Tome Signal Consolidator 48 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and signal metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints. The target consolidator value is the maximum value in the sequence after removing the two maximum elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Consolidator 48"
WHY DOES IT MATTER?
Identifying top‑k elements is a recurring pattern in performance‑critical code.
OPTIMIZATION CHALLENGE
Reducing from O(n log n) to O(n) while keeping memory constant is the key win.
REAL-WORLD CONNECTION
It mirrors priority‑queue handling in task schedulers where only the highest‑priority jobs matter.
Prefer a fixed‑size min‑heap or three variables; avoid dynamic structures that grow with input size.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to finding the third largest element in an unsorted array. A naive sort (O(n log n)) or scanning for max twice (O(n) each) wastes time and memory, especially for massive streams where only constant extra space is permissible. The optimal paradigm leverages a min‑heap of fixed size three (or three variables) to keep track of the top three values while traversing the data once, guaranteeing O(n) time and O(1) auxiliary space. This approach exploits the heap property: the smallest of the three candidates is always at the root, allowing efficient replacement when a larger element appears.
Interview Questions on This Problem
Q1How would you find the third maximum element in a single pass without sorting?
Maintain three variables (or a size‑3 min‑heap) for the largest, second, and third values, updating them as you iterate. Each element is compared at most three times, yielding O(n) time.
Q2Why is a full sort overkill for this problem?
Sorting costs O(n log n) even though we only need three order statistics. A linear‑time selection or bounded heap avoids unnecessary work.
Q3What edge cases must you handle when the array has fewer than three distinct elements?
If the length is <3, the answer is undefined or you return the maximum after removing available maxes. Duplicate values also require careful comparison to avoid counting the same element twice.
Examples
Input
[1, 2, 3, 4, 5]
Output
3
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we construct a max heap and then remove the maximum element twice, giving output 3 as the target consolidator value
Input
[10, 20, 30, 40, 50]
Output
30
Explanation: Step-by-step: with input [10, 20, 30, 40, 50], we construct a max heap and then remove the maximum element twice, giving output 30 as the target consolidator value
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min‑heap of size three (or three variables) to keep the three largest values while iterating, achieving O(n) time and O(1) space.
Brute Force Approach
Sort the entire array and pick the element at index n‑3, which is O(n log n) time and O(1) extra space.
Verified Code Solutions
function solution(nums) {
let maxHeap = new MaxHeap();
for (let num of nums) {
maxHeap.insert(num);
}
maxHeap.removeMax();
maxHeap.removeMax();
return maxHeap.getMax();
}class Solution {
public:
int solution(vector<int>& nums) {
priority_queue<int> maxHeap;
for (int num : nums) {
maxHeap.push(num);
}
maxHeap.pop();
maxHeap.pop();
return maxHeap.top();
}
};class Solution {
public int solution(int[] nums) {
PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
for (int num : nums) {
maxHeap.add(num);
}
maxHeap.poll();
maxHeap.poll();
return maxHeap.peek();
}
}import heapq
def solution(nums):
maxHeap = [-num for num in nums]
heapq.heapify(maxHeap)
heapq.heappop(maxHeap)
heapq.heappop(maxHeap)
return -maxHeap[0]function solution(nums) {
let maxHeap = new MaxHeap();
for (let num of nums) {
maxHeap.insert(num);
}
maxHeap.removeMax();
maxHeap.removeMax();
return maxHeap.getMax();
}Asked in Top Tech Interviews
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