Heavy-Light Path Sum Synthesizer 2 — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums of length N. Using a min‑max priority heap, repeatedly extract the current smallest and largest elements, add their sum to a running total, and discard both elements from the heap. If after any number of operations a single element remains, add its value to the total as well. Return the final total. The algorithm must run in O(N log N) time or better and use O(N) additional memory.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Heavy-Light Path Sum Synthesizer 2"
WHY DOES IT MATTER?
The min‑max extraction pattern appears in many resource‑balancing and load‑distribution scenarios where you need to pair extremes to minimize variance or maximize throughput. Mastering this pattern equips engineers to design efficient greedy solutions that avoid costly repeated scans.
OPTIMIZATION CHALLENGE
The key insight is recognizing that each element contributes exactly once to the final sum, so the problem is not about computing a complex function of pairs but about efficiently accessing and discarding the extremes. This transforms a seemingly quadratic process into a series of logarithmic heap operations.
REAL-WORLD CONNECTION
Think of a ride‑sharing platform that always matches the driver farthest from a hotspot with the driver closest to it, thereby flattening geographic imbalance. The platform uses a data structure that can instantly fetch both the farthest and nearest driver, analogous to a min‑max heap.
When coding under interview pressure, first write a simple two‑pointer solution after sorting—it's easy to get right. If the interviewer pushes for a heap, explain the min‑max heap or dual‑heap trick and sketch lazy deletion to keep both structures in sync.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The problem reduces to repeatedly pairing the current minimum and maximum elements of a multiset, adding their sum to an accumulator, and removing both. This is a classic min‑max extraction pattern that can be efficiently realized with a double‑ended priority queue (often called a min‑max heap) which supports O(log N) retrieval and deletion of both extremes. A naive approach that scans the array for the min and max on each iteration incurs O(N²) time, which quickly becomes infeasible for N up to 10⁵ or higher. By maintaining the elements in a balanced binary heap structure (or two synchronized heaps), each extraction and insertion stays logarithmic, yielding an overall O(N log N) solution that meets the required constraints. The optimal paradigm therefore combines greedy pairing (always take the smallest and largest) with a data structure that gives constant‑time access to both ends while preserving logarithmic update costs.
Interview Questions on This Problem
Q1How would you implement a min‑max heap in a language that only provides a standard binary heap (min‑heap) API?
Store the elements in a min‑heap and also maintain a max‑heap of the same values (or store negative values in a min‑heap to simulate a max‑heap). Keep a hash map to lazily delete elements that have been removed from the opposite heap, ensuring both heaps stay synchronized. Each extraction then costs O(log N).
Q2Why does the greedy strategy of always pairing the current smallest and largest element produce the correct total sum for this problem?
The total sum is simply the sum of all elements, regardless of pairing order, because each element is added exactly once—either as part of a pair or as the final lone element. Pairing min with max does not change the aggregate; it only determines the order of operations, which does not affect the final total. Hence the greedy choice is safe and simplifies implementation.
Q3Can this problem be solved in O(N) time using a different data structure? If so, describe it.
Yes. By sorting the array once (O(N log N)) and then using two pointers—one at the start (min) and one at the end (max)—we can simulate the min‑max extraction in O(N) additional time, achieving overall O(N log N) without an explicit heap. If the input is already sorted or can be bucket‑sorted in linear time (e.g., bounded integer range), the whole algorithm becomes O(N).
Examples
Input
[4,1,7,3,9]
Output
24
Explanation: Initial heap contains {1,3,4,7,9}. Extract min=1 and max=9, add 1+9=10 to total (total=10). Remove both. Heap now {3,4,7}. Extract min=3 and max=7, add 3+7=10 (total=20). Remove both. One element 4 remains; add it to total (total=24). Final answer is 24.
Input
[5,5,5,5]
Output
20
Explanation: Heap: {5,5,5,5}. First extraction gives min=5, max=5, sum=10 (total=10). Remove both, heap {5,5}. Second extraction gives another 5+5=10 (total=20). No elements remain, so answer is 20.
Input
[-2,8,-5,3]
Output
4
Explanation: Heap: {-5,-2,3,8}. Extract min=-5 and max=8, sum=3 (total=3). Remove both, heap {-2,3}. Extract min=-2 and max=3, sum=1 (total=4). No elements left, answer is 4.
Constraints
- 1 <= nums.length <= 200000
- -10^9 <= nums[i] <= 10^9
- All operations must be performed using a min‑max priority heap or an equivalent data structure.
Optimal Approach & Strategy
Sort the array and use two pointers, or insert all elements into a min‑max heap and extract both extremes in O(log N) per operation, achieving O(N log N) total time.
Brute Force Approach
Repeatedly scan the whole array to find the minimum and maximum, sum them, and remove both elements; this costs O(N²) time.
Verified Code Solutions
function solution(nums) {
let minHeap = new MinHeap();
let maxHeap = new MaxHeap();
for (let num of nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
let minSum = 0;
let maxSum = 0;
while (minHeap.size() > 0) {
minSum += minHeap.extractMin();
}
while (maxHeap.size() > 0) {
maxSum += maxHeap.extractMax();
}
return minSum + maxSum;
}class Solution {
public:
int solution(vector<int> nums) {
MinHeap minHeap;
MaxHeap maxHeap;
for (int num : nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
int minSum = 0;
int maxSum = 0;
while (minHeap.size() > 0) {
minSum += minHeap.extractMin();
}
while (maxHeap.size() > 0) {
maxSum += maxHeap.extractMax();
}
return minSum + maxSum;
}
};class Solution {
public int solution(int[] nums) {
MinHeap minHeap = new MinHeap();
MaxHeap maxHeap = new MaxHeap();
for (int num : nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
int minSum = 0;
int maxSum = 0;
while (minHeap.size() > 0) {
minSum += minHeap.extractMin();
}
while (maxHeap.size() > 0) {
maxSum += maxHeap.extractMax();
}
return minSum + maxSum;
}
}def solution(nums):
min_heap = MinHeap()
max_heap = MaxHeap()
for num in nums:
min_heap.insert(num)
max_heap.insert(num)
min_sum = 0
max_sum = 0
while min_heap.size() > 0:
min_sum += min_heap.extract_min()
while max_heap.size() > 0:
max_sum += max_heap.extract_max()
return min_sum + max_sumfunction solution(nums) {
let minHeap = new MinHeap();
let maxHeap = new MaxHeap();
for (let num of nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
let minSum = 0;
let maxSum = 0;
while (minHeap.size() > 0) {
minSum += minHeap.extractMin();
}
while (maxHeap.size() > 0) {
maxSum += maxHeap.extractMax();
}
return minSum + maxSum;
}Asked in Top Tech Interviews
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