Sensor Packet Evaluator 18 — Problem Statement & Solution Guide

Binary SearchMedium2D Grid DP

TimeO(log n)

SpaceO(1)

Problem Description

A distributed sensor network transmits a sorted sequence of integer packet identifiers. The system administrator needs to audit the data stream to identify the count of packets that appear strictly after the first identifier exceeding a specific threshold K. Given a non-decreasing array of integers representing the packet IDs and a target integer K, compute the number of elements located to the right of the first element that is strictly greater than K. If no element in the sequence exceeds K, the result is 0.

**Input**

- The first line contains an integer n, representing the total number of packets.

- The second line contains n space-separated integers, representing the sorted packet identifiers.

- The third line contains an integer K, the threshold value.

**Output**

- Print a single integer representing the count of packets strictly to the right of the first packet ID greater than K.

DSA Pattern Breakdown

"Sensor Packet Evaluator 18"

medium

WHY DOES IT MATTER?

Upper‑bound binary search is a fundamental pattern for answering "how many elements satisfy a condition" in sorted data. It converts a potentially linear count into a logarithmic query, enabling scalable analytics on massive logs, time‑series, or sensor streams.

OPTIMIZATION CHALLENGE

The key insight is to treat the problem as a search for an insertion point rather than a value match. By maintaining a half‑open interval and moving the high pointer to mid when arr[mid] > K, we shrink the search space until low equals the desired index.

REAL-WORLD CONNECTION

Imagine a distributed logging service that stores timestamps in order. To know how many logs occurred after a certain cutoff time, you locate the first timestamp greater than the cutoff and count the remaining entries—exactly the upper‑bound problem.

When coding binary search, always write it as a reusable function (e.g., upperBound(arr, K)). Test it against edge cases like empty arrays, all elements ≤ K, and all elements > K to avoid off‑by‑one bugs during an interview.

COMPLEXITY AT A GLANCE

⏱ Time:O(log n)

💾 Space:O(1)

Core Theory — Why This Approach?

Binary search exploits the monotonic property of a sorted array to locate a target position in logarithmic time. In this problem we are not looking for an exact match but for the first element that exceeds a threshold K; this is a classic "upper bound" query. By repeatedly halving the search interval we can pinpoint the smallest index i such that arr[i] > K, and the answer is simply the number of elements to the right, i.e., n - i. A naive linear scan would examine each element until the condition is met, leading to O(n) time, which becomes prohibitive for large streams of sensor packets that can reach millions of entries. The optimal paradigm therefore combines the sorted guarantee with binary search to achieve O(log n) time while using O(1) extra space.

The binary‑search implementation must be careful with edge cases: when all elements are ≤ K the upper bound does not exist, and when the first element itself is > K the answer equals the full length of the array. By initializing low = 0 and high = n and using a half‑open interval [low, high), we can converge on the correct insertion point without overflow. This approach is widely used in standard libraries (e.g., C++'s std::upper_bound) and forms a building block for many range‑query problems in competitive programming and system design.

Interview Questions on This Problem

Q1How would you find the number of elements greater than K in a non‑decreasing array without scanning the entire array?

Use binary search to locate the first index where arr[i] > K (upper bound). The count is then array.length - i. If no such index exists, the count is 0.

Q2What modifications are needed if the array can contain duplicate values and you must count elements strictly greater than K?

The upper‑bound binary search already skips duplicates of K because it searches for the first element > K. No extra handling is required; duplicates of K are naturally excluded from the count.

Q3Explain why a linear scan might still be acceptable in a real‑time sensor system despite its O(n) complexity.

If the data size per audit window is bounded (e.g., a few thousand packets) or the system processes only a single query per batch, the constant factors of a linear scan may be lower than the overhead of setting up binary search, making it acceptable. However, for high‑throughput streams or multiple queries, the logarithmic guarantee of binary search becomes essential.

Examples

Example 1

Input

5
1 3 5 7 9
4

Output

Explanation: The array is [1, 3, 5, 7, 9]. The threshold K is 4. The first element strictly greater than 4 is 5, located at index 2. The elements strictly to the right of index 2 are 7 and 9. Wait, the problem asks for elements to the right of the *first* element greater than K. The first element > 4 is 5 (index 2). The elements to the right of index 2 are at indices 3 and 4. That is 2 elements. Let me re-read carefully: 'how many elements of nums lie strictly to the right of the first element that is greater than K'. Let's re-evaluate Example 1. Array: [1, 3, 5, 7, 9] K: 4 First element > 4 is 5 (index 2). Elements strictly to the right of index 2 are 7 (index 3) and 9 (index 4). Count is 2. Let's create a new example to be safe and clear. Example 1: Input: 5 1 3 5 7 9 4 Output: 2 Explanation: The first element greater than 4 is 5 at index 2. The elements to the right are 7 and 9. Count is 2.

Example 2

Input

6
2 2 4 4 6 8
3

Output

Explanation: The array is [2, 2, 4, 4, 6, 8]. The threshold K is 3. The first element strictly greater than 3 is 4, located at index 2. The elements strictly to the right of index 2 are 4 (index 3), 6 (index 4), and 8 (index 5). The count is 3. Wait, let me check indices. Index 0:2, 1:2, 2:4, 3:4, 4:6, 5:8. First > 3 is at index 2. Right of index 2 are indices 3, 4, 5. Count is 3. Let's adjust the example to be clearer. Input: 6 2 2 4 4 6 8 3 Output: 3

Example 3

Input

4
10 20 30 40
50

Output

Explanation: The array is [10, 20, 30, 40]. The threshold K is 50. No element in the array is strictly greater than 50. Therefore, the count is 0.

Constraints

1 <= n <= 10^5
-10^9 <= nums[i] <= 10^9
-10^9 <= K <= 10^9
The array nums is sorted in non-decreasing order.

Optimal Approach & Strategy

Apply binary search to locate the upper bound index where arr[i] > K, then compute n - i. This runs in O(log n) time with O(1) extra space.

Brute Force Approach

Iterate from the start of the array until you find the first element > K, then return the remaining length. This scans up to O(n) elements in the worst case.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(log n)

/**
 * @param {number[]} packets
 * @param {number} K
 * @return {number}
 */
var countAfterThreshold = function(packets, K) {
    const n = packets.length;
    let left = 0, right = n - 1;
    let firstGreaterIndex = n;
    
    while (left <= right) {
        const mid = Math.floor((left + right) / 2);
        if (packets[mid] > K) {
            firstGreaterIndex = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    
    return n - firstGreaterIndex;
};

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int countAfterThreshold(vector<int>& packets, int K) {
        int n = packets.size();
        int left = 0, right = n - 1;
        int firstGreaterIndex = n;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (packets[mid] > K) {
                firstGreaterIndex = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        
        return n - firstGreaterIndex;
    }
};

import java.util.*;

class Solution {
    public int countAfterThreshold(int[] packets, int K) {
        int n = packets.length;
        int left = 0, right = n - 1;
        int firstGreaterIndex = n;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (packets[mid] > K) {
                firstGreaterIndex = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        
        return n - firstGreaterIndex;
    }
}

from typing import List

class Solution:
    def countAfterThreshold(self, packets: List[int], K: int) -> int:
        n = len(packets)
        left, right = 0, n - 1
        first_greater_index = n
        
        while left <= right:
            mid = (left + right) // 2
            if packets[mid] > K:
                first_greater_index = mid
                right = mid - 1
            else:
                left = mid + 1
        
        return n - first_greater_index

/**
 * @param {number[]} packets
 * @param {number} K
 * @return {number}
 */
var countAfterThreshold = function(packets, K) {
    const n = packets.length;
    let left = 0, right = n - 1;
    let firstGreaterIndex = n;
    
    while (left <= right) {
        const mid = Math.floor((left + right) / 2);
        if (packets[mid] > K) {
            firstGreaterIndex = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    
    return n - firstGreaterIndex;
};

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Sensor Packet Evaluator 18 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Sensor Packet Evaluator 18 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor