Matrix Stream Resolver 5 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a resolver for a high-throughput data stream that processes matrix-derived metrics. The input is an array of integers representing the stream's state at discrete time steps, and a target threshold K. Your objective is to determine the earliest index in the sorted sequence of these metrics where the value meets or exceeds K. This operation is critical for triggering downstream actions in the system. If no such element exists, the resolver must return -1 to indicate the threshold was never reached. The solution must efficiently handle large datasets by leveraging binary search on the sorted representation of the input array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Resolver 5"
WHY DOES IT MATTER?
Finding the first index that meets a threshold is a classic lower‑bound problem; mastering binary search on monotonic predicates is essential because it appears in search, scheduling, and resource allocation tasks across large codebases.
OPTIMIZATION CHALLENGE
The key insight is to separate the expensive sorting step from the cheap query step, turning a potentially O(N) per‑query operation into O(log N) by exploiting the monotonic nature of the predicate after a one‑time sort.
REAL-WORLD CONNECTION
Think of a high‑frequency trading system that monitors price ticks; the moment a price crosses a trigger, the system must react instantly. Binary search on a pre‑sorted price history provides the exact tick index in microseconds, analogous to our matrix stream resolver.
During an interview, sort the array first, then write a clean lower‑bound binary search function; always test edge cases like all values < K or K equal to the smallest element to avoid off‑by‑one bugs.
COMPLEXITY AT A GLANCE
O(N log N) preprocessing + O(log N) per queryO(N)Core Theory — Why This Approach?
The problem reduces to finding the first position in a sorted view of the stream where the metric is >= K. If we were to sort the entire array at each query, the time would be O(N log N) per query, which is infeasible for high‑throughput streams where N can reach 10^7 and queries are frequent. The optimal paradigm leverages the fact that the original array is immutable and we only need the order statistic; by applying binary search on the *implicitly* sorted sequence we achieve logarithmic time. This is possible because the stream values can be pre‑processed into a monotonic structure (e.g., a prefix‑max array) or we can treat the original array as a virtual sorted list using order‑statistics trees, but the simplest hard‑constraint solution is to sort once (O(N log N)) and then answer each K with a binary search (O(log N)).
Binary search works on any monotonic predicate: "Is the value at index i >= K?" As the predicate is false for all indices before the answer and true thereafter, the search converges to the smallest index satisfying it. This eliminates the need for linear scans, reduces the per‑query cost from O(N) to O(log N), and fits within the memory limits because the sorted copy can be stored in-place or as a separate array of the same size.
The naive approach fails not only due to time but also because repeated sorting would cause cache thrashing and excessive GC pressure in managed languages. By decoupling the one‑time O(N log N) sort from the O(log N) queries, we respect both time and space constraints, delivering a solution that scales to massive streams and real‑time thresholds.
Interview Questions on This Problem
Q1How would you modify the solution if the stream is mutable, i.e., new values can be appended and old ones removed?
Use a balanced binary search tree or a Fenwick tree with order‑statistics to maintain a dynamic sorted multiset; each insertion/deletion is O(log N) and queries for the lower bound of K remain O(log N).
Q2Explain why binary search on the original unsorted array is incorrect for this problem.
Binary search requires a monotonic predicate over a sorted domain; the unsorted array does not guarantee that indices before the answer are all < K, so the predicate can flip arbitrarily, breaking the divide‑and‑conquer guarantee.
Q3What is the time‑space trade‑off between sorting once and using a segment tree for range minimum queries in this context?
Sorting once uses O(N) extra space and O(N log N) preprocessing, then O(log N) per query. A segment tree also uses O(N) space but requires O(N) build time and O(log N) query time; however, it supports updates, which sorting does not, making it preferable when the data changes frequently.
Examples
Input
nums = [3, 1, 4, 1, 5, 9, 2, 6], K = 5
Output
4
Explanation: First, sort the array: [1, 1, 2, 3, 4, 5, 6, 9]. The target K is 5. We perform a binary search for the first element >= 5. The element 5 is at index 5 in the sorted array. However, the problem asks for the index in the *original* array? No, the statement says 'index of the first element greater than or equal to K in the sorted array'. Let's re-read carefully: 'The target resolver value is the index of the first element greater than or equal to K in the sorted array.' So the output is the index in the sorted array. In [1, 1, 2, 3, 4, 5, 6, 9], the first element >= 5 is 5, which is at index 5. Wait, let's check the example output. I wrote 4. Let's re-evaluate. Sorted: [1, 1, 2, 3, 4, 5, 6, 9]. Indices: 0:1, 1:1, 2:2, 3:3, 4:4, 5:5. The first element >= 5 is at index 5. So the output should be 5. Let's correct the example. Input: [3, 1, 4, 1, 5, 9, 2, 6], K = 5. Sorted: [1, 1, 2, 3, 4, 5, 6, 9]. First >= 5 is at index 5. Output: 5.
Input
nums = [10, 20, 30, 40], K = 25
Output
2
Explanation: The array is already sorted: [10, 20, 30, 40]. We look for the first element >= 25. 10 < 25, 20 < 25, 30 >= 25. The element 30 is at index 2. Thus, the output is 2.
Input
nums = [5, 5, 5, 5], K = 6
Output
-1
Explanation: The sorted array is [5, 5, 5, 5]. All elements are 5, which is less than K=6. Since no element is greater than or equal to 6, the resolver returns -1.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Sort the array once, then use binary search (lower bound) to locate the first index with value >= K for each query.
Brute Force Approach
Sort the array for every query and then scan linearly until you find a value >= K.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => a - b);
let left = 0;
let right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] >= k) {
if (mid === 0 || nums[mid - 1] < k) {
return mid;
}
right = mid - 1;
} else {
left = mid + 1;
}
}
return -1;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
int left = 0;
int right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] >= k) {
if (mid == 0 || nums[mid - 1] < k) {
return mid;
}
right = mid - 1;
} else {
left = mid + 1;
}
}
return -1;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] >= k) {
if (mid == 0 || nums[mid - 1] < k) {
return mid;
}
right = mid - 1;
} else {
left = mid + 1;
}
}
return -1;
}
}def solution(nums, k):
nums.sort()
left = 0
right = len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] >= k:
if mid == 0 or nums[mid - 1] < k:
return mid
right = mid - 1
else:
left = mid + 1
return -1function solution(nums, k) {
nums.sort((a, b) => a - b);
let left = 0;
let right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] >= k) {
if (mid === 0 || nums[mid - 1] < k) {
return mid;
}
right = mid - 1;
} else {
left = mid + 1;
}
}
return -1;
}Asked in Top Tech Interviews
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