Network Protocol Validator 7 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a validation routine for a distributed network protocol. The system receives a sorted array of integer latency measurements, latencies, and a target threshold, target. Your goal is to determine the index of the first latency value that is strictly greater than target. This index represents the boundary where the protocol switches from a 'stable' state to a 'congested' state. If no such value exists, return -1. If all values are less than or equal to the target, return the length of the array. You must solve this using binary search to ensure O(log n) time complexity, as the dataset can be extremely large.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Validator 7"
WHY DOES IT MATTER?
Finding the first greater element is a classic boundary‑search pattern used in range queries.
OPTIMIZATION CHALLENGE
The challenge is reducing a linear scan to logarithmic time by leveraging sorted order.
REAL-WORLD CONNECTION
It mirrors locating the cut‑off point where response times exceed SLA thresholds in monitoring dashboards.
Initialize low=0, high=n and use mid=(low+high)/2; move high=mid on >= and low=mid+1 otherwise.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
Binary search exploits the monotonic property of sorted arrays to eliminate half of the remaining candidates at each step, achieving logarithmic time. By repeatedly comparing the middle element with the target, we can decide whether the desired index lies to the left or right, converging on the first element strictly greater than the target.
A naive linear scan checks each element sequentially, which is O(n) and becomes prohibitive for large latency logs common in distributed systems. The optimal paradigm—upper bound binary search—maintains two pointers defining a search window and narrows it until the lower bound points to the smallest index where latency > target, guaranteeing O(log n) time with O(1) extra space.
Interview Questions on This Problem
Q1How does upper bound binary search differ from standard binary search for a value?
Upper bound seeks the first element greater than the target, not equality. It adjusts the high pointer on >= comparisons to ensure the final low index points to the correct boundary.
Q2What edge cases must be handled when implementing this search?
When all elements are ≤ target, the answer should be the array length (no greater element). When the first element itself is > target, the answer is index 0.
Q3Why is O(log n) preferable to O(n) in high‑throughput network monitoring?
Latency logs can contain millions of entries; O(log n) reduces query time from seconds to microseconds. This enables real‑time threshold alerts without overwhelming the system.
Examples
Input
latencies = [1, 3, 5, 7, 9], target = 4
Output
2
Explanation: The array is sorted. We search for the first element > 4. Index 0 (1) <= 4. Index 1 (3) <= 4. Index 2 (5) > 4. Thus, the first index where the condition is met is 2.
Input
latencies = [10, 20, 30, 40], target = 5
Output
0
Explanation: The first element, 10, is already greater than the target 5. Therefore, the index is 0.
Input
latencies = [1, 2, 3, 4], target = 10
Output
-1
Explanation: All elements in the array are less than or equal to 10. There is no element strictly greater than the target. According to the problem statement, we return the length of the array, which is 4.
Input
latencies = [5, 5, 5, 5], target = 5
Output
-1
Explanation: The condition is 'strictly greater than'. Since all elements are equal to 5, none are greater than 5. The boundary is at the end of the array, so we return the length, 4.
Constraints
- 1 <= latencies.length <= 10^5
- 1 <= latencies[i] <= 10^9
- 1 <= target <= 10^9
- latencies is sorted in non-decreasing order
Optimal Approach & Strategy
Perform an upper‑bound binary search: maintain low and high pointers, compute mid, and move high to mid when latencies[mid] > target, otherwise move low to mid+1.
Brute Force Approach
Iterate from the start of the array until you encounter a value > target and return its index.
Verified Code Solutions
function findFirstGreater(latencies, target) {
let left = 0;
let right = latencies.length - 1;
let ans = -1;
while (left <= right) {
const mid = Math.floor((left + right) / 2);
if (latencies[mid] > target) {
ans = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return ans;
}#include <vector>
using namespace std;
int findFirstGreater(const vector<int>& latencies, int target) {
int left = 0, right = (int)latencies.size() - 1;
int ans = -1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (latencies[mid] > target) {
ans = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return ans;
}
public class Solution {
public int findFirstGreater(int[] latencies, int target) {
int left = 0;
int right = latencies.length - 1;
int ans = -1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (latencies[mid] > target) {
ans = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return ans;
}
}
def find_first_greater(latencies, target):
left, right = 0, len(latencies) - 1
ans = -1
while left <= right:
mid = (left + right) // 2
if latencies[mid] > target:
ans = mid
right = mid - 1
else:
left = mid + 1
return ans
function findFirstGreater(latencies, target) {
let left = 0;
let right = latencies.length - 1;
let ans = -1;
while (left <= right) {
const mid = Math.floor((left + right) / 2);
if (latencies[mid] > target) {
ans = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return ans;
}
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