Kth Maximum Partition Analyzer 8 — Problem Statement & Solution Guide
Problem Description
In a distributed network topology, edge weights represent the cost of establishing connections between nodes. To optimize bandwidth allocation, the system must identify the most expensive connections that form the backbone of the network. Given an array of integers representing these edge weights, determine the sum of the k largest values. This metric is critical for calculating the total overhead of the highest-priority links in a spanning structure.
You are provided with an array weights of length n and an integer k. Your task is to compute the sum of the k largest elements in weights. If the array contains fewer than k elements, return the sum of all elements. The solution should efficiently handle large datasets where sorting the entire array may be suboptimal, though for this specific problem variant, a partial selection approach is expected.
Input: An array weights of integers and an integer k.
Output: A single integer representing the sum of the k largest values in weights.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Kth Maximum Partition Analyzer 8"
WHY DOES IT MATTER?
Top‑k selection appears in load‑balancing, recommendation engines, and financial risk calculations where only the most significant contributors matter, making it a high‑frequency pattern in performance‑critical systems.
OPTIMIZATION CHALLENGE
The key insight is that you don't need a total order—maintaining a bounded min‑heap (or using QuickSelect) lets you discard irrelevant elements early, shrinking both time and space from O(n log n) to O(n log k) or O(n).
REAL-WORLD CONNECTION
In a distributed CDN, the k most expensive links determine the backbone cost; optimizing their sum directly influences budgeting and capacity planning, mirroring the algorithmic need to isolate the heaviest edges.
When coding under pressure, first write the heap‑based solution because it’s deterministic and easy to debug; only switch to QuickSelect if k is tiny relative to n and you need the absolute fastest average‑case runtime.
COMPLEXITY AT A GLANCE
O(n log k)O(k)Core Theory — Why This Approach?
The task of finding the sum of the k largest elements is a classic selection problem. A naïve solution would sort the entire array, which costs O(n log n) time, and then sum the last k entries; while correct, it wastes work when n is large and k is small. More efficient paradigms exploit the fact that we only need the top‑k subset, not a full ordering. Using a min‑heap of size k we can maintain the current k largest values while scanning the array in O(n log k) time, because each insertion or replacement costs O(log k). An alternative is the QuickSelect algorithm, which partitions the array around a pivot and recursively works on the side containing the k‑th largest element, achieving expected O(n) time and O(1) extra space. The optimal approach therefore hinges on reducing unnecessary comparisons and memory overhead by focusing solely on the required top‑k slice.
Interview Questions on This Problem
Q1How would you compute the sum of the k largest numbers in an unsorted array with O(n log k) time and O(k) extra space?
Maintain a min‑heap of size k. Iterate through the array, pushing each element onto the heap; if the heap size exceeds k, pop the smallest element. After processing all elements, the heap contains the k largest values, which you sum.
Q2Explain how QuickSelect can be adapted to find the sum of the k largest elements and discuss its average‑case complexity.
QuickSelect partitions the array around a pivot, placing larger elements on one side. Recursively apply it to the side that contains the k‑th largest element until the pivot lands at the (n‑k)th index. Then sum the elements from that index to the end. The average‑case time is O(n) with O(1) extra space, though worst‑case degrades to O(n²) without randomization.
Q3What edge cases must you guard against when implementing the k‑largest‑sum routine in production code?
Handle cases where k is zero, k exceeds the array length (return sum of all elements), and arrays containing negative numbers or duplicates. Also ensure the accumulator uses a wide type (e.g., 64‑bit) to avoid overflow.
Examples
Input
weights = [5, 1, 9, 3, 7], k = 2
Output
16
Explanation: The array is [5, 1, 9, 3, 7]. Sorting in descending order yields [9, 7, 5, 3, 1]. The top 2 values are 9 and 7. Their sum is 9 + 7 = 16.
Input
weights = [10, 10, 10, 10], k = 3
Output
30
Explanation: The array is [10, 10, 10, 10]. All elements are identical. The top 3 values are 10, 10, and 10. Their sum is 10 + 10 + 10 = 30.
Input
weights = [-2, -5, -1, -8], k = 2
Output
-3
Explanation: The array is [-2, -5, -1, -8]. Sorting in descending order yields [-1, -2, -5, -8]. The top 2 values are -1 and -2. Their sum is -1 + (-2) = -3.
Input
weights = [42, 17, 88, 5, 99, 31], k = 4
Output
260
Explanation: The array is [42, 17, 88, 5, 99, 31]. Sorting in descending order yields [99, 88, 42, 31, 17, 5]. The top 4 values are 99, 88, 42, and 31. Their sum is 99 + 88 + 42 + 31 = 260.
Constraints
- 1 <= weights.length <= 10^5
- -10^9 <= weights[i] <= 10^9
- 1 <= k <= weights.length
- The sum of the k largest elements will fit within a 64-bit integer.
Optimal Approach & Strategy
Use a min‑heap of size k while iterating through the array, keeping only the k largest values; after the pass, sum the heap contents. This runs in O(n log k) time and O(k) extra space.
Brute Force Approach
Sort the entire array in descending order and sum the first k elements. This works but costs O(n log n) time even when k is much smaller than n.
Verified Code Solutions
function solution(nums, k) {
if (!Array.isArray(nums) || k <= 0) {
return 0;
}
let maxElements = nums.slice(0, k).sort((a, b) => b - a);
return maxElements.reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (nums.empty() || k <= 0) {
return 0;
}
vector<int> maxElements(nums.begin(), nums.begin() + k);
sort(maxElements.begin(), maxElements.end());
reverse(maxElements.begin(), maxElements.end());
int sum = 0;
for (int num : maxElements) {
sum += num;
}
return sum;
}
}class Solution {
public int solution(int[] nums, int k) {
if (nums == null || k <= 0) {
return 0;
}
int[] maxElements = Arrays.copyOfRange(nums, 0, k);
Arrays.sort(maxElements);
for (int i = 0; i < maxElements.length; i++) {
maxElements[i] = -maxElements[i]; // reverse sort
}
int sum = 0;
for (int num : maxElements) {
sum += num;
}
return sum;
}
}def solution(nums, k):
if not isinstance(nums, list) or k <= 0:
return 0
max_elements = sorted(nums[:k], reverse=True)
return sum(max_elements)function solution(nums, k) {
if (!Array.isArray(nums) || k <= 0) {
return 0;
}
let maxElements = nums.slice(0, k).sort((a, b) => b - a);
return maxElements.reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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