Protocol Pipeline Resolver 26 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and pipeline metrics, construct an optimal algorithm to evaluate and compute the target resolver value under given operational constraints. The algorithm should iterate through the input array and count the number of elements that are less than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Pipeline Resolver 26"
WHY DOES IT MATTER?
Efficient range counting underpins many search‑heavy services.
OPTIMIZATION CHALLENGE
Transforming a linear scan into O(log N) via ordering cuts runtime by orders of magnitude.
REAL-WORLD CONNECTION
Databases use indexed columns to binary‑search sorted rows for range queries.
Always sort once and reuse the sorted view for all subsequent threshold checks.
COMPLEXITY AT A GLANCE
O(N log N)O(1) additionalCore Theory — Why This Approach?
Counting elements ≤ K is a classic selection problem. A naïve double‑loop that compares each element with every other leads to O(N²) time, which quickly becomes infeasible for large N. The optimal paradigm leverages ordering: by sorting the array once (O(N log N)) we can then apply binary search to locate the first element greater than K, yielding the count as the index of that element. This reduces the per‑query cost to O(log N) and, when only a single query exists, the overall complexity remains O(N log N) with O(1) extra space beyond the sort.
Interview Questions on This Problem
Q1Why is sorting followed by binary search preferable to scanning the array for each query?
Sorting creates a monotonic structure that enables O(log N) look‑ups, whereas scanning is O(N) per query. For multiple queries the amortized cost drops dramatically.
Q2What binary‑search variant is used to count elements ≤ K?
Upper‑bound (first element > K) is used; its index equals the count of ≤ K. It can be implemented with standard library functions or a custom loop.
Q3How does the algorithm behave with duplicate values equal to K?
All duplicates are included because the upper‑bound stops at the first element greater than K. Thus the count correctly reflects every occurrence of K.
Examples
Input
[1, 2, 3, 4, 5]
Output
3
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and K = 3, we iterate through the array and count the numbers less than or equal to 3. The numbers 1, 2, and 3 satisfy this condition, so the output is 3.
Input
[10, 20, 30, 40, 50]
Output
5
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50] and K = 50, we iterate through the array and count the numbers less than or equal to 50. All numbers in the array satisfy this condition, so the output is 5.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort the array and perform an upper‑bound binary search to get the count in O(log N) after O(N log N) preprocessing.
Brute Force Approach
Iterate through the array and increment a counter for each element ≤ K, resulting in O(N) time.
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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