Node Payload Tracker 31 — Problem Statement & Solution Guide
Problem Description
Node Payload Tracker 31
You are given a list of integers and a target value. Your task is to compute the sum of all elements in the list that are strictly greater than the target. The input consists of the number of elements, the elements themselves, and the target value. The output is a single integer representing the required sum. If no element exceeds the target, the sum is zero.
The problem can be solved efficiently by iterating through the array once, adding each element that satisfies the condition to an accumulator. This approach runs in linear time and uses constant additional space.
Input format:
- The first line contains an integer n, the number of elements.
- The second line contains n space‑separated integers.
- The third line contains the target integer.
Output format:
- A single integer: the sum of all elements greater than the target.
The solution should handle negative numbers and large ranges of values as specified in the constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Tracker 31"
WHY DOES IT MATTER?
Finding a threshold‑based aggregate quickly is a common sub‑problem in search‑heavy applications.
OPTIMIZATION CHALLENGE
The key is reducing a linear scan to logarithmic time by leveraging order and pre‑computed aggregates.
REAL-WORLD CONNECTION
Databases use indexed range scans and pre‑aggregated statistics to answer queries like "sum of sales > $10k" efficiently.
Always verify the input is sorted or sort once, then build a cumulative array; reuse it for multiple queries to amortize the O(n) cost.
COMPLEXITY AT A GLANCE
O(log n) per query after O(n) preprocessingO(n) for the suffix‑sum arrayCore Theory — Why This Approach?
Binary search exploits the monotonic ordering of a sorted array to locate the boundary where elements transition from ≤ target to > target in O(log n) time, eliminating the need to examine each element. By pairing this with a prefix‑sum (or suffix‑sum) array, we can retrieve the sum of all elements beyond that boundary in O(1) after an O(n) preprocessing pass, yielding an overall query cost that scales logarithmically rather than linearly. Naïve linear scans require O(n) time per query, which becomes prohibitive for large n or multiple queries, especially when the input size reaches millions. The optimal paradigm therefore combines sorting (if not already sorted), binary search for the split point, and cumulative sums to achieve the best possible asymptotic performance.
Interview Questions on This Problem
Q1How does binary search guarantee O(log n) time on a sorted array?
Each iteration halves the search interval, reducing the problem size exponentially. After at most log₂n steps the interval collapses to the target position.
Q2Why might you prefer a suffix‑sum array over recomputing the sum each time?
A suffix‑sum stores the total of all elements from each index to the end, allowing constant‑time retrieval of any tail sum. This avoids repeated O(n) aggregation for each query.
Q3What edge cases must you handle when summing elements greater than a target?
You must correctly handle an empty array, all elements ≤ target (result 0), and potential integer overflow for large sums.
Examples
Input
5 1 5 3 7 2 3
Output
12
Explanation: Elements greater than 3 are 5 and 7. Their sum is 5 + 7 = 12.
Input
4 -10 0 10 20 5
Output
30
Explanation: Elements greater than 5 are 10 and 20. Their sum is 10 + 20 = 30.
Input
6 100 200 300 400 500 600 600
Output
0
Explanation: No element is greater than 600, so the sum is 0.
Constraints
- 1 <= n <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= target <= 1000000000
Optimal Approach & Strategy
Sort the list, binary‑search for the first > target, then use a suffix‑sum array to retrieve the required sum in O(1).
Brute Force Approach
Iterate through the entire list, adding each element that is > target; O(n) time, O(1) extra space.
Verified Code Solutions
function solution(nums, target) { return nums.filter(num => num > target).reduce((a, b) => a + b, 0); }class Solution { public: int solution(vector<int>& nums, int target) { int sum = 0; for (int num : nums) { if (num > target) sum += num; } return sum; } };class Solution { public int solution(int[] nums, int target) { int sum = 0; for (int num : nums) { if (num > target) sum += num; } return sum; } }def solution(nums, target): return sum(num for num in nums if num > target)function solution(nums, target) { return nums.filter(num => num > target).reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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