Payload Cipher Detector 24 — 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 evaluate and compute the target detector value, which is the sum of elements greater than K minus the sum of elements less than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Cipher Detector 24"
WHY DOES IT MATTER?
Binary search on a sorted structure turns a linear decision into logarithmic time, crucial for large data sets.
OPTIMIZATION CHALLENGE
The key is to reduce repeated O(N) scans to a single O(N log N) preprocessing step followed by O(log N) queries.
REAL-WORLD CONNECTION
Databases use indexed B‑trees to locate range boundaries before aggregating rows, mirroring this pattern.
Cache the total sum and prefix array once; reuse them for every K to avoid recomputation.
COMPLEXITY AT A GLANCE
O(N log N) preprocessing + O(log N) per queryO(N) for the sorted array and prefix sumsCore Theory — Why This Approach?
The problem reduces to partitioning the array around a threshold K and computing two aggregates: the sum of values strictly greater than K and the sum of values less than or equal to K. By sorting the array, the boundary where values cross K becomes a monotonic index, allowing binary search to locate it in O(log N) time, after which prefix‑sum arrays give O(1) range‑sum queries.
A naïve O(N) scan per query works for a single K but fails when the same logic must be reused for many K values or when the input size reaches 10^5‑10^6, because repeated linear scans blow up to O(N·Q). The optimal paradigm combines a one‑time O(N log N) sort with O(N) prefix sums, turning each K evaluation into a logarithmic search plus constant‑time arithmetic, which is the hallmark of the binary‑search‑on‑sorted‑array pattern.
Interview Questions on This Problem
Q1Why is sorting the array a prerequisite for applying binary search in this problem?
Binary search relies on a monotonic ordering; sorting guarantees that all elements ≤ K appear before those > K. Without ordering, the index of the split point cannot be found in logarithmic time.
Q2How does a prefix‑sum array enable O(1) computation of the required sums after the split index is known?
The prefix‑sum at position i stores the total of the first i elements, so the sum of elements ≤ K is prefix[split] and the sum of elements > K is total‑prefix[split]. This eliminates the need for another linear scan.
Q3If the input contains negative numbers, does the algorithm change?
No; the ordering and prefix‑sum logic remain valid because they depend only on relative positions, not on sign. The final subtraction works identically with negative contributions.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
6
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 3, we first calculate the sum of elements greater than K, which is 9 (4 + 5). Then, we calculate the sum of elements less than or equal to K, which is 6 (1 + 2 + 3). Finally, we subtract the sum of elements less than or equal to K from the sum of elements greater than K, giving output 9 - 6 = 3.
Input
[10, 20, 30, 40, 50], 30
Output
80
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K = 30, we first calculate the sum of elements greater than K, which is 90 (40 + 50). Then, we calculate the sum of elements less than or equal to K, which is 90 (10 + 20 + 30). Finally, we subtract the sum of elements less than or equal to K from the sum of elements greater than K, giving output 90 - 90 = 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort once, compute prefix sums, then binary‑search the split index for each K to get the result in O(log N).
Brute Force Approach
Iterate through the array, adding to two accumulators based on the comparison with K, which is O(N) per query.
Verified Code Solutions
function solution(nums, K) {
let sumGreater = 0;
let sumLessOrEqual = 0;
for (let num of nums) {
if (num > K) {
sumGreater += num;
} else {
sumLessOrEqual += num;
}
}
return sumGreater - sumLessOrEqual;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sumGreater = 0;
int sumLessOrEqual = 0;
for (int num : nums) {
if (num > K) {
sumGreater += num;
} else {
sumLessOrEqual += num;
}
}
return sumGreater - sumLessOrEqual;
}
};class Solution {
public int solution(int[] nums, int K) {
int sumGreater = 0;
int sumLessOrEqual = 0;
for (int num : nums) {
if (num > K) {
sumGreater += num;
} else {
sumLessOrEqual += num;
}
}
return sumGreater - sumLessOrEqual;
}
}def solution(nums, K):
sum_greater = 0
sum_less_or_equal = 0
for num in nums:
if num > K:
sum_greater += num
else:
sum_less_or_equal += num
return sum_greater - sum_less_or_equalfunction solution(nums, K) {
let sumGreater = 0;
let sumLessOrEqual = 0;
for (let num of nums) {
if (num > K) {
sumGreater += num;
} else {
sumLessOrEqual += num;
}
}
return sumGreater - sumLessOrEqual;
}Asked in Top Tech Interviews
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