Sequential Parity Sequence — Problem Statement & Solution Guide
Problem Description
In a distributed data processing pipeline, a stream of integer metrics is received sequentially. To optimize downstream aggregation, the system requires a specific parity-based transformation of this stream. You are tasked with implementing the core logic that processes an array of integers and computes the 'Sequential Parity Sequence' sum.
The algorithm operates as follows: Iterate through the input array from left to right. For each element, determine its parity (0 for even, 1 for odd). Maintain a running accumulator that starts at 0. At each step, add the current element's parity value to the accumulator. The final result is the total sum of these parity values after processing the entire sequence.
Given an array nums of length N, return the integer sum of the parities of all elements in the array. Note that the parity of a negative integer is determined by its absolute value's divisibility by 2 (i.e., -3 is odd, -4 is even).
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sequential Parity Sequence"
WHY DOES IT MATTER?
Stream partitioning by parity or attributes is a fundamental pattern in streaming architectures where distinct data types require different downstream routing or transformation logic while maintaining arrival sequence.
OPTIMIZATION CHALLENGE
The key challenge is preserving relative order within parity partitions without performing quadratic search-and-shift operations on dynamic arrays.
REAL-WORLD CONNECTION
This mimics real-world network traffic shaping and QoS (Quality of Service) systems where high-priority and low-priority network packets are ingested into separate queue buffers and serviced sequentially according to strict fairness or alternating scheduling algorithms.
Always explicitly check queue empty conditions before dequeuing during stream processing to prevent underflow errors or null pointer exceptions.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The 'Sequential Parity Sequence' problem explores stream processing under structural constraints where incoming data must be grouped, reordered, or evaluated based on numerical parity (even vs. odd values). In streaming and distributed pipelines, data often arrives out of optimal order. Using FIFO queues for separate parity tracks allows linear-time re-sequencing without sacrificing temporal order within each parity subset.
A naive approach might repeatedly scan the input array or sort elements using customized comparator functions, incurring O(N log N) or O(N^2) overhead per batch. This becomes prohibitively expensive when handling high-throughput real-time streams where latency constraints are strict. Furthermore, array-shifting operations during iterative extractions introduce hidden quadratic runtimes.
The optimal approach leverages dual Queues to maintain order invariants. By buffering even and odd numbers into separate FIFO queues as they arrive, we decouple ingestion from processing. This allows O(1) enqueuing and dequeuing, enabling linear processing complexity O(N) and optimal memory utilization O(N) for stream aggregation.
Interview Questions on This Problem
Q1How would you handle unbounded data streams where storing all elements in memory is impossible?
In an unbounded stream setting, we enforce fixed-capacity circular queues or a sliding window mechanism. When a queue reaches threshold capacity, we can either trigger batch aggregation downstream or evict metrics based on drop policies while keeping track of running aggregated metric states.
Q2Why use FIFO queues over priority queues for preserving sequential parity ordering?
FIFO queues guarantee strictly O(1) time complexity for insertion and removal while inherently preserving the original temporal arrival order of elements within each parity track. Priority queues introduce an unnecessary O(log K) overhead per operation and are intended for priority ordering rather than sequential FIFO processing.
Q3How does this pattern scale in a multi-threaded consumer-producer environment?
In concurrent systems, standard queues are replaced with lock-free or bounded blocking queues (e.g., ArrayBlockingQueue in Java or ConcurrentQueue in C++). Producers append to even and odd queues concurrently, and worker threads consume from both queues in a thread-safe manner without global lock contention.
Examples
Input
nums = [1, 2, 3, 4, 5]
Output
3
Explanation: Process elements sequentially: 1. Element 1 is odd (parity 1). Sum = 0 + 1 = 1. 2. Element 2 is even (parity 0). Sum = 1 + 0 = 1. 3. Element 3 is odd (parity 1). Sum = 1 + 1 = 2. 4. Element 4 is even (parity 0). Sum = 2 + 0 = 2. 5. Element 5 is odd (parity 1). Sum = 2 + 1 = 3. Final output is 3.
Input
nums = [10, 20, 30]
Output
0
Explanation: Process elements sequentially: 1. Element 10 is even (parity 0). Sum = 0 + 0 = 0. 2. Element 20 is even (parity 0). Sum = 0 + 0 = 0. 3. Element 30 is even (parity 0). Sum = 0 + 0 = 0. Final output is 0.
Input
nums = [-1, -2, -3, -4]
Output
2
Explanation: Process elements sequentially: 1. Element -1 is odd (parity 1). Sum = 0 + 1 = 1. 2. Element -2 is even (parity 0). Sum = 1 + 0 = 1. 3. Element -3 is odd (parity 1). Sum = 1 + 1 = 2. 4. Element -4 is even (parity 0). Sum = 2 + 0 = 2. Final output is 2.
Input
nums = [7, 7, 7]
Output
3
Explanation: Process elements sequentially: 1. Element 7 is odd (parity 1). Sum = 0 + 1 = 1. 2. Element 7 is odd (parity 1). Sum = 1 + 1 = 2. 3. Element 7 is odd (parity 1). Sum = 2 + 1 = 3. Final output is 3.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The input array will not be empty.
- Time complexity must be O(N) where N is the length of the array.
- Space complexity must be O(1) excluding the input storage.
Optimal Approach & Strategy
Separate the input sequence into two FIFO queues for even and odd elements in a single O(N) pass. Merge or aggregate elements by popping from the appropriate queue based on target sequence rules in O(1) time per step.
Brute Force Approach
Repeatedly scan the array to search for the next element matching the required parity, extracting it and shifting remaining elements. This leads to an O(N^2) time complexity due to repeated linear scans and dynamic array resizing operations.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += (num % 2 === 0) ? 0 : 1;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += (num % 2 == 0) ? 0 : 1;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += (num % 2 == 0) ? 0 : 1;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += (num % 2 == 0) and 0 or 1
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += (num % 2 === 0) ? 0 : 1;
}
return sum;
}Asked in Top Tech Interviews
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