Heavy-Light Path Sum Synthesizer 3 — Problem Statement & Solution Guide

HeapHardMin-Max Priority Heap Queue

TimeO(n log n)

SpaceO(n)

Problem Description

You are tasked with processing a sequence of integer values using a specialized Min-Max Priority Heap Queue structure. The objective is to compute the cumulative sum of all elements in the input array while strictly adhering to the operational semantics of a Min-Max heap. In this context, the 'Min-Max' aspect implies that the heap must maintain the ability to efficiently access both the minimum and maximum elements, although for the purpose of this specific summation task, the primary operation is the extraction and accumulation of all values.

Given an array of integers nums, you must initialize a Min-Max Priority Heap Queue with these values. Then, you must repeatedly extract elements from the heap until it is empty, adding each extracted value to a running total. The final result is this running total. Although the summation of an array is mathematically straightforward, this problem tests your ability to correctly implement or simulate the internal mechanics of a Min-Max heap, including the sifting down/up operations that maintain the heap property during extraction. The challenge lies in ensuring that the extraction order and the heap maintenance logic are correctly applied, even though the final sum is invariant to the order of extraction.

Your function should accept the array nums and return the integer sum of all its elements, derived through the complete process of heap initialization and full extraction. This problem serves as a rigorous test of understanding priority queue data structures, specifically the dual-heap or min-max heap variant, where the structure must support efficient retrieval of both extremes if required, but here is used to demonstrate the full lifecycle of heap operations.

DSA Pattern Breakdown

"Heavy-Light Path Sum Synthesizer 3"

hard

WHY DOES IT MATTER?

The Min‑Max heap pattern is essential when an algorithm must balance two extremes—extracting both the smallest and largest elements efficiently. It eliminates the need for two separate heaps or repeated scans, reducing both time and space overhead.

OPTIMIZATION CHALLENGE

The key insight is that by structuring the heap to alternate min and max levels, we can perform both extractions in O(log n) time, avoiding the O(n) cost of linear scans and the O(n log n) cost of two separate heaps.

REAL-WORLD CONNECTION

In high‑frequency trading platforms, order books maintain the best bid (max) and best ask (min) prices. A Min‑Max heap can model this dual‑access requirement, allowing rapid updates as orders are added or removed.

When explaining this pattern in an interview, emphasize the level‑alternating property and show how a single array representation supports both operations. Demonstrate with a small example to illustrate the parent‑grandparent comparisons.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log n)

💾 Space:O(n)

Core Theory — Why This Approach?

A Min‑Max heap is a complete binary tree that supports both “extract‑min” and “extract‑max” in O(log n) time. Unlike a standard binary heap, which only guarantees fast access to one extreme, a Min‑Max heap alternates levels: even levels store minima, odd levels store maxima, and each node is compared only with its grandparents. This structure allows us to retrieve the smallest and largest elements in constant time while maintaining the heap property after each insertion or deletion.

Naïve approaches that simply sum the array or use a single‑direction heap fail on large inputs because they either ignore the required Min‑Max semantics or incur O(n) per operation. For example, repeatedly scanning the array to find the min or max would be O(n²). A single‑direction heap would require two passes—one to find all minima and another for maxima—doubling the cost and violating the problem’s constraint of using a Min‑Max heap.

The optimal paradigm is to build a Min‑Max heap in O(n) time using Floyd’s algorithm, then repeatedly perform extract‑min and extract‑max while accumulating the popped values. Each extraction costs O(log n), and we perform 2n extractions (n minima and n maxima) to visit every element exactly once, yielding an overall O(n log n) time and O(n) space solution that respects the heap’s dual‑access semantics.

Interview Questions on This Problem

Q1How does a Min‑Max heap differ from a standard binary heap, and why is it useful for problems requiring both minimum and maximum extraction?

A Min‑Max heap alternates levels between min and max nodes, allowing O(1) access to both extremes and O(log n) updates. This is useful when an algorithm must frequently remove both the smallest and largest elements, such as in median maintenance or this cumulative sum problem.

Q2What is the time complexity of building a Min‑Max heap from an unsorted array, and how does it compare to inserting elements one by one?

Building a Min‑Max heap can be done in O(n) time using Floyd’s algorithm, whereas inserting n elements individually would cost O(n log n). The linear build is achieved by heapifying from the bottom up, reducing the number of percolations.

Q3In a distributed system that processes streaming data, how could a Min‑Max heap be employed to maintain real‑time statistics, and what challenges might arise?

A Min‑Max heap can keep track of the current minimum and maximum in a sliding window or stream, enabling quick updates as new data arrives. Challenges include handling duplicate values, ensuring thread safety, and managing memory when the stream is unbounded.

Examples

Example 1

Input

nums = [3, 1, 4, 1, 5, 9, 2, 6]

Output

Explanation: 1. Initialize Min-Max Heap with [3, 1, 4, 1, 5, 9, 2, 6]. 2. Extract min (1), sum = 1. Heap restructures. 3. Extract min (1), sum = 2. Heap restructures. 4. Extract min (2), sum = 4. Heap restructures. 5. Extract min (3), sum = 7. Heap restructures. 6. Extract min (4), sum = 11. Heap restructures. 7. Extract min (5), sum = 16. Heap restructures. 8. Extract min (6), sum = 22. Heap restructures. 9. Extract min (9), sum = 31. Heap is empty. 10. Return 31.

Example 2

Input

nums = [-5, 10, -2, 7, 0]

Output

Explanation: 1. Initialize Min-Max Heap with [-5, 10, -2, 7, 0]. 2. Extract min (-5), sum = -5. Heap restructures. 3. Extract min (-2), sum = -7. Heap restructures. 4. Extract min (0), sum = -7. Heap restructures. 5. Extract min (7), sum = 0. Heap restructures. 6. Extract min (10), sum = 10. Heap is empty. 7. Return 10.

Example 3

Input

nums = [1000000, -1000000, 500000, -500000]

Output

Explanation: 1. Initialize Min-Max Heap with [1000000, -1000000, 500000, -500000]. 2. Extract min (-1000000), sum = -1000000. Heap restructures. 3. Extract min (-500000), sum = -1500000. Heap restructures. 4. Extract min (500000), sum = -1000000. Heap restructures. 5. Extract min (1000000), sum = 0. Heap is empty. 6. Return 0.

Example 4

Input

nums = [42]

Output

Explanation: 1. Initialize Min-Max Heap with [42]. 2. Extract min (42), sum = 42. Heap is empty. 3. Return 42.

Constraints

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

Optimal Approach & Strategy

Build a Min‑Max heap in O(n) time, then perform 2n extract‑min and extract‑max operations, each costing O(log n), while accumulating the popped values. This yields O(n log n) time and O(n) space.

Brute Force Approach

Insert all numbers into a simple list, then repeatedly scan the list to find the minimum and maximum, remove them, and add to the sum. This takes O(n²) time because each scan is linear.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n log n)

function solution(nums) {
      class MinMaxHeap {
         constructor() {
            this.heap = [];
         }

         insert(val) {
            this.heap.push(val);
            this.heapifyUp(this.heap.length - 1);
         }

         extractMin() {
            if (this.heap.length === 0) return null;
            if (this.heap.length === 1) return this.heap.pop();

            const min = this.heap[0];
            this.heap[0] = this.heap.pop();
            this.heapifyDown(0);
            return min;
         }

         heapifyUp(index) {
            if (index <= 0) return;
            const parentIndex = Math.floor((index - 1) / 2);
            if (this.heap[parentIndex] > this.heap[index]) {
               this.swap(parentIndex, index);
               this.heapifyUp(parentIndex);
            }
         }

         heapifyDown(index) {
            const leftChildIndex = 2 * index + 1;
            const rightChildIndex = 2 * index + 2;
            let smallest = index;

            if (leftChildIndex < this.heap.length && this.heap[leftChildIndex] < this.heap[smallest]) {
               smallest = leftChildIndex;
            }

            if (rightChildIndex < this.heap.length && this.heap[rightChildIndex] < this.heap[smallest]) {
               smallest = rightChildIndex;
            }

            if (smallest !== index) {
               this.swap(smallest, index);
               this.heapifyDown(smallest);
            }
         }

         swap(i, j) {
            const temp = this.heap[i];
            this.heap[i] = this.heap[j];
            this.heap[j] = temp;
         }
      }

      const minMaxHeap = new MinMaxHeap();
      for (const num of nums) {
         minMaxHeap.insert(num);
      }

      let sum = 0;
      while (true) {
         const min = minMaxHeap.extractMin();
         if (min === null) break;
         sum += min;
      }

      return sum;
   }

class MinMaxHeap {
   public:
      std::vector<int> heap;
      int size;

      MinMaxHeap(int capacity) {
         heap.resize(capacity);
         size = 0;
      }

      void insert(int val) {
         if (size == heap.size()) {
            resize();
         }
         heap[size] = val;
         heapifyUp(size);
         size++;
      }

      int extractMin() {
         if (size == 0) {
            return INT_MIN;
         }
         int min = heap[0];
         heap[0] = heap[size - 1];
         size--;
         heapifyDown(0);
         return min;
      }

   private:
      void heapifyUp(int index) {
         if (index <= 0) {
            return;
         }
         int parentIndex = (index - 1) / 2;
         if (heap[parentIndex] > heap[index]) {
            swap(parentIndex, index);
            heapifyUp(parentIndex);
         }
      }

      void heapifyDown(int index) {
         int leftChildIndex = 2 * index + 1;
         int rightChildIndex = 2 * index + 2;
         int smallest = index;

         if (leftChildIndex < size && heap[leftChildIndex] < heap[smallest]) {
            smallest = leftChildIndex;
         }

         if (rightChildIndex < size && heap[rightChildIndex] < heap[smallest]) {
            smallest = rightChildIndex;
         }

         if (smallest != index) {
            swap(smallest, index);
            heapifyDown(smallest);
         }
      }

      void swap(int i, int j) {
         int temp = heap[i];
         heap[i] = heap[j];
         heap[j] = temp;
      }

      void resize() {
         std::vector<int> newHeap(heap.size() * 2);
         std::copy(heap.begin(), heap.end(), newHeap.begin());
         heap = newHeap;
      }
   };

   class Solution {
   public:
      int solution(std::vector<int>& nums) {
         MinMaxHeap minMaxHeap(nums.size());
         for (int num : nums) {
            minMaxHeap.insert(num);
         }

         int sum = 0;
         while (true) {
            int min = minMaxHeap.extractMin();
            if (min == INT_MIN) {
               break;
            }
            sum += min;
         }

         return sum;
      }
   };

class MinMaxHeap {
      private int[] heap;
      private int size;

      public MinMaxHeap(int capacity) {
         heap = new int[capacity];
         size = 0;
      }

      public void insert(int val) {
         if (size == heap.length) {
            resize();
         }
         heap[size] = val;
         heapifyUp(size);
         size++;
      }

      public int extractMin() {
         if (size == 0) {
            return Integer.MIN_VALUE;
         }
         int min = heap[0];
         heap[0] = heap[size - 1];
         size--;
         heapifyDown(0);
         return min;
      }

      private void heapifyUp(int index) {
         if (index <= 0) {
            return;
         }
         int parentIndex = (index - 1) / 2;
         if (heap[parentIndex] > heap[index]) {
            swap(parentIndex, index);
            heapifyUp(parentIndex);
         }
      }

      private void heapifyDown(int index) {
         int leftChildIndex = 2 * index + 1;
         int rightChildIndex = 2 * index + 2;
         int smallest = index;

         if (leftChildIndex < size && heap[leftChildIndex] < heap[smallest]) {
            smallest = leftChildIndex;
         }

         if (rightChildIndex < size && heap[rightChildIndex] < heap[smallest]) {
            smallest = rightChildIndex;
         }

         if (smallest != index) {
            swap(smallest, index);
            heapifyDown(smallest);
         }
      }

      private void swap(int i, int j) {
         int temp = heap[i];
         heap[i] = heap[j];
         heap[j] = temp;
      }

      private void resize() {
         int[] newHeap = new int[heap.length * 2];
         System.arraycopy(heap, 0, newHeap, 0, heap.length);
         heap = newHeap;
      }
   }

public class Solution {
   public int solution(int[] nums) {
      MinMaxHeap minMaxHeap = new MinMaxHeap(nums.length);
      for (int num : nums) {
         minMaxHeap.insert(num);
      }

      int sum = 0;
      while (true) {
         int min = minMaxHeap.extractMin();
         if (min == Integer.MIN_VALUE) {
            break;
         }
         sum += min;
      }

      return sum;
   }
}

class MinMaxHeap:
      def __init__(self):
         self.heap = []

      def insert(self, val):
         self.heap.append(val)
         self.heapify_up(len(self.heap) - 1)

      def extract_min(self):
         if len(self.heap) == 0:
            return None
         if len(self.heap) == 1:
            return self.heap.pop()

         min_val = self.heap[0]
         self.heap[0] = self.heap.pop()
         self.heapify_down(0)
         return min_val

      def heapify_up(self, index):
         if index <= 0:
            return
         parent_index = (index - 1) // 2
         if self.heap[parent_index] > self.heap[index]:
            self.swap(parent_index, index)
            self.heapify_up(parent_index)

      def heapify_down(self, index):
         left_child_index = 2 * index + 1
         right_child_index = 2 * index + 2
         smallest = index

         if left_child_index < len(self.heap) and self.heap[left_child_index] < self.heap[smallest]:
            smallest = left_child_index

         if right_child_index < len(self.heap) and self.heap[right_child_index] < self.heap[smallest]:
            smallest = right_child_index

         if smallest != index:
            self.swap(smallest, index)
            self.heapify_down(smallest)

      def swap(self, i, j):
         temp = self.heap[i]
         self.heap[i] = self.heap[j]
         self.heap[j] = temp

def solution(nums):
   min_max_heap = MinMaxHeap()
   for num in nums:
      min_max_heap.insert(num)

   total_sum = 0
   while True:
      min_val = min_max_heap.extract_min()
      if min_val is None:
         break
      total_sum += min_val

   return total_sum

function solution(nums) {
      class MinMaxHeap {
         constructor() {
            this.heap = [];
         }

         insert(val) {
            this.heap.push(val);
            this.heapifyUp(this.heap.length - 1);
         }

         extractMin() {
            if (this.heap.length === 0) return null;
            if (this.heap.length === 1) return this.heap.pop();

            const min = this.heap[0];
            this.heap[0] = this.heap.pop();
            this.heapifyDown(0);
            return min;
         }

         heapifyUp(index) {
            if (index <= 0) return;
            const parentIndex = Math.floor((index - 1) / 2);
            if (this.heap[parentIndex] > this.heap[index]) {
               this.swap(parentIndex, index);
               this.heapifyUp(parentIndex);
            }
         }

         heapifyDown(index) {
            const leftChildIndex = 2 * index + 1;
            const rightChildIndex = 2 * index + 2;
            let smallest = index;

            if (leftChildIndex < this.heap.length && this.heap[leftChildIndex] < this.heap[smallest]) {
               smallest = leftChildIndex;
            }

            if (rightChildIndex < this.heap.length && this.heap[rightChildIndex] < this.heap[smallest]) {
               smallest = rightChildIndex;
            }

            if (smallest !== index) {
               this.swap(smallest, index);
               this.heapifyDown(smallest);
            }
         }

         swap(i, j) {
            const temp = this.heap[i];
            this.heap[i] = this.heap[j];
            this.heap[j] = temp;
         }
      }

      const minMaxHeap = new MinMaxHeap();
      for (const num of nums) {
         minMaxHeap.insert(num);
      }

      let sum = 0;
      while (true) {
         const min = minMaxHeap.extractMin();
         if (min === null) break;
         sum += min;
      }

      return sum;
   }

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Heavy-Light Path Sum Synthesizer 3 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Heap

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Heavy-Light Path Sum Synthesizer 3 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor