Matrix Stream Validator 48 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a high-throughput data stream represented as a one-dimensional array of integers, where each integer corresponds to a specific metric value. The system requires a validation check against a global threshold K. Your objective is to determine the exact count of elements in the stream that strictly exceed this threshold. While a linear scan is trivial, the problem context implies a scenario where the data structure or access pattern might benefit from optimized search techniques or monotonic properties, though for this specific counting task, the core requirement is precise enumeration of values greater than K.
Given an array stream of length N and an integer K, return the number of indices i such that stream[i] > K. The solution must be efficient and handle large input sizes within strict time limits. Although the topic is tagged as Binary Search and the pattern as Monotonic Stack, the fundamental operation here is a filtered count. In a real-world 'Matrix Stream Validator' context, this might be a sub-routine where the stream is sorted or has specific monotonic properties, but for this problem, you must implement the most robust and correct counting mechanism.
Input: An array of integers stream and an integer K.
Output: An integer representing the count of elements in stream that are strictly greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Validator 48"
WHY DOES IT MATTER?
Binary search transforms a linear counting problem into a logarithmic boundary search, drastically reducing runtime on large datasets. It is essential in systems where every millisecond counts, such as real‑time analytics or fraud detection pipelines.
OPTIMIZATION CHALLENGE
The key insight is to convert the counting requirement into a boundary‑finding problem: locate the first element > K and compute the count as n − index. This reduces both time and space complexity.
REAL-WORLD CONNECTION
Consider a database index: to find all records with a score above a threshold, the index allows a binary search to jump directly to the first qualifying record, then a simple offset gives the count. This mirrors the stream validator’s need to quickly determine how many metrics exceed K.
When explaining the solution, emphasize that the array’s sorted nature is the lever. Show the interviewer the pseudo‑code for lower_bound and how the final count is derived, highlighting the O(log n) time and O(1) space.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
Binary search is a divide‑and‑conquer technique that operates on sorted data. By repeatedly halving the search interval, it locates a target or a boundary in logarithmic time, O(log n). In this problem, the array is sorted in non‑decreasing order, so the boundary between elements ≤ K and > K can be found by a single binary search. The naive linear scan examines every element, yielding O(n) time, which becomes prohibitive when the stream contains millions of metrics. The optimal paradigm reduces the number of comparisons dramatically by exploiting order: once the first element greater than K is found, all subsequent elements are also greater, allowing us to compute the count as n − index.
The key insight is that the count of elements > K equals the total length minus the index of the first element that exceeds K. Binary search guarantees that this index is found in log n steps, and because we never need to inspect the rest of the array, we avoid the linear cost. This pattern is a classic example of transforming a counting problem into a boundary‑finding problem, which is a powerful tool in algorithm design.
Moreover, binary search’s cache‑friendly nature and constant‑time arithmetic make it suitable for high‑throughput systems where latency matters. By reducing the number of memory accesses, we lower the chance of cache misses, which is critical in distributed data pipelines that process streams in real time.
Interview Questions on This Problem
Q1How would you modify the binary search if the array could contain duplicate values and you need the count of elements strictly greater than K?
The binary search should locate the first index where the value is greater than K. If duplicates of K exist, the search must skip over them. Once that index is found, the count is simply n minus that index. The implementation uses a standard lower_bound style search with a condition arr[mid] <= K to move the left pointer up.
Q2A fintech platform processes transaction amounts in real time. They need to flag the top 5% of transactions exceeding a threshold. How would you adapt the algorithm to find the threshold value itself?
First, determine the rank r = ceil(0.05 × n). Then perform a binary search on the value domain (e.g., min to max transaction) to find the smallest value v such that count of elements > v is ≤ r. This is a selection problem that can be solved with a binary search on values, each step counting with the same O(log n) method, yielding O(log V × log n) overall, where V is the value range.
Q3During an interview at a high‑growth startup, you’re asked to explain why binary search is preferred over a hash‑based approach for this problem. What would you say?
Hashing would require O(n) space to store all elements or a frequency map, and it doesn’t exploit the sorted property of the data. Binary search uses O(1) extra space and runs in O(log n) time, which is far more efficient for large streams where memory and latency are critical. Additionally, hashing would not provide a direct way to count elements greater than K without scanning the entire hash table.
Examples
Input
stream = [12, 4, 7, 20, 3, 15], K = 10
Output
3
Explanation: Iterate through the stream: 12 > 10 (count=1), 4 <= 10, 7 <= 10, 20 > 10 (count=2), 3 <= 10, 15 > 10 (count=3). The final count is 3.
Input
stream = [5, 5, 5, 5], K = 5
Output
0
Explanation: All elements are equal to K. Since the condition is strictly greater than (>, not >=), no elements satisfy the condition. The count remains 0.
Input
stream = [-100, -50, 0, 50, 100], K = -1
Output
2
Explanation: Check each element against -1: -100 <= -1, -50 <= -1, 0 > -1 (count=1), 50 > -1 (count=2), 100 > -1 (count=3). Wait, 0 > -1 is true. Let's re-verify. -100 (no), -50 (no), 0 (yes, 0 > -1), 50 (yes), 100 (yes). The count is 3. Correction: The output should be 3. Let's adjust the example to be clearer or fix the output. Let's use K = 0. stream = [-100, -50, 0, 50, 100], K = 0. -100 (no), -50 (no), 0 (no, 0 is not > 0), 50 (yes), 100 (yes). Output 2. This is better.
Constraints
- 1 <= stream.length <= 10^5
- -10^9 <= stream[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Perform a binary search to find the first index where the element exceeds K, then compute the count as n minus that index. This runs in O(log n) time with O(1) extra space.
Brute Force Approach
Iterate through each element of the array, increment a counter whenever the element is greater than K. This takes O(n) time and O(1) space.
Verified Code Solutions
function solution(nums, K) {
let count = 0;
for (let num of nums) {
if (num > K) {
count++;
}
}
return count;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int count = 0;
for (int num : nums) {
if (num > K) {
count++;
}
}
return count;
}
};class Solution {
public int solution(int[] nums, int K) {
int count = 0;
for (int num : nums) {
if (num > K) {
count++;
}
}
return count;
}
}def solution(nums, K):
count = 0
for num in nums:
if num > K:
count += 1
return countfunction solution(nums, K) {
let count = 0;
for (let num of nums) {
if (num > K) {
count++;
}
}
return count;
}Asked in Top Tech Interviews
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