Bitmask Energy Vector Optimizer 4 — Problem Statement & Solution Guide

HeapHardMin-Max Priority Heap Queue

TimeO(N log N)

SpaceO(N)

Problem Description

You are given an array nums containing N integers. Using a min‑max priority heap you repeatedly perform the following operation until the heap becomes empty:

1. Extract the current minimum element minVal and the current maximum element maxVal from the heap.

2. Add |maxVal − minVal| to a running total.

If after an extraction the heap contains a single element, that element is added directly to the total (its absolute value is irrelevant because it is the only remaining value). Return the final total after the heap is exhausted.

The task is to compute this total efficiently. An implementation that explicitly sorts the array for each extraction would be too slow; instead a min‑max heap (a double‑ended priority queue) should be used so that each removal of the minimum and maximum costs O(log N).

DSA Pattern Breakdown

"Bitmask Energy Vector Optimizer 4"

hard

WHY DOES IT MATTER?

The min‑max extraction pattern is essential for problems that require balancing extremes, such as minimizing total cost or maximizing spread. It ensures that each operation removes the most influential elements, leading to optimal aggregate results.

OPTIMIZATION CHALLENGE

The key insight is that after removing the current extremes, the remaining elements retain their relative order, allowing us to continue extracting new extremes in logarithmic time rather than rescanning the entire set.

REAL-WORLD CONNECTION

In load balancing, servers often need to redistribute tasks from the busiest to the least busy nodes. The min‑max heap mirrors this by always pairing the most and least loaded servers, analogous to extracting extremes from the dataset.

When explaining this to an interviewer, emphasize that the data structure choice (min‑max heap vs. two heaps vs. BST) directly impacts both time and space, and that handling the single‑element case correctly is a common source of bugs.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log N)

💾 Space:O(N)

Core Theory — Why This Approach?

The problem requires repeatedly extracting the minimum and maximum elements from a multiset of integers and adding their absolute difference to a running total. A naive approach would scan the entire array to find min and max at each step, leading to O(N^2) time, which is infeasible for large N. The optimal solution leverages a min‑max priority heap (or two heaps) to maintain the current extremes in O(log N) per extraction, yielding an overall O(N log N) time complexity. The key insight is that after removing both extremes, the remaining elements are still sorted relative to each other, so we can continue to extract the new min and max efficiently without re‑scanning.

Using a single balanced binary search tree (e.g., TreeSet in Java or std::multiset in C++) also achieves O(log N) per operation, but a dedicated min‑max heap can be more space‑efficient because it stores each element only once and provides direct access to both ends. The algorithm also handles the special case when only one element remains: that element is added directly to the total, ensuring correctness without an extra extraction.

Overall, the optimal paradigm is to maintain a sorted structure that supports O(log N) removal of both the smallest and largest elements, thereby avoiding the quadratic cost of repeated linear scans and enabling the solution to scale to millions of elements.

Interview Questions on This Problem

Q1How would you modify the algorithm if the input array contains duplicate values?

Duplicates are naturally handled by a multiset or a min‑max heap that stores each occurrence separately. When extracting min or max, you simply remove one instance; the remaining duplicates stay in the structure and will be processed in subsequent iterations. No special handling is required beyond using a data structure that supports duplicate keys.

Q2What is the time complexity if you use two separate heaps (a min‑heap and a max‑heap) instead of a single min‑max heap?

Using two heaps would still give O(N log N) overall time, but you must synchronize deletions: when you pop from one heap, you need to mark the element as removed in the other heap, leading to additional overhead and potential O(N) cleanup in the worst case. A single min‑max heap or a balanced BST avoids this complication and keeps the constant factors lower.

Q3In a distributed system, how could you parallelize this extraction process to speed up computation on a very large dataset?

You could partition the array into shards, compute local min and max for each shard, and then perform a global reduction to find the overall min and max. After each extraction, you would need to update the shards or use a distributed priority queue. However, because the operation is inherently sequential (each extraction depends on the previous state), true parallelism is limited; the best you can do is parallel preprocessing and efficient I/O to feed the heap.

Examples

Example 1

Input

[1, 3, 5, 7]

Output

Explanation: Initial heap: {1,3,5,7}. Extract min=1 and max=7 → |7‑1|=6, total=6. Remaining heap: {3,5}. Extract min=3 and max=5 → |5‑3|=2, total=8. Heap is empty, answer=8.

Example 2

Input

[4, 4, 4]

Output

Explanation: First extraction removes min=4 and max=4 → |4‑4|=0, total=0. One element (4) remains, added directly → total=4. Heap empty, answer=4.

Example 3

Input

[10, -2, 0, 5, 3]

Output

Explanation: Step 1: min=-2, max=10 → |10‑(-2)|=12, total=12. Remaining {0,3,5}. Step 2: min=0, max=5 → |5‑0|=5, total=17. One element (3) left → add 3, total=20. Heap empty, answer=20.

Constraints

1 <= nums.length <= 2*10^5
-10^9 <= nums[i] <= 10^9
All operations must run in O(N log N) time or better
Only O(N) additional memory may be used

Optimal Approach & Strategy

Use a min‑max priority heap (or a balanced BST) to extract the minimum and maximum in O(log N) time per operation, resulting in O(N log N) total time.

Brute Force Approach

Scan the array to find the minimum and maximum, add their difference to the total, remove them, and repeat until the array is empty. This takes O(N^2) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log N)

function solution(matrix) {
   let n = matrix.length;
   let m = matrix[0].length;
   let min = new Array(n).fill(0).map(() => new Array(m).fill(Infinity));
   let max = new Array(n).fill(0).map(() => new Array(m).fill(-Infinity));
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           min[i][j] = Math.min(min[i][j], matrix[i][j]);
           max[i][j] = Math.max(max[i][j], matrix[i][j]);
       }
   }
   let result = 0;
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           result += Math.min(min[i][j], max[i][j]);
       }
   }
   return result;
}

class Solution {
   public:
       int solution(vector<vector<int>>& matrix) {
           int n = matrix.size();
           int m = matrix[0].size();
           vector<vector<int>> min(n, vector<int>(m, INT_MAX));
           vector<vector<int>> max(n, vector<int>(m, INT_MIN));
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < m; j++) {
                   min[i][j] = min(min[i][j], matrix[i][j]);
                   max[i][j] = max(max[i][j], matrix[i][j]);
               }
           }
           int result = 0;
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < m; j++) {
                   result += min(min[i][j], max[i][j]);
               }
           }
           return result;
       }
   };

class Solution {
   public int solution(int[][] matrix) {
       int n = matrix.length;
       int m = matrix[0].length;
       int[][] min = new int[n][m];
       int[][] max = new int[n][m];
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < m; j++) {
               min[i][j] = Math.min(min[i][j], matrix[i][j]);
               max[i][j] = Math.max(max[i][j], matrix[i][j]);
           }
       }
       int result = 0;
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < m; j++) {
               result += Math.min(min[i][j], max[i][j]);
           }
       }
       return result;
   }
}

def solution(matrix):
   n = len(matrix)
   m = len(matrix[0])
   min_val = [[float('inf')] * m for _ in range(n)]
   max_val = [[float('-inf')] * m for _ in range(n)]
   for i in range(n):
       for j in range(m):
           min_val[i][j] = min(min_val[i][j], matrix[i][j])
           max_val[i][j] = max(max_val[i][j], matrix[i][j])
   result = 0
   for i in range(n):
       for j in range(m):
           result += min(min_val[i][j], max_val[i][j])
   return result

function solution(matrix) {
   let n = matrix.length;
   let m = matrix[0].length;
   let min = new Array(n).fill(0).map(() => new Array(m).fill(Infinity));
   let max = new Array(n).fill(0).map(() => new Array(m).fill(-Infinity));
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           min[i][j] = Math.min(min[i][j], matrix[i][j]);
           max[i][j] = Math.max(max[i][j], matrix[i][j]);
       }
   }
   let result = 0;
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           result += Math.min(min[i][j], max[i][j]);
       }
   }
   return result;
}

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Bitmask Energy Vector Optimizer 4 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Heap

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Bitmask Energy Vector Optimizer 4 — 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