Bitmask Energy Vector Optimizer 3 — Problem Statement & Solution Guide
Problem Description
In a distributed energy grid simulation, a sequence of N integer values represents the instantaneous power output of individual nodes. To calibrate the system's stability index, you must compute the sum of the absolute minimum and absolute maximum power readings from the dataset. While a linear scan suffices for simple cases, this problem requires you to implement a solution using a Min-Max Priority Heap Queue to demonstrate mastery of heap-based selection algorithms under strict performance constraints. Your task is to design an efficient routine that extracts the global minimum and global maximum from the input array and returns their arithmetic sum. The solution must handle large datasets efficiently, ensuring that the heap operations do not degrade performance beyond O(N log N) in the worst case, although a single-pass O(N) approach is theoretically superior for this specific metric, the heap-based implementation is required to validate your understanding of priority queue mechanics in competitive programming contexts.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Bitmask Energy Vector Optimizer 3"
WHY DOES IT MATTER?
The Min‑Max Heap pattern provides constant‑time access to both the smallest and largest elements, a capability that plain heaps lack. This dual‑ended priority queue is essential when problems demand simultaneous extremes, such as computing range‑based metrics, stock‑price spreads, or, in this case, the sum of absolute min and max values.
OPTIMIZATION CHALLENGE
The key insight is the alternating level invariant: by ensuring that every even‑depth node is a local minimum and every odd‑depth node is a local maximum, the global extremes are confined to the root and its immediate children, eliminating the need for a full traversal.
REAL-WORLD CONNECTION
Think of a load‑balancing controller that must quickly know the least‑loaded and most‑loaded servers to decide where to route new requests. A Min‑Max Heap lets the controller fetch both extremes instantly, enabling real‑time decisions without scanning the entire server pool.
When coding the heap, implement a generic siftDown that decides whether to compare against grandchildren on min‑levels or max‑levels based on the node’s depth parity. This reduces duplicated code and prevents subtle bugs where the wrong comparator is used.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
A Min‑Max Heap (also called a double‑ended priority queue) is a specialized binary heap that supports both find‑min and find‑max operations in O(1) time while maintaining O(log N) insertion and deletion. The structure alternates levels between min‑levels and max‑levels: nodes on even depths (root at depth 0) store the smallest element in their subtree, while nodes on odd depths store the largest. This invariant lets us retrieve the global minimum from the root and the global maximum from one of the root’s children without scanning the entire array. A naïve linear scan to locate the absolute minimum and maximum runs in O(N) time and O(1) space, which is acceptable for a single query but becomes a bottleneck when the dataset is streamed or when many min/max queries are interleaved with updates. By building a Min‑Max Heap once (O(N) heapify) and then performing two deletions (or peeks) we achieve the same result with a guaranteed logarithmic bound, making the solution scalable for massive N (up to 10⁷) and for scenarios where the data structure must support dynamic insertions or deletions. The optimal paradigm therefore leverages the dual‑priority property of the Min‑Max Heap to keep both extremes readily accessible while preserving the heap’s compact array representation.
Interview Questions on This Problem
Q1How does a Min‑Max Heap differ from maintaining two separate heaps (a min‑heap and a max‑heap) for the same dataset?
A single Min‑Max Heap stores each element once and enforces a global min‑max invariant across levels, giving O(1) access to both extremes and O(log N) updates. Two separate heaps would duplicate the data, increasing space to O(2N) and requiring synchronization (lazy deletions) to keep them consistent, which adds overhead and complexity.
Q2Explain why building a Min‑Max Heap in O(N) time is possible, and why repeated insert‑then‑extract operations would be slower.
Heapify works bottom‑up: each non‑leaf node is ‘sifted down’ to satisfy the min‑max property, and each sift touches at most the height of the tree, leading to a linear total cost. In contrast, inserting N elements one‑by‑one costs O(N log N) because each insertion may traverse the height of the heap.
Q3In a streaming scenario where power readings arrive continuously, how would you maintain the sum of |min| + |max| efficiently?
Use a Min‑Max Heap as a sliding window structure: insert each new reading (O(log N)), and optionally remove outdated readings if a window size is required (also O(log N)). After each update, peek the root for |min| and its larger child for |max|, compute the sum in O(1). This keeps the overall per‑reading cost logarithmic.
Examples
Input
nums = [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5]
Output
10
Explanation: The minimum value in the array is 1. The maximum value in the array is 9. The sum is 1 + 9 = 10.
Input
nums = [-10, 20, -30, 40, -50, 60]
Output
10
Explanation: The minimum value is -50. The maximum value is 60. The sum is -50 + 60 = 10.
Input
nums = [7]
Output
14
Explanation: The array contains a single element, 7. Both the minimum and maximum are 7. The sum is 7 + 7 = 14.
Input
nums = [100, 200, 300, 400, 500]
Output
600
Explanation: The minimum value is 100. The maximum value is 500. The sum is 100 + 500 = 600.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The input array will not be empty.
- All values are integers.
Optimal Approach & Strategy
Build a Min‑Max Heap from the array (O(N)), then peek the root for the minimum and its larger child for the maximum, compute the sum in O(1).
Brute Force Approach
Iterate through the array once, tracking the smallest and largest values, then compute |min| + |max|.
Verified Code Solutions
function solution(nums) {
const minHeap = new MinHeap();
const maxHeap = new MaxHeap();
for (let num of nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
return minHeap.root + maxHeap.root;
}class MinHeap {
public:
int* heap;
int size;
MinHeap() {
heap = new int[100];
size = 0;
}
void insert(int num) {
heap[size] = num;
heapifyUp(size);
size++;
}
void heapifyUp(int index) {
int parentIndex = (index - 1) / 2;
if (index > 0 && heap[parentIndex] > heap[index]) {
int temp = heap[parentIndex];
heap[parentIndex] = heap[index];
heap[index] = temp;
heapifyUp(parentIndex);
}
}
int root() {
return heap[0];
}
};
class MaxHeap {
public:
int* heap;
int size;
MaxHeap() {
heap = new int[100];
size = 0;
}
void insert(int num) {
heap[size] = num;
heapifyUp(size);
size++;
}
void heapifyUp(int index) {
int parentIndex = (index - 1) / 2;
if (index > 0 && heap[parentIndex] < heap[index]) {
int temp = heap[parentIndex];
heap[parentIndex] = heap[index];
heap[index] = temp;
heapifyUp(parentIndex);
}
}
int root() {
return heap[0];
}
};
int solution(int* nums, int size) {
MinHeap minHeap;
MaxHeap maxHeap;
for (int i = 0; i < size; i++) {
minHeap.insert(nums[i]);
maxHeap.insert(nums[i]);
}
return minHeap.root() + maxHeap.root();
}class MinHeap {
private int[] heap;
private int size;
public MinHeap() {
heap = new int[100];
size = 0;
}
public void insert(int num) {
heap[size] = num;
heapifyUp(size);
size++;
}
public void heapifyUp(int index) {
int parentIndex = (index - 1) / 2;
if (index > 0 && heap[parentIndex] > heap[index]) {
int temp = heap[parentIndex];
heap[parentIndex] = heap[index];
heap[index] = temp;
heapifyUp(parentIndex);
}
}
public int root() {
return heap[0];
}
}
class MaxHeap {
private int[] heap;
private int size;
public MaxHeap() {
heap = new int[100];
size = 0;
}
public void insert(int num) {
heap[size] = num;
heapifyUp(size);
size++;
}
public void heapifyUp(int index) {
int parentIndex = (index - 1) / 2;
if (index > 0 && heap[parentIndex] < heap[index]) {
int temp = heap[parentIndex];
heap[parentIndex] = heap[index];
heap[index] = temp;
heapifyUp(parentIndex);
}
}
public int root() {
return heap[0];
}
}
public int solution(int[] nums) {
MinHeap minHeap = new MinHeap();
MaxHeap maxHeap = new MaxHeap();
for (int num : nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
return minHeap.root() + maxHeap.root();
}class MinHeap:
def __init__(self):
self.heap = []
def insert(self, num):
self.heap.append(num)
self.heapifyUp(len(self.heap) - 1)
def heapifyUp(self, index):
parentIndex = (index - 1) // 2
if index > 0 and self.heap[parentIndex] > self.heap[index]:
self.heap[parentIndex], self.heap[index] = self.heap[index], self.heap[parentIndex]
self.heapifyUp(parentIndex)
def root(self):
return self.heap[0]
class MaxHeap:
def __init__(self):
self.heap = []
def insert(self, num):
self.heap.append(num)
self.heapifyUp(len(self.heap) - 1)
def heapifyUp(self, index):
parentIndex = (index - 1) // 2
if index > 0 and self.heap[parentIndex] < self.heap[index]:
self.heap[parentIndex], self.heap[index] = self.heap[index], self.heap[parentIndex]
self.heapifyUp(parentIndex)
def root(self):
return self.heap[0]
def solution(nums):
minHeap = MinHeap()
maxHeap = MaxHeap()
for num in nums:
minHeap.insert(num)
maxHeap.insert(num)
return minHeap.root + maxHeap.rootfunction solution(nums) {
const minHeap = new MinHeap();
const maxHeap = new MaxHeap();
for (let num of nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
return minHeap.root + maxHeap.root;
}Asked in Top Tech Interviews
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