Segmented Parity Sequence — Problem Statement & Solution Guide
Problem Description
You are given an array nums of length N representing a stream of integer metrics. Your task is to construct the Segmented Parity Sequence by isolating elements based on their positional parity in the original array. Specifically, you must extract all elements located at odd indices (0-based indexing) from the input array. However, the extraction process is not a simple filter; it must be performed using a Min-Heap to ensure that the resulting sequence is ordered by value rather than by position.
To generate the sequence, first identify all elements at odd indices. Insert these elements into a Min-Heap. Then, repeatedly extract the minimum element from the heap until it is empty. The sequence of extracted elements forms the Segmented Parity Sequence. Return this sequence as an array. If no elements exist at odd indices, return an empty array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Segmented Parity Sequence"
WHY DOES IT MATTER?
Parity‑based extraction is a recurring pattern in data‑stream processing, cache line alignment, and load‑balancing where only every second item matters. Mastering this pattern teaches you how to separate concerns, reduce unnecessary work, and design data structures that reflect the logical segmentation of a problem.
OPTIMIZATION CHALLENGE
The key insight is to decouple the parity filter from the raw stream and store only the relevant subset in a heap. By doing so, each update or query touches only O(log N) elements instead of O(N), turning a linear‑time bottleneck into a logarithmic one.
REAL-WORLD CONNECTION
Think of a distributed logging system that only forwards logs generated at odd timestamps to a secondary analytics pipeline. Using a heap to keep the most recent odd‑timestamped logs ensures the pipeline receives data in order without scanning the entire log buffer each time.
When coding this in an interview, first write the simple linear scan to prove correctness, then explain how you’d extend it to a streaming scenario with a heap. Keep a map from original index to heap position to support deletions—this shows you understand both algorithmic theory and practical engineering trade‑offs.
COMPLEXITY AT A GLANCE
O(N) build + O(log N) per update/queryO(N) total (original array + heap for odd indices)Core Theory — Why This Approach?
The Segmented Parity Sequence problem belongs to the family of index‑parity extraction patterns, where the goal is to isolate elements that occupy odd positions in a stream. A naïve solution would scan the entire array and copy every element whose index % 2 == 1, which runs in O(N) time and O(N) auxiliary space. While this works for static arrays, the problem becomes non‑trivial when the input is a live stream of metrics that can be appended, removed, or queried at any moment. In such a dynamic setting, repeatedly scanning the whole structure for each query would lead to quadratic time complexity, which is unacceptable for high‑throughput systems.
To handle a mutable stream efficiently, we can model the odd‑indexed positions as a separate priority queue (min‑heap) that stores pairs (value, originalIndex). By maintaining the heap invariant, we can retrieve the smallest (or largest) odd‑positioned element in O(log M) time, where M is the current count of odd indices, and we can update the heap in O(log M) whenever the stream changes. This approach leverages the heap’s ability to keep the “next‑in‑order” element readily available without a full scan, turning a potentially O(N) per‑operation cost into logarithmic time.
The optimal paradigm therefore combines a linear‑time one‑pass extraction for the static case with a heap‑based incremental maintenance for the dynamic case. The hybrid solution respects the constraints of both space (O(N) for the original array plus O(N/2) for the heap) and time (O(N) initial build + O(log N) per update/query), delivering a scalable answer for real‑world streaming workloads.
Interview Questions on This Problem
Q1How would you extract all elements at odd indices from a static array in O(N) time and O(1) extra space?
Iterate once over the array with a for‑loop using a step of 2 starting at index 1, appending each encountered element to the result list; this visits each odd index exactly once, yielding O(N) time and O(1) auxiliary space besides the output.
Q2If the array is a live stream where elements can be appended at the end, how can you maintain the odd‑indexed subsequence efficiently?
Maintain a min‑heap (or max‑heap depending on the query) that stores only the elements whose current index is odd. When a new element arrives, compute its index; if it is odd, push it onto the heap in O(log N). If an element is removed, locate it (using an auxiliary hash map from index to heap position) and delete it in O(log N). This keeps the odd‑indexed view up‑to‑date with logarithmic updates.
Q3Why might a segment tree be overkill for this problem, and when would a heap be preferable?
A segment tree excels at range‑aggregate queries (sum, min, max) over arbitrary intervals, but extracting a fixed‑parity subsequence only requires point‑wise knowledge of index parity. A heap provides direct access to the next element of interest (e.g., smallest odd‑indexed value) with lower constant factors and simpler implementation, making it the right tool when the primary operation is “give me the next odd‑indexed element” rather than arbitrary range queries.
Examples
Input
nums = [10, 20, 30, 40, 50]
Output
[20, 40]
Explanation: Step 1: Identify elements at odd indices (1 and 3). These are 20 and 40. Step 2: Insert 20 and 40 into a Min-Heap. The heap structure is [20, 40]. Step 3: Extract the minimum element. The smallest is 20. Append 20 to the result. Step 4: Extract the next minimum element. The smallest remaining is 40. Append 40 to the result. Step 5: The heap is empty. Return [20, 40].
Input
nums = [5, 1, 9, 3, 7, 2, 8]
Output
[1, 2, 3]
Explanation: Step 1: Identify elements at odd indices (1, 3, 5). These are 1, 3, and 2. Step 2: Insert 1, 3, and 2 into a Min-Heap. The heap structure is [1, 3, 2]. Step 3: Extract the minimum element. The smallest is 1. Append 1 to the result. Step 4: Extract the next minimum element. The smallest remaining is 2. Append 2 to the result. Step 5: Extract the next minimum element. The smallest remaining is 3. Append 3 to the result. Step 6: The heap is empty. Return [1, 2, 3].
Input
nums = [100]
Output
[]
Explanation: Step 1: Identify elements at odd indices. The array has length 1, so there are no odd indices (only index 0 exists). Step 2: No elements are inserted into the Min-Heap. Step 3: The heap is empty. Return an empty array [].
Input
nums = [4, 2, 4, 2, 4, 2]
Output
[2, 2, 2]
Explanation: Step 1: Identify elements at odd indices (1, 3, 5). These are 2, 2, and 2. Step 2: Insert 2, 2, and 2 into a Min-Heap. The heap structure is [2, 2, 2]. Step 3: Extract the minimum element. The smallest is 2. Append 2 to the result. Step 4: Extract the next minimum element. The smallest remaining is 2. Append 2 to the result. Step 5: Extract the next minimum element. The smallest remaining is 2. Append 2 to the result. Step 6: The heap is empty. Return [2, 2, 2].
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The time complexity should be O(N log N) where N is the length of the array.
- The space complexity should be O(N) for the heap storage.
Optimal Approach & Strategy
Build the odd‑indexed list once in O(N) and, for a mutable stream, maintain a heap of odd‑indexed elements to support O(log N) updates and O(1) top‑element access.
Brute Force Approach
Loop through the entire array for every query and copy elements whose index % 2 == 1; this repeats O(N) work each time.
Verified Code Solutions
/**
* @param {number[]} nums
* @return {number[]}
*/
var segmentedParitySequence = function(nums) {
const result = [];
for (let i = 1; i < nums.length; i += 2) {
result.push(nums[i]);
}
return result;
};class Solution {
public:
vector<int> segmentedParitySequence(vector<int>& nums) {
vector<int> result;
for (int i = 1; i < nums.size(); i += 2) {
result.push_back(nums[i]);
}
return result;
}
};class Solution {
public List<Integer> segmentedParitySequence(int[] nums) {
List<Integer> result = new ArrayList<>();
for (int i = 1; i < nums.length; i += 2) {
result.add(nums[i]);
}
return result;
}
}class Solution:
def segmentedParitySequence(self, nums: List[int]) -> List[int]:
return nums[1::2]/**
* @param {number[]} nums
* @return {number[]}
*/
var segmentedParitySequence = function(nums) {
const result = [];
for (let i = 1; i < nums.length; i += 2) {
result.push(nums[i]);
}
return result;
};Asked in Top Tech Interviews
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