Bitmask Energy Vector Architect — Problem Statement & Solution Guide

HeapHardMin-Max Priority Heap Queue

TimeO(N log N)

SpaceO(1)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Heap and solve the Bitmask Energy Vector Architect 2 problem optimally.

TopicHeap

PatternMin-Max Priority Heap Queue

TimeO(N log N)

SpaceO(1)

Problem Description

You are given an array nums of length N. Repeatedly perform the following operation until the array becomes empty: • If the array contains at least two elements, remove the current minimum value minVal and the current maximum value maxVal (and they may be the same element when the array size is two). Add the product minVal × maxVal to a running total. • If exactly one element remains, remove it and add its value (as a product with itself) to the total. After all removals, output the final total. The process must be carried out using a data structure that can retrieve both the minimum and maximum efficiently (a min‑max priority heap). The algorithm is greedy because at each step the extreme values are chosen to maximise the contribution of the current pair.

DSA Pattern Breakdown

"Bitmask Energy Vector Architect"

hard

WHY DOES IT MATTER?

Pairing extremes is a classic greedy pattern that guarantees optimality for problems where the objective depends on the product or difference of two elements. By always combining the smallest with the largest, we avoid suboptimal pairings that could arise from arbitrary removal orders.

OPTIMIZATION CHALLENGE

The key insight is that after sorting, the min and max are fixed at the two ends of the array, allowing us to compute all required products in a single linear scan instead of repeatedly searching for extremes.

REAL-WORLD CONNECTION

In distributed systems, load balancing often involves pairing the most heavily loaded node with the least loaded one to minimize overall latency. Similarly, in resource allocation, matching scarce resources with abundant demand mirrors the min-max pairing strategy.

During interviews, emphasize that sorting once and using two pointers is both time and space efficient. Mention that this approach is stable, easy to implement, and avoids the pitfalls of heap-based solutions.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log N)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to repeatedly pairing the smallest and largest remaining elements and summing their products. A naive approach would scan the array to find the min and max at each step, remove them, and repeat until the array is empty. This results in an O(N^2) time complexity because each scan is linear and we perform it N/2 times, which is infeasible for large N.

A more efficient strategy is to observe that the order in which we remove elements does not affect the final sum as long as we always pair the current minimum with the current maximum. By sorting the array once in O(N log N) time, we can then use two pointers—one starting at the beginning (the smallest) and one at the end (the largest)—to traverse the array in a single pass. At each step we multiply the values at the two pointers, add the product to the total, and move both pointers inward. This two-pointer technique guarantees O(N) additional time after sorting and uses only constant extra space if the sort is performed in-place.

Sorting is optimal because it provides a global view of the array’s order, eliminating the need for repeated min/max extraction. While a min-heap and max-heap approach also achieves O(N log N) time, it incurs higher constant factors and additional memory overhead. Therefore, sorting followed by a linear two-pointer sweep is the preferred paradigm for this problem.

Interview Questions on This Problem

Q1How would you handle an input array of size up to 10^7 while keeping memory usage low?

I would sort the array in-place using an efficient algorithm like quicksort or introsort that operates on the original array to avoid extra memory. After sorting, I would apply the two-pointer technique to compute the sum in a single linear pass, ensuring O(1) auxiliary space and O(N log N) time.

Q2Explain the time complexity if you used a min-heap and a max-heap instead of sorting.

Building both heaps takes O(N) time. Each extraction of min and max is O(log N), and we perform N/2 such extractions, leading to O(N log N) total time. However, the constant factors are higher compared to a single sort and two-pointer pass.

Q3What changes, if any, are needed when the array contains negative numbers?

The algorithm remains unchanged because the smallest (most negative) and largest (most positive) values are still correctly identified after sorting. The product of a negative and a positive number will be negative, and the sum will reflect that. Care must be taken with integer overflow when multiplying large magnitude negatives.

Examples

Example 1

Input

4
1 3 5 7

Output

Explanation: Step 1: min=1, max=7 → product=7, total=7, remaining [3,5]. Step 2: min=3, max=5 → product=15, total=7+15=22, array empty. Final answer is 22.

Example 2

Input

5
2 2 2 2 2

Output

Explanation: Step 1: min=2, max=2 → product=4, total=4, remaining [2,2,2]. Step 2: min=2, max=2 → product=4, total=8, remaining [2]. Step 3: only 2 left → product=2, total=10. Output 10.

Example 3

Input

3
-1 4 0

Output

-4

Explanation: Step 1: min=-1, max=4 → product=-4, total=-4, remaining [0]. Step 2: only 0 left → product=0, total stays -4. Output -4.

Constraints

1 <= N <= 2*10^5
-10^9 <= nums[i] <= 10^9
The answer may exceed 64‑bit signed integer range; use 128‑bit or arbitrary‑precision arithmetic.
All operations must run in O(N log N) time or better.

Optimal Approach & Strategy

Sort the array once in O(N log N) time, then use two pointers to traverse from both ends, multiplying and summing in a single O(N) pass. This uses only constant extra space if sorting is in-place.

Brute Force Approach

Find the minimum and maximum by scanning the array, remove them, multiply and add to the total, and repeat until the array is empty. This requires O(N^2) time because each scan is linear and we perform it N/2 times.

Code Solutions

Language:

JavaScript Solution

Time: O(N log N)

function solution(nums) {
   nums.sort((a, b) => a - b);
   let sum = 0;
   for (let i = 0; i < nums.length - 1; i++) {
       sum += nums[i];
       nums.splice(nums.indexOf(Math.max(...nums)), 1);
   }
   return sum;
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       sort(nums.begin(), nums.end());
       int sum = 0;
       for (int i = 0; i < nums.size() - 1; i++) {
           sum += nums[i];
           nums.erase(remove(nums.begin(), nums.end(), *max_element(nums.begin(), nums.end())), nums.end());
       }
       return sum;
   }
}

class Solution {
   public int solution(int[] nums) {
       Arrays.sort(nums);
       int sum = 0;
       for (int i = 0; i < nums.length - 1; i++) {
           sum += nums[i];
           nums = removeMax(nums);
       }
       return sum;
   }
   private int[] removeMax(int[] nums) {
       int maxIndex = Arrays.asList(nums).indexOf(Collections.max(nums));
       int[] newNums = new int[nums.length - 1];
       System.arraycopy(nums, 0, newNums, 0, maxIndex);
       System.arraycopy(nums, maxIndex + 1, newNums, maxIndex, nums.length - maxIndex - 1);
       return newNums;
   }
}

def solution(nums):
   nums.sort()
   sum = 0
   for i in range(len(nums) - 1):
       sum += nums[i]
       nums.remove(max(nums))
   return sum

function solution(nums) {
   nums.sort((a, b) => a - b);
   let sum = 0;
   for (let i = 0; i < nums.length - 1; i++) {
       sum += nums[i];
       nums.splice(nums.indexOf(Math.max(...nums)), 1);
   }
   return sum;
}

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