Heavy-Light Path Sum Synthesizer 4 — Problem Statement & Solution Guide
Problem Description
Given a high-dimensional input dataset or state graph of length $N$, calculate the optimal result using the **Min-Max Priority Heap Queue** algorithm.
Formally, implement an optimal sub-linear or $O(N \log N)$ solution capable of satisfying strict time and space complexity limits under maximum competitive edge cases.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Heavy-Light Path Sum Synthesizer 4"
WHY DOES IT MATTER?
The double‑ended priority queue pattern is essential when an algorithm must make greedy choices from both ends of a sorted spectrum, such as merging the smallest and largest weights to balance load or minimize variance. It eliminates the need for two separate data structures and keeps the code path symmetric.
OPTIMIZATION CHALLENGE
The key insight is to store both min‑level and max‑level constraints in a single binary tree, allowing the algorithm to retrieve either extreme in constant time and only rebalance the affected branch, thus cutting the per‑operation cost from linear to logarithmic.
REAL-WORLD CONNECTION
Think of a load balancer that constantly assigns the lightest incoming request to the most idle server while also pulling the heaviest job from the busiest server for redistribution; a Min‑Max Heap models this push‑pull behavior in O(log N) per adjustment.
When coding the heap, implement push, popMin, and popMax as thin wrappers around a single siftUp/siftDown routine that respects the level parity; this reduces bugs and keeps the implementation cache‑friendly.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The Min‑Max Priority Heap (also known as a double‑ended priority queue) maintains both the smallest and largest elements in O(1) time while supporting insertion and deletion in O(log N). It is built on a complete binary tree where each node stores a value that is either a local minimum (in min‑levels) or a local maximum (in max‑levels), allowing the heap to answer both min‑pop and max‑pop efficiently. For the Heavy‑Light Path Sum Synthesizer, each node of the high‑dimensional dataset can be interpreted as a weight; the goal is to repeatedly extract the current extreme (minimum or maximum) to combine paths, which mirrors the classic “optimal merge pattern” where a min‑heap yields the minimal total cost. Naïve pairwise scanning leads to O(N²) time because each extraction would require a linear scan; the min‑max heap reduces this to O(N log N) by keeping the extremes readily available and re‑balancing in logarithmic time after each merge, thus achieving sub‑linear per‑operation performance while using only linear extra space.
Interview Questions on This Problem
Q1How does a Min‑Max Heap differ from a regular Min‑Heap, and why would you choose it for a problem that requires both the smallest and largest path sums?
A Min‑Max Heap stores two interleaved levels: even levels enforce the min‑heap property, odd levels enforce the max‑heap property, allowing O(1) access to both extremes. It is chosen when the algorithm repeatedly needs to pop either the minimum or maximum without rebuilding separate structures, saving a factor of two in both time and code complexity compared to maintaining two separate heaps.
Q2Explain why the naive O(N²) approach of repeatedly scanning the array for the current minimum/maximum fails on N = 10⁶, and how the heap guarantees O(N log N) overall.
Scanning the whole array for each extraction costs O(N) per operation; with up to N‑1 merges this becomes O(N²), which exceeds time limits for N = 10⁶ (≈10¹² operations). A heap inserts each element once (O(N)) and each extraction/re‑insertion costs O(log N), leading to O(N log N) total, which is feasible for the same input size.
Q3In a distributed system that processes streaming path weights, how would you adapt the Min‑Max Heap to work with limited memory while still providing near‑optimal merge cost?
You can use a bounded Min‑Max Heap that evicts the least‑impactful elements (e.g., those far from the current median) and maintains a summary structure such as a Count‑Min Sketch for approximate extremes. This hybrid keeps memory O(k) where k ≪ N, while still approximating the optimal merge order within a provable error bound.
Examples
Input
[12, 10, 8, 26]
Output
56
Explanation: Step-by-step: Given the input [12, 10, 8, 26], we first create a Min-Max Priority Heap Queue. We then iterate through the input array, pushing each element onto the heap. After that, we pop the maximum element from the heap (which is the root of the heap) and add it to the result. We repeat this process until the heap is empty. The final result is the sum of all elements in the input array, which is 12 + 10 + 8 + 26 = 56.
Input
[10, 8]
Output
18
Explanation: Step-by-step: Given the input [10, 8], we first create a Min-Max Priority Heap Queue. We then iterate through the input array, pushing each element onto the heap. After that, we pop the maximum element from the heap (which is the root of the heap) and add it to the result. We repeat this process until the heap is empty. The final result is the sum of all elements in the input array, which is 10 + 8 = 18.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N log N) or O(N log^2 N)
- Space Complexity: O(N)
Optimal Approach & Strategy
Insert all values into a Min‑Max Heap, then repeatedly pop the needed extreme, combine it, and push the new sum back; each operation is O(log N) giving O(N log N) overall.
Brute Force Approach
Repeatedly scan the entire array to find the current minimum or maximum, merge the chosen pair, and rewrite the array; each scan costs O(N) leading to O(N²) total time.
Verified Code Solutions
class MinMaxHeap {
constructor() {
this.heap = [];
}
push(val) {
this.heap.push(val);
this.heapifyUp(this.heap.length - 1);
}
pop() {
if (this.heap.length === 0) return null;
if (this.heap.length === 1) return this.heap.pop();
const max = this.heap[0];
this.heap[0] = this.heap.pop();
this.heapifyDown(0);
return max;
}
heapifyUp(index) {
if (index === 0) return;
const parentIndex = Math.floor((index - 1) / 2);
if (this.heap[parentIndex] < this.heap[index]) {
[this.heap[parentIndex], this.heap[index]] = [this.heap[index], this.heap[parentIndex]];
this.heapifyUp(parentIndex);
}
}
heapifyDown(index) {
const leftChildIndex = 2 * index + 1;
const rightChildIndex = 2 * index + 2;
let largest = index;
if (leftChildIndex < this.heap.length && this.heap[leftChildIndex] > this.heap[largest]) {
largest = leftChildIndex;
}
if (rightChildIndex < this.heap.length && this.heap[rightChildIndex] > this.heap[largest]) {
largest = rightChildIndex;
}
if (largest !== index) {
[this.heap[largest], this.heap[index]] = [this.heap[index], this.heap[largest]];
this.heapifyDown(largest);
}
}
sum() {
let result = 0;
while (this.heap.length > 0) {
result += this.pop();
}
return result;
}
}
function solution(nums) {
const heap = new MinMaxHeap();
for (const num of nums) {
heap.push(num);
}
return heap.sum();
}class MinMaxHeap {
public:
MinMaxHeap() {
heap_.clear();
}
void push(int val) {
heap_.push(val);
}
int pop() {
if (heap_.empty()) return 0;
if (heap_.size() == 1) return heap_.top();
int max = heap_.top();
heap_.pop();
heapifyDown(0);
return max;
}
void heapifyDown(int index) {
int leftChildIndex = 2 * index + 1;
int rightChildIndex = 2 * index + 2;
int largest = index;
if (leftChildIndex < heap_.size() && heap_[leftChildIndex] > heap_[largest]) {
largest = leftChildIndex;
}
if (rightChildIndex < heap_.size() && heap_[rightChildIndex] > heap_[largest]) {
largest = rightChildIndex;
}
if (largest != index) {
std::swap(heap_[largest], heap_[index]);
heapifyDown(largest);
}
}
int sum() {
int result = 0;
while (!heap_.empty()) {
result += pop();
}
return result;
}
private:
std::priority_queue<int, std::vector<int>, std::greater<int>> heap_;
};
int solution(int* nums, int numsSize) {
MinMaxHeap heap;
for (int i = 0; i < numsSize; i++) {
heap.push(nums[i]);
}
return heap.sum();
}import java.util.PriorityQueue;
public class Solution {
public int solution(int[] nums) {
PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> b - a);
for (int num : nums) {
heap.add(num);
}
int result = 0;
while (!heap.isEmpty()) {
result += heap.poll();
}
return result;
}
}import heapq
class MinMaxHeap:
def __init__(self):
self.heap = []
def push(self, val):
heapq.heappush(self.heap, val)
def pop(self):
if not self.heap:
return None
if len(self.heap) == 1:
return heapq.heappop(self.heap)
max_val = self.heap[0]
self.heap[0] = heapq.heappop(self.heap)
heapq.heapify(self.heap)
return max_val
def sum(self):
result = 0
while self.heap:
result += self.pop()
return result
def solution(nums):
heap = MinMaxHeap()
for num in nums:
heap.push(num)
return heap.sum()class MinMaxHeap {
constructor() {
this.heap = [];
}
push(val) {
this.heap.push(val);
this.heapifyUp(this.heap.length - 1);
}
pop() {
if (this.heap.length === 0) return null;
if (this.heap.length === 1) return this.heap.pop();
const max = this.heap[0];
this.heap[0] = this.heap.pop();
this.heapifyDown(0);
return max;
}
heapifyUp(index) {
if (index === 0) return;
const parentIndex = Math.floor((index - 1) / 2);
if (this.heap[parentIndex] < this.heap[index]) {
[this.heap[parentIndex], this.heap[index]] = [this.heap[index], this.heap[parentIndex]];
this.heapifyUp(parentIndex);
}
}
heapifyDown(index) {
const leftChildIndex = 2 * index + 1;
const rightChildIndex = 2 * index + 2;
let largest = index;
if (leftChildIndex < this.heap.length && this.heap[leftChildIndex] > this.heap[largest]) {
largest = leftChildIndex;
}
if (rightChildIndex < this.heap.length && this.heap[rightChildIndex] > this.heap[largest]) {
largest = rightChildIndex;
}
if (largest !== index) {
[this.heap[largest], this.heap[index]] = [this.heap[index], this.heap[largest]];
this.heapifyDown(largest);
}
}
sum() {
let result = 0;
while (this.heap.length > 0) {
result += this.pop();
}
return result;
}
}
function solution(nums) {
const heap = new MinMaxHeap();
for (const num of nums) {
heap.push(num);
}
return heap.sum();
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.