left side without re‑examining the whole window, reducing both time and auxiliary space.
REAL-WORLD CONNECTION
Think of a security checkpoint where a guard must see all required ID badges before allowing a convoy to pass; the guard opens the gate as soon as the convoy contains every badge, then tightens the convoy to the smallest possible size—mirroring window expansion and contraction.
During an interview, keep two hash maps: one for target frequencies and one for current window counts, and maintain a counter of how many distinct characters have met their required count—this lets you check window validity in O(1).
COMPLEXITY AT A GLANCE
O(N + M)O(K)Core Theory — Why This Approach?
The problem is a classic instance of the minimum window substring, which can be modeled as a sliding‑window over the source stream while maintaining a frequency map of required characters from the target identifier set. A naive solution would recompute the presence of all target characters for every possible window, leading to O(N^2) time on a stream of length N, which is infeasible for large‑scale sensor data. The optimal paradigm leverages two pointers (left and right) that expand the window to include required characters and contract it to discard unnecessary prefix characters, updating counts in O(1) per movement. This approach ensures each character is visited at most twice, yielding linear time complexity and a compact hash‑map or array for character frequencies, which is the essence of the sliding‑window technique for substring coverage problems.
Interview Questions on This Problem
Q1How would you modify the minimum window substring algorithm to handle Unicode characters beyond the ASCII range?
Use a hash map (e.g., unordered_map<char32_t, int>) instead of a fixed‑size array to store frequencies, and ensure the sliding window updates the map based on code points; the rest of the algorithm remains unchanged.
Q2Explain how you could adapt the solution to return all minimal windows instead of just the shortest one.
After finding a minimal window, continue sliding the left pointer while maintaining the required counts; each time the window becomes valid again, record its boundaries. This may increase time to O(N + K) where K is the number of minimal windows, but still linear in the input size.
Q3In a distributed sensor network, why might you prefer a streaming version of this algorithm over batch processing?
A streaming version processes characters on the fly with O(1) amortized update cost and constant memory per window, enabling real‑time detection of integrity windows without storing the entire stream, which is critical for low‑latency, high‑throughput systems.
Examples
Input
s = "ADOBECODEBANC", t = "ABC"
Output
"BANC"
Explanation: The target 'ABC' requires one 'A', one 'B', and one 'C'. The window 'BANC' (indices 9-12) is the shortest substring containing all these characters. 'B' is at index 9, 'A' at 10, 'N' at 11, 'C' at 12. Length is 4. Other windows like 'ADOBEC' are longer.
Input
s = "a", t = "a"
Output
"a"
Explanation: The input string is exactly the target. The window covers the entire string, which is the only possible valid window. Length is 1.
Input
s = "aa", t = "aa"
Output
"aa"
Explanation: The target requires two 'a's. The input has two 'a's. The entire string is the minimum window. Length is 2.
Input
s = "ab", t = "c"
Output
""
Explanation: The target 'c' is not present in the input 'ab'. No valid window exists, so return an empty string.
Constraints
- 1 <= s.length <= 10^5
- 1 <= t.length <= 10^5
- s and t consist of only uppercase and lowercase English letters.
- The answer is guaranteed to be unique.
Optimal Approach & Strategy
Use a sliding window with two pointers and a frequency map to expand until the window is valid, then contract to minimize length, updating the answer on the fly.
Brute Force Approach
Generate every possible substring, check if it contains all target characters, and keep the shortest valid one.
Verified Code Solutions
function solution(nums) {
let maxSum = -Infinity;
let minLen = Infinity;
let windowSum = 0;
let left = 0;
for (let right = 0; right < nums.length; right++) {
windowSum += nums[right];
while (windowSum > maxSum) {
maxSum = windowSum;
if (right - left + 1 < minLen) {
minLen = right - left + 1;
}
windowSum -= nums[left++];
}
}
return minLen;
}class Solution {
public:
int solution(vector<int>& nums) {
int maxSum = INT_MIN;
int minLen = INT_MAX;
int windowSum = 0;
int left = 0;
for (int right = 0; right < nums.size(); right++) {
windowSum += nums[right];
while (windowSum > maxSum) {
maxSum = windowSum;
if (right - left + 1 < minLen) {
minLen = right - left + 1;
}
windowSum -= nums[left++];
}
}
return minLen;
}
};class Solution {
public int solution(int[] nums) {
int maxSum = Integer.MIN_VALUE;
int minLen = Integer.MAX_VALUE;
int windowSum = 0;
int left = 0;
for (int right = 0; right < nums.length; right++) {
windowSum += nums[right];
while (windowSum > maxSum) {
maxSum = windowSum;
if (right - left + 1 < minLen) {
minLen = right - left + 1;
}
windowSum -= nums[left++];
}
}
return minLen;
}
}def solution(nums):
max_sum = float('-inf')
min_len = float('inf')
window_sum = 0
left = 0
for right in range(len(nums)):
window_sum += nums[right]
while window_sum > max_sum:
max_sum = window_sum
if right - left + 1 < min_len:
min_len = right - left + 1
window_sum -= nums[left]
left += 1
return min_lenfunction solution(nums) {
let maxSum = -Infinity;
let minLen = Infinity;
let windowSum = 0;
let left = 0;
for (let right = 0; right < nums.length; right++) {
windowSum += nums[right];
while (windowSum > maxSum) {
maxSum = windowSum;
if (right - left + 1 < minLen) {
minLen = right - left + 1;
}
windowSum -= nums[left++];
}
}
return minLen;
}Asked in Top Tech Interviews
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