Payload Sequence Optimizer 23 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and sequence metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints. The algorithm should select the K largest numbers in the array and return their sum, handling edge cases where K is greater than the length of the array or the input array contains non-numeric values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Optimizer 23"
WHY DOES IT MATTER?
Selecting top‑K elements is a common sub‑problem in ranking, recommendation, and resource allocation systems.
OPTIMIZATION CHALLENGE
The key is reducing work from O(N log N) to O(N log K) or O(N) by avoiding full sorts.
REAL-WORLD CONNECTION
Think of a streaming service picking the K most‑watched movies to display on the homepage.
When K is known early, allocate a fixed‑size min‑heap and update it in‑place to keep cache locality high.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The naive solution sorts the entire array and then sums the last K elements, which costs O(N log N) time. For very large N (up to 10^7) this becomes a bottleneck, especially when memory constraints prevent storing a full copy of the data.
A more optimal paradigm uses a min‑heap of size K to keep track of the K largest values seen so far. Each insertion or replacement costs O(log K), yielding an overall O(N log K) time and O(K) extra space, which scales dramatically better when K ≪ N.
Interview Questions on This Problem
Q1Why is a min‑heap preferred over sorting when K is much smaller than N?
A min‑heap maintains only K elements, so each operation is O(log K) instead of O(log N). This reduces both time and memory usage when K ≪ N.
Q2How would you handle the case where K exceeds the array length?
Clamp K to the array size before processing, effectively summing all elements. This avoids out‑of‑bounds errors and matches the problem’s specification.
Q3Can you achieve O(N) time for this problem, and if so, how?
Yes, using the QuickSelect algorithm to find the K‑th largest element partitions the array in linear average time. After partitioning, a single pass sums the K largest values.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
12
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 3, we first sort the array in descending order to get [5, 4, 3, 2, 1]. Then, we select the first 3 elements, which are indeed the 3 largest numbers. However, the problem statement asks for the sum of the K largest numbers, not the sum of the first K elements. Therefore, we should return the sum of the 3 largest numbers, which is 5 + 4 + 3 = 12.
Input
[10, 20, 30, 40, 50], 5
Output
150
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K = 5, we first sort the array in descending order to get [50, 40, 30, 20, 10]. Then, we select all elements in the array, since K is equal to the length of the array. Therefore, we return the sum of all elements in the array, which is 50 + 40 + 30 + 20 + 10 = 150.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a min‑heap of size K while scanning the array, pushing larger elements and popping the smallest when the heap exceeds K.
Brute Force Approach
Sort the entire array descending and sum the first K elements; if K > N, sum all elements.
Verified Code Solutions
function solution(nums, k) {
if (k >= nums.length) {
return nums.reduce((a, b) => a + b, 0);
}
nums.sort((a, b) => b - a);
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k >= nums.size()) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
sort(nums.rbegin(), nums.rend());
int largestK[k];
copy(nums.begin(), nums.begin() + k, largestK);
int sum = 0;
for (int num : largestK) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k >= nums.length) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
Arrays.sort(nums);
int[] largestK = new int[k];
System.arraycopy(nums, 0, largestK, 0, k);
int sum = 0;
for (int num : largestK) {
sum += num;
}
return sum;
}
}def solution(nums, k):
if k >= len(nums):
return sum(nums)
nums.sort(reverse=True)
return sum(nums[:k])function solution(nums, k) {
if (k >= nums.length) {
return nums.reduce((a, b) => a + b, 0);
}
nums.sort((a, b) => b - a);
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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