Tome Signal Tracker 40 — Problem Statement & Solution Guide

QueueMediumBFS / Union Find

TimeO(N)

SpaceO(1)

Problem Description

Tome Signal Tracker 40

You are given a queue of tomes, each tome carrying an integer signal value. A monitoring system records the signal values in the order the tomes arrive. For a given threshold K, the system must compute the total signal strength contributed by all tomes whose signal value exceeds K. The result is the sum of all such values.

Input

The first line contains two integers N and K, where N is the number of tomes in the queue and K is the threshold value. The second line contains N space‑separated integers representing the signal values of the tomes in arrival order.

Output

Print a single integer: the sum of all signal values that are strictly greater than K.

The task is to process the queue efficiently, ensuring that the algorithm runs in linear time with respect to N and uses only constant additional space beyond the input array.

DSA Pattern Breakdown

"Tome Signal Tracker 40"

medium

WHY DOES IT MATTER?

Linear‑time aggregation over a queue is a foundational pattern for streaming analytics, where you must compute metrics on-the-fly without revisiting data. Mastering this pattern prevents over‑engineering solutions that waste time and memory.

OPTIMIZATION CHALLENGE

The key insight is recognizing that the predicate (value > K) is stateless and does not depend on other elements, so a single pass suffices. Any attempt to sort, bucket, or use nested loops adds unnecessary overhead.

REAL-WORLD CONNECTION

Think of a network router that monitors packet sizes and needs to report the total bandwidth used by packets larger than a certain MTU. The router processes packets in arrival order, summing only those that exceed the threshold, exactly like this queue problem.

During an interview, write the loop that dequeues each element, checks the condition, updates the sum, and immediately re‑enqueues if the original queue must be preserved. This demonstrates both correctness and respect for the data structure's contract.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to a single-pass aggregation over a linear data structure. Since the queue preserves arrival order, we can process each tome exactly once, checking whether its signal exceeds the threshold K and, if so, adding it to a running total. A naive solution might attempt to sort the values or use nested loops to compare each element against every other, which inflates the time complexity to O(N log N) or O(N^2) and quickly becomes infeasible for large N. The optimal paradigm leverages the fact that the condition (value > K) is independent of other elements, allowing a straightforward linear scan that runs in O(N) time and O(1) auxiliary space.

Interview Questions on This Problem

Q1How would you modify the solution if the queue needed to support real‑time queries for the sum of values greater than K after each insertion?

Maintain a running sum of values > K as elements are enqueued. For each new element, compare it to K; if it exceeds K, add it to the sum. This keeps query time O(1) while insertion remains O(1).

Q2What changes are required if the threshold K can change dynamically between queries?

Store the values in a balanced BST or a Fenwick tree keyed by the signal value, along with prefix sums. Each query then becomes a range‑sum query for (K, ∞), achievable in O(log N) time, while insertions stay O(log N).

Q3Explain how you would handle the problem if the input stream is too large to fit in memory.

Process the stream in a streaming fashion: read each value, compare to K, and update a 64‑bit accumulator. No storage of individual elements is needed, guaranteeing O(1) memory regardless of stream size.

Examples

Example 1

Input

5 3
1 4 5 2 3

Output

Explanation: Only the values 4 and 5 are greater than 3. Their sum is 4+5=9.

Example 2

Input

7 0
-1 0 1 2 3 4 5

Output

Explanation: Values greater than 0 are 1, 2, 3, 4, 5. Their sum is 1+2+3+4+5=15.

Example 3

Input

4 10
11 12 13 14

Output

Explanation: All four values exceed 10. Sum is 11+12+13+14=50.

Example 4

Input

6 5
5 5 5 5 5 5

Output

Explanation: No value is greater than 5, so the sum is 0.

Constraints

1 <= N <= 100000
-1000000000 <= nums[i] <= 1000000000
-1000000000 <= K <= 1000000000
The answer fits in a 64‑bit signed integer.

Optimal Approach & Strategy

Iterate through the queue once, adding each value > K to a running sum – O(N) time and O(1) extra space.

Brute Force Approach

Sort the entire list of signals and then binary‑search for the first value > K, summing the tail – O(N log N) time and O(N) space.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

/**
 * @param {number[]} tomes
 * @param {number} K
 * @return {number}
 */
function solve(tomes, K) {
    let total = 0;
    for (let signal of tomes) {
        if (signal > K) {
            total += signal;
        }
    }
    return total;
}

// Example usage
const tomes = [1, 4, 5, 2, 3];
const K = 3;
console.log(solve(tomes, K));

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

class Solution {
public:
    int solve(vector<int>& tomes, int K) {
        int total = 0;
        for (int signal : tomes) {
            if (signal > K) {
                total += signal;
            }
        }
        return total;
    }
};

int main() {
    int n, K;
    cin >> n >> K;
    vector<int> tomes(n);
    for (int i = 0; i < n; ++i) {
        cin >> tomes[i];
    }
    Solution sol;
    cout << sol.solve(tomes, K) << endl;
    return 0;
}

import java.util.*;

class Solution {
    public int solve(int[] tomes, int K) {
        int total = 0;
        for (int signal : tomes) {
            if (signal > K) {
                total += signal;
            }
        }
        return total;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int K = sc.nextInt();
        int[] tomes = new int[n];
        for (int i = 0; i < n; i++) {
            tomes[i] = sc.nextInt();
        }
        Solution sol = new Solution();
        System.out.println(sol.solve(tomes, K));
    }
}

from typing import List

class Solution:
    def solve(self, tomes: List[int], K: int) -> int:
        total = 0
        for signal in tomes:
            if signal > K:
                total += signal
        return total

if __name__ == "__main__":
    n, K = map(int, input().split())
    tomes = list(map(int, input().split()))
    sol = Solution()
    print(sol.solve(tomes, K))

/**
 * @param {number[]} tomes
 * @param {number} K
 * @return {number}
 */
function solve(tomes, K) {
    let total = 0;
    for (let signal of tomes) {
        if (signal > K) {
            total += signal;
        }
    }
    return total;
}

// Example usage
const tomes = [1, 4, 5, 2, 3];
const K = 3;
console.log(solve(tomes, K));

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Tome Signal Tracker 40 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Queue

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Tome Signal Tracker 40 — 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