Tome Cache Synthesizer 49 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing a distributed cache system that stores data in a linked list structure. The cache maintains a sequence of integer values representing memory block identifiers. To improve retrieval efficiency for high-priority requests, you need to identify the top k most significant blocks. Given a singly linked list of integers and an integer k, determine the sum of the k largest values present in the list. If the list contains fewer than k elements, return the sum of all elements. The solution must efficiently process the linked list without converting it entirely into an array if possible, leveraging the structural properties of the list and the specific pattern of selecting maximums.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Cache Synthesizer 49"
WHY DOES IT MATTER?
Selecting top‑k items is a core reduction pattern for streaming and memory‑constrained data.
OPTIMIZATION CHALLENGE
The challenge is to cut the sorting cost from O(n log n) to O(n log k) by limiting stored state.
REAL-WORLD CONNECTION
Databases use similar heap‑based top‑k queries to serve ranked results without scanning the entire table.
Initialize the heap with the first k nodes, then reuse the same node objects to avoid extra allocations.
COMPLEXITY AT A GLANCE
O(n log k)O(k)Core Theory — Why This Approach?
The naive solution scans the list and stores every element, then sorts to extract the k largest, which costs O(n log n) time and O(n) extra space—prohibitive for massive streams where n can be in the millions. The optimal paradigm leverages a size‑k min‑heap (or quick‑select partition) to maintain only the current top k elements while traversing the list once, guaranteeing O(n log k) time and O(k) auxiliary space, which scales gracefully with large inputs.
By keeping the smallest of the top k at the heap root, each new node can be compared in O(1) and, if larger, replace the root in O(log k). This incremental selection avoids full sorting and eliminates the need to store the entire list, aligning with the streaming nature of linked‑list caches where random access is unavailable.
Interview Questions on This Problem
Q1How would you find the k largest values in a singly linked list without converting it to an array?
Maintain a min‑heap of size k while traversing; push the first k nodes, then for each subsequent node, replace the heap root if the node's value is larger.
Q2What is the time and space complexity of using a min‑heap for this problem?
Time complexity is O(n log k) and auxiliary space is O(k), since the heap never grows beyond k elements.
Q3Why is quick‑select less favorable than a heap for a linked list implementation?
Quick‑select requires random access to partition elements, which a singly linked list cannot provide efficiently, leading to O(n²) in the worst case.
Examples
Input
head = [12, 5, 8, 20, 3], k = 2
Output
28
Explanation: The values in the linked list are 12, 5, 8, 20, and 3. The two largest values are 20 and 12. Their sum is 20 + 12 = 28.
Input
head = [1, 1, 1, 1], k = 3
Output
3
Explanation: The list contains four 1s. The three largest values are 1, 1, and 1. Their sum is 1 + 1 + 1 = 3.
Input
head = [100, -50, 200, -10, 300], k = 1
Output
300
Explanation: The list contains 100, -50, 200, -10, and 300. The single largest value is 300. The sum is 300.
Input
head = [7, 7, 7], k = 5
Output
21
Explanation: The list has only 3 elements, which is less than k=5. Therefore, we sum all elements: 7 + 7 + 7 = 21.
Constraints
- 1 <= length of linked list <= 10^5
- -10^9 <= node.val <= 10^9
- 1 <= k <= 10^5
- The linked list is guaranteed to be non-circular.
Optimal Approach & Strategy
Use a size‑k min‑heap during a single pass to keep only the current top k values.
Brute Force Approach
Collect all node values into an array, sort descending, and take the first k elements.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length) {
k = nums.length;
}
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k > nums.size()) {
k = nums.size();
}
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length) {
k = nums.length;
}
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
if k > len(nums):
k = len(nums)
nums.sort(reverse=True)
return sum(nums[:k])function solution(nums, k) {
if (k > nums.length) {
k = nums.length;
}
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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