Payload Cipher Extractor 3 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and cipher metrics, and an integer K, construct an optimal algorithm to compute the sum of all elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Cipher Extractor 3"
WHY DOES IT MATTER?
The "prefix‑sum / DP accumulation" pattern is essential because it transforms repetitive aggregation into a single linear pass, eliminating redundant calculations and enabling constant‑time updates for each element.
OPTIMIZATION CHALLENGE
The key insight is to treat the answer as a cumulative state that can be updated incrementally; recognizing that the predicate (value > K) is stateless allows us to add or skip each element in O(1) time, collapsing an otherwise quadratic process.
REAL-WORLD CONNECTION
Think of a network firewall that continuously tallies the total size of packets exceeding a security threshold. Instead of re‑scanning all past packets each time, it updates a running total as each packet arrives—mirroring the DP accumulation technique.
During an interview, write the DP recurrence first (dp[i] = dp[i‑1] + contribution) and then immediately simplify it to a single variable; this shows you can spot unnecessary space and keep the solution lean.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The core of this problem lies in recognizing that the sum of elements satisfying a simple predicate (value > K) can be expressed as a prefix‑sum style dynamic programming relation. For each index i we can define dp[i] = dp[i‑1] + (arr[i] > K ? arr[i] : 0). This recurrence builds the answer in a single left‑to‑right pass, turning an O(N²) brute‑force double loop into O(N) time. Naïve approaches that iterate over every pair of elements or recompute the sum for each query explode when N reaches 10⁶ or when the interview platform runs many test cases. By treating the cumulative contribution as a DP state, we avoid repeated work and keep memory to O(1) beyond the input array. When the problem is extended to multiple K queries, sorting the array and pre‑computing suffix sums (another DP variant) yields O(N log N) preprocessing and O(log N) per query, which is the optimal paradigm for large‑scale data streams.
Interview Questions on This Problem
Q1How would you modify the solution if you were given Q queries, each asking for the sum of elements greater than a different K?
Sort the array once (O(N log N)) and build a suffix sum array where suffix[i] = sum of elements from i to N‑1. For each query K, binary‑search the first index where arr[idx] > K and answer = suffix[idx]; this gives O(log N) per query.
Q2Explain why a segment tree or Fenwick tree might be overkill for a single‑pass sum‑greater‑than‑K problem.
Both data structures support range queries and point updates in O(log N), but the problem only requires a single static aggregation over the whole array. A linear scan with a DP accumulator achieves O(N) time and O(1) extra space, which is asymptotically better and far simpler to implement.
Q3In a streaming context where elements arrive one‑by‑one, how can you maintain the sum of values > K without storing the entire stream?
Maintain a running total variable. For each incoming element x, if x > K add x to the total; otherwise ignore it. This uses O(1) memory and O(1) amortized time per element, perfectly fitting the streaming model.
Examples
Input
[10, 20, 30, 40, 50], 3
Output
150
Explanation: Step-by-step: with input array [10, 20, 30, 40, 50] and K = 3, we sum all elements greater than K. Since all elements are greater than 3, we sum them all: 10 + 20 + 30 + 40 + 50 = 150.
Input
[5, 5, 5, 5, 5], 5
Output
0
Explanation: Step-by-step: with input array [5, 5, 5, 5, 5] and K = 5, we sum all elements greater than K. Since no elements are greater than 5, the sum is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single linear scan, adding only elements > K to a running total, achieving O(N) time and O(1) extra space.
Brute Force Approach
Iterate over every element for each possible K, recomputing the sum each time, leading to O(N²) time.
Verified Code Solutions
function solution(nums, k) { let sum = 0; for (let num of nums) { if (num > k) { sum += num; } } return sum; }class Solution { public: int solution(vector<int>& nums, int k) { int sum = 0; for (int num : nums) { if (num > k) { sum += num; } } return sum; } };class Solution { public int solution(int[] nums, int k) { int sum = 0; for (int num : nums) { if (num > k) { sum += num; } } return sum; } }def solution(nums, k): return sum(num for num in nums if num > k)function solution(nums, k) { let sum = 0; for (let num of nums) { if (num > k) { sum += num; } } return sum; }Asked in Top Tech Interviews
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