Tome Signal Validator 24 — Problem Statement & Solution Guide
Problem Description
You are provided with an array of integers representing signal amplitudes and an integer K. Your task is to compute the 'validator value' by identifying the first K elements in the array that are strictly greater than K. Sum these selected elements to produce the final result. If fewer than K elements satisfy the condition, sum all available qualifying elements. The selection process must traverse the array from left to right, ensuring the 'first' occurrence criterion is strictly adhered to. This problem models a scenario where early high-magnitude signals are prioritized for validation, and the threshold K acts as both the selection criterion and the count limit.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Validator 24"
WHY DOES IT MATTER?
This pattern is essential for problems involving 'first N' or 'top N' items based on a simple predicate. It teaches the importance of early termination and avoiding over-engineering. Many candidates instinctively reach for complex data structures like heaps or balanced BSTs, but recognizing that a simple linear scan suffices demonstrates algorithmic maturity and efficiency awareness.
OPTIMIZATION CHALLENGE
The key insight is the early termination condition. Once the count of selected elements reaches K, the loop must break immediately. Failing to do so results in unnecessary O(N) traversal even when the answer is determined in O(K) time. Additionally, ensuring the comparison is 'strictly greater than' (>) and not 'greater than or equal to' (>=) is a subtle but critical optimization in correctness.
REAL-WORLD CONNECTION
This is analogous to a network packet filter that drops packets after the first K valid packets are logged for debugging. The system doesn't need to analyze the entire traffic stream; it only needs to capture the initial burst of valid data. This is common in intrusion detection systems (IDS) where early detection is critical.
In interviews, explicitly state your early termination logic. Say, 'I will break the loop as soon as I have collected K elements to optimize for the best-case scenario.' This shows you are thinking about performance boundaries, not just correctness. Also, clarify the behavior when fewer than K elements qualify, as this is a common point of ambiguity.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem 'Tome Signal Validator 24' is a classic instance of a linear scan with early termination, a fundamental pattern in greedy algorithms where the optimal solution is found by making locally optimal choices at each step. The core logic dictates that we must traverse the array sequentially to identify elements strictly greater than K. The 'greedy' aspect lies in the immediate acceptance of any qualifying element until the quota of K elements is met. This approach is optimal because the problem does not require global optimization (like finding the K largest elements) but rather a specific count of the first K valid items. Any deviation from a single-pass linear scan, such as sorting or using a heap, would introduce unnecessary overhead and violate the 'first K' constraint, which implies order preservation.
Interview Questions on This Problem
Q1At a fintech platform processing high-frequency trading signals, how would you modify this algorithm to handle a stream of data where the array is not fully available in memory?
You would implement a streaming version of the algorithm. Maintain a counter for qualifying elements and a running sum. As each new signal arrives, check if it is strictly greater than K. If it is, increment the counter and add to the sum. If the counter reaches K, you can stop processing further signals for this validation cycle or switch to a 'passive' mode if the requirement is only for the first K. This maintains O(1) space complexity and O(1) time per element, suitable for real-time systems.
Q2In a high-growth engineering startup, if the array size is 10^9 and K is 10^5, what are the memory and time implications of this approach compared to a sorting-based approach?
The linear scan approach has O(N) time complexity and O(1) space complexity (excluding input storage). A sorting-based approach would require O(N log N) time and potentially O(N) extra space for the sort. For N=10^9, the linear scan is vastly superior. Furthermore, if the qualifying elements are sparse, the linear scan might terminate early if we modify the logic to stop after finding K elements, potentially reducing the effective time complexity to O(K) in the best case, whereas sorting must process the entire array.
Q3At a global product company, how would you handle the edge case where K is 0 or negative in this validator logic?
If K is 0 or negative, the condition 'strictly greater than K' might still be valid for positive numbers, but the requirement to select 'first K elements' becomes undefined or trivial. Typically, if K <= 0, the sum should be 0 because no elements can be selected. The interviewer expects you to validate input constraints first. If K is 0, return 0 immediately. If K is negative, clarify the business logic, but standard practice is to treat it as 0 selections, resulting in a sum of 0.
Examples
Input
nums = [1, 5, 3, 7, 2, 8], K = 3
Output
20
Explanation: Traverse the array: 1 is not > 3. 5 is > 3 (1st element, sum=5). 3 is not > 3. 7 is > 3 (2nd element, sum=12). 2 is not > 3. 8 is > 3 (3rd element, sum=20). We have found K=3 elements. Return 20.
Input
nums = [2, 2, 2, 2], K = 5
Output
0
Explanation: Traverse the array: No element is strictly greater than 5. The count of qualifying elements is 0, which is less than K. Sum remains 0. Return 0.
Input
nums = [10, 1, 10, 1, 10], K = 2
Output
20
Explanation: Traverse the array: 10 is > 2 (1st element, sum=10). 1 is not > 2. 10 is > 2 (2nd element, sum=20). We have found K=2 elements. Stop processing. Return 20.
Input
nums = [4, 5, 6], K = 4
Output
15
Explanation: Traverse the array: 4 is not > 4. 5 is > 4 (1st element, sum=5). 6 is > 4 (2nd element, sum=11). End of array. Only 2 elements found, which is less than K=4. Return the sum of all qualifying elements: 11. Wait, let me re-check. 5+6=11. Correct.
Constraints
- 1 <= nums.length <= 10^5
- 1 <= nums[i] <= 10^9
- 1 <= K <= 10^5
Optimal Approach & Strategy
Iterate through the array once, maintaining a count of selected elements and a running sum. Add an element to the sum and increment the count only if it is strictly greater than K and the count is less than K. Break the loop immediately when the count reaches K or the array ends. This uses O(1) space and allows for early termination.
Brute Force Approach
Iterate through the entire array, collect all elements strictly greater than K into a list, then take the first K elements from that list and sum them. This approach uses O(N) extra space for the list and always traverses the entire array, even if the first K elements are found early.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
sum += nums[i];
K++;
}
if (K === nums.length) break;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > K) {
sum += nums[i];
K++;
}
if (K == nums.size()) break;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > K) {
sum += nums[i];
K++;
}
if (K == nums.length) break;
}
return sum;
}
}def solution(nums, K):
sum = 0
for i in range(len(nums)):
if nums[i] > K:
sum += nums[i]
K += 1
if K == len(nums):
break
return sumfunction solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
sum += nums[i];
K++;
}
if (K === nums.length) break;
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.