Heavy-Light Path Sum Resolver — Problem Statement & Solution Guide

HeapHardMin-Max Priority Heap Queue

TimeO(N log N)

SpaceO(N)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Heap and solve the Heavy-Light Path Sum Resolver 3 problem optimally.

TopicHeap

PatternMin-Max Priority Heap Queue

TimeO(N log N)

SpaceO(N)

Problem Description

You are tasked with processing a sequence of N integers representing sensor readings from a distributed network. The goal is to compute the total sum of all elements in the sequence, but with a specific constraint: you must utilize a Min-Max Priority Heap Queue to manage the data flow. Although a standard linear scan would suffice for summation, this problem requires you to simulate the insertion of each element into a heap structure that supports both minimum and maximum extraction, ensuring that the heap property is maintained at every step. The final result is the arithmetic sum of all N elements, derived after all insertions are complete. This exercise tests your ability to integrate heap-based data structures into seemingly simple aggregation tasks, emphasizing the overhead and structural integrity of maintaining a dual-priority queue.

Input: An array of integers nums of length N.

Output: A single integer representing the sum of all elements in nums.

The algorithm must explicitly push each element into a Min-Max Heap. While the sum can be computed in O(N) time, the requirement to use the heap structure implies that you must handle the internal re-balancing and priority ordering associated with such a data structure. The final sum is independent of the heap's internal ordering but must be calculated as part of the processing pipeline that involves heap operations.

DSA Pattern Breakdown

"Heavy-Light Path Sum Resolver"

hard

WHY DOES IT MATTER?

The Min‑Max Heap pattern is essential when a problem demands simultaneous access to both extremes of a dynamic dataset while still supporting frequent insertions or deletions, a scenario common in load‑balancing, financial tick data, and real‑time monitoring.

OPTIMIZATION CHALLENGE

The key insight is to avoid a second pass for the sum: maintain a running total variable while inserting each element into the heap, thereby achieving the required O(N log N) time without extra O(N) traversals.

REAL-WORLD CONNECTION

Think of a distributed sensor network where each node pushes its reading to a central aggregator. The aggregator must constantly know the hottest (max) and coolest (min) sensor while also keeping a running total for analytics—exactly what a Min‑Max Heap with an accumulated sum provides.

During an interview, insert the element, percolate it according to Min‑Max rules, and immediately add its value to a ‘totalSum’ variable. This shows you understand both the data‑structure invariants and how to keep auxiliary aggregates efficiently.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log N)

💾 Space:O(N)

Core Theory — Why This Approach?

A Min‑Max Priority Heap (also called a double‑ended priority queue) stores elements such that both the minimum and maximum can be accessed in O(1) time while insertions and deletions cost O(log N). Internally it is a binary heap where each level alternates between min‑level and max‑level constraints, guaranteeing that the root holds the global minimum and one of its children holds the global maximum. When the problem asks to compute the total sum while *simulating* insertions into such a structure, the algorithm must respect the heap’s ordering invariants, which forces every element to be percolated up or down, incurring a logarithmic cost per element.

A naïve linear scan that simply adds the numbers runs in O(N) time and O(1) space, but it completely sidesteps the required heap operations. In real‑world scenarios—e.g., streaming sensor data where you must constantly query the current min, max, and cumulative sum—the heap provides the necessary ordering without sacrificing the ability to maintain an aggregate. The optimal paradigm therefore combines a Min‑Max Heap for ordering with an auxiliary variable that accumulates the sum during each insertion, achieving O(N log N) total time while preserving O(N) auxiliary space for the heap.

Interview Questions on This Problem

Q1How would you modify a standard binary heap to support O(1) retrieval of both the minimum and maximum elements?

Implement a Min‑Max Heap where nodes at even depths satisfy the min‑heap property and nodes at odd depths satisfy the max‑heap property; the root holds the minimum and one of its children holds the maximum, allowing O(1) access to both.

Q2Given a stream of N integers, you need to output the running sum after each insertion while also being able to query the current min and max in O(log N). Which data structure fits and why?

A Min‑Max Heap fits because it maintains ordering for min and max queries in O(1) and supports insertions in O(log N); a separate variable can keep the running sum, updated in O(1) per insertion.

Q3Why might a naïve O(N) summation be rejected in a system that processes millions of sensor readings per second?

Because the system also requires real‑time min/max queries; a plain scan cannot provide those without additional passes, leading to higher latency. Using a heap ensures each new reading is integrated with logarithmic overhead while preserving instant access to extremal values.

Examples

Example 1

Input

nums = [3, 1, 4, 1, 5]

Output

Explanation: Step 1: Initialize an empty Min-Max Heap and a sum variable `total = 0`. Step 2: Insert 3 into the heap. Heap: [3]. total = 3. Step 3: Insert 1 into the heap. Heap re-balances to maintain min-max property. Heap: [1, 3]. total = 4. Step 4: Insert 4 into the heap. Heap: [1, 3, 4]. total = 8. Step 5: Insert 1 into the heap. Heap: [1, 1, 4, 3]. total = 9. Step 6: Insert 5 into the heap. Heap: [1, 1, 4, 3, 5]. total = 14. Final Output: 14.

Example 2

Input

nums = [-2, 0, 2, -4, 4]

Output

Explanation: Step 1: Initialize heap and `total = 0`. Step 2: Insert -2. Heap: [-2]. total = -2. Step 3: Insert 0. Heap: [-2, 0]. total = -2. Step 4: Insert 2. Heap: [-2, 0, 2]. total = 0. Step 5: Insert -4. Heap: [-4, -2, 2, 0]. total = -4. Step 6: Insert 4. Heap: [-4, -2, 2, 0, 4]. total = 0. Final Output: 0.

Example 3

Input

nums = [100, 200, 300]

Output

Explanation: Step 1: Initialize heap and `total = 0`. Step 2: Insert 100. Heap: [100]. total = 100. Step 3: Insert 200. Heap: [100, 200]. total = 300. Step 4: Insert 300. Heap: [100, 200, 300]. total = 600. Final Output: 600.

Example 4

Input

nums = [7, 7, 7, 7]

Output

Explanation: Step 1: Initialize heap and `total = 0`. Step 2: Insert 7. Heap: [7]. total = 7. Step 3: Insert 7. Heap: [7, 7]. total = 14. Step 4: Insert 7. Heap: [7, 7, 7]. total = 21. Step 5: Insert 7. Heap: [7, 7, 7, 7]. total = 28. Final Output: 28.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
The sum of all elements will fit within a 64-bit signed integer.

Optimal Approach & Strategy

Insert each element into a Min‑Max Heap (O(log N) per insert) while updating a running sum – total O(N log N) time, O(N) space for the heap.

Brute Force Approach

Scan the array once, adding each element to a sum variable – O(N) time, O(1) space.

Code Solutions

Language:

JavaScript Solution

Time: O(N log N)

function solution(nums) {
   let minHeap = new MinHeap();
   let maxHeap = new MaxHeap();
   let sum = 0;

   for (let num of nums) {
      minHeap.insert(num);
      maxHeap.insert(num);
   }

   while (minHeap.size() > 0) {
      sum += minHeap.extractMin();
   }

   while (maxHeap.size() > 0) {
      sum -= maxHeap.extractMax();
   }

   return sum;
}

class MinHeap {
public:
    void insert(int num) {}
};
class MaxHeap {
public:
    void insert(int num) {}
};
class Solution {
public:
    int solution(vector<int>& nums) {
        MinHeap minHeap;
        MaxHeap maxHeap;
        int sum = 0;
        
        for (int num : nums) {
            minHeap.insert(num);
            maxHeap.insert(num);
            sum += num;
        }
        
        return sum;
    }
};

import java.util.PriorityQueue;
public class Solution {
    public int solution(int[] nums) {
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
        int sum = 0;
        
        for (int num : nums) {
            minHeap.add(num);
            maxHeap.add(num);
            sum += num;
        }
        
        return sum;
    }
}

import heapq

def solution(nums):
    minHeap = []
    maxHeap = []
    sum = 0
    
    for num in nums:
        heapq.heappush(minHeap, num)
        heapq.heappush(maxHeap, -num)
        sum += num
    
    return sum

function solution(nums) {
   let minHeap = new MinHeap();
   let maxHeap = new MaxHeap();
   let sum = 0;

   for (let num of nums) {
      minHeap.insert(num);
      maxHeap.insert(num);
   }

   while (minHeap.size() > 0) {
      sum += minHeap.extractMin();
   }

   while (maxHeap.size() > 0) {
      sum -= maxHeap.extractMax();
   }

   return sum;
}

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