Segmented Cycle Metric — Problem Statement & Solution Guide
Problem Description
You are given an array of length N representing numerical values. Your task is to compute the sum of the elements in the array after sorting them in ascending order.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Segmented Cycle Metric"
WHY DOES IT MATTER?
Sorting before aggregation is a recurring pattern in data pipelines, enabling deterministic calculations like median, percentile, or cumulative distribution functions. Mastering heap‑based sorting equips engineers to handle large, unsorted streams where O(N log N) guarantees are essential.
OPTIMIZATION CHALLENGE
The key insight is to avoid repeated linear scans for the next smallest element; instead, maintain a structure that gives O(log N) access to the minimum, turning an O(N^2) process into O(N log N) while using only constant extra memory.
REAL-WORLD CONNECTION
Think of a priority queue in a distributed task scheduler: tasks (array elements) are enqueued with a priority (value) and the scheduler always picks the smallest (or highest) priority next, analogous to extracting sorted elements to compute a running total.
During an interview, build the heap in‑place using the standard heapify algorithm, then loop over the array extracting the root and swapping it to the end—this not only sorts but also lets you accumulate the sum without extra passes.
COMPLEXITY AT A GLANCE
O(N log N)O(1) auxiliaryCore Theory — Why This Approach?
Sorting is a fundamental operation that imposes a total order on a collection of items, enabling deterministic aggregation such as prefix sums or cumulative metrics. The naive approach of repeatedly scanning for the minimum element to build a sorted list incurs O(N^2) time, which quickly becomes prohibitive for N in the order of 10^5 or higher. By leveraging a binary heap—a complete binary tree where each parent node respects the heap property—we can insert all N elements in O(N) time (heapify) and then extract the minimum N times in O(N log N) total, yielding an optimal sorting routine known as heap sort. This paradigm not only guarantees O(N log N) worst‑case performance regardless of input distribution, but also operates in‑place with only O(1) auxiliary space, making it ideal for memory‑constrained environments and for interview settings where deterministic bounds are prized.
Interview Questions on This Problem
Q1How would you compute the sum of an array after sorting it in ascending order for N up to 10^6, and why is a heap‑based sort preferable to a naïve selection sort in this scenario?
Insert all elements into a min‑heap (or heapify the array) in O(N) time, then repeatedly extract the smallest element, adding it to a running total. Each extraction costs O(log N), so the total time is O(N log N) with O(1) extra space. A naïve selection sort would require O(N^2) comparisons, which is infeasible for N = 10^6.
Q2In a fintech platform, you need to compute the sorted‑order sum of transaction amounts in real‑time as new transactions stream in. Which data structure helps maintain the sorted order efficiently, and how would you update the sum?
A balanced binary heap (or a self‑balancing BST like a treap) can maintain the order of incoming amounts. Insert each new transaction into the heap in O(log N) and adjust the cumulative sum by adding the new value; if you need the sum of the sorted prefix at any point, you can also maintain a Fenwick tree alongside the heap for O(log N) prefix‑sum queries.
Q3A high‑growth startup asks you to design a service that returns the sum of the k smallest elements after sorting a massive list. How would you extend the heap solution to answer this query efficiently?
Build a min‑heap of the entire list (O(N)). Then extract the minimum k times, accumulating the sum; each extraction is O(log N), so the query runs in O(k log N). For repeated queries, keep the heap intact and use a secondary max‑heap of size k to maintain the k smallest elements in O(N log k) preprocessing and O(1) query time.
Examples
Input
[2, 5, 6]
Output
13
Explanation: Step-by-step: with input [2, 5, 6], we first sort the array in ascending order, resulting in [2, 5, 6]. Then, we calculate the sum of the elements, which is 2 + 5 + 6 = 13.
Input
[6, 8]
Output
14
Explanation: Step-by-step: with input [6, 8], we first sort the array in ascending order, resulting in [6, 8]. Then, we calculate the sum of the elements, which is 6 + 8 = 14.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
Heapify the array into a min‑heap (O(N)) and then extract the minimum N times, accumulating the sum; each extraction is O(log N), yielding O(N log N) overall.
Brute Force Approach
Repeatedly scan the unsorted array to find the minimum, remove it, and add it to the sum—this costs O(N^2) time.
Verified Code Solutions
function solution(nums) {
nums.sort((a, b) => a - b);
return nums.reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums) {
sort(nums.begin(), nums.end());
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};import java.util.Arrays;
class Solution {
public int solution(int[] nums) {
Arrays.sort(nums);
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
nums.sort()
return sum(nums)function solution(nums) {
nums.sort((a, b) => a - b);
return nums.reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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