Galactic Expedition Data Processor — Problem Statement & Solution Guide
Problem Description
You are given two independent telemetry streams, streamA and streamB, each represented as an array of integers. The values within each stream are unordered and may contain duplicates. Your task is to produce a single array that contains all measurements from both streams, sorted in non‑decreasing order. The resulting array should preserve the multiplicity of each value.
Input: Two arrays of integers, streamA and streamB.
Output: One array containing all elements from streamA and streamB, sorted in ascending order.
The solution must handle large inputs efficiently, using an algorithm with linearithmic time complexity or better.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Expedition Data Processor"
WHY DOES IT MATTER?
This pattern is the foundation of the Merge Sort algorithm and is critical for understanding how to combine sorted data streams. It teaches the two-pointer technique, which is ubiquitous in array and linked list problems. Understanding the difference between sorting unsorted data and merging sorted data is a key conceptual hurdle for junior engineers.
OPTIMIZATION CHALLENGE
The key insight is recognizing that if the inputs were sorted, the problem would be linear. Since they are not, the bottleneck is the sorting step. The optimization challenge is to choose the most efficient sorting algorithm for the individual arrays (e.g., Timsort for stability and performance on partially sorted data) and then perform a stable merge to preserve the relative order of equal elements if required.
REAL-WORLD CONNECTION
This is analogous to merging two sorted log files from different servers. In distributed systems, data is often partitioned and sorted locally before being merged globally. The 'unsorted' aspect here represents the initial ingestion phase where data arrives in arbitrary order, requiring a local sort before the global merge can occur efficiently.
In an interview, explicitly state that you are assuming the arrays are unsorted. If the interviewer clarifies they are sorted, pivot immediately to the two-pointer merge. If they remain unsorted, discuss the trade-offs of sorting strategies. Mentioning 'stable sort' is a bonus point, as it ensures that equal elements from streamA appear before those from streamB if that is a requirement.
COMPLEXITY AT A GLANCE
O(N log N + M log M)O(N + M)Core Theory — Why This Approach?
The problem of merging two unsorted arrays into a single sorted array is fundamentally a sorting problem. The naive approach involves concatenating the two arrays and applying a general-purpose sorting algorithm like QuickSort or MergeSort. While this works, it treats the input as a single unstructured dataset, ignoring the fact that we are dealing with two distinct sources. For large inputs, the overhead of a full sort on $N+M$ elements is $O((N+M)\log(N+M))$. However, if the input arrays were already sorted, we could merge them in linear time $O(N+M)$ using a two-pointer technique. Since the inputs are unsorted, we must first establish order within each stream or use a global sorting strategy.
Interview Questions on This Problem
Q1At a fintech platform processing high-frequency transaction logs, you have two unsorted arrays of transaction IDs. How would you merge them into a single sorted list for audit purposes, and what is the trade-off between sorting each array individually versus concatenating and sorting once?
Sorting each array individually takes $O(N\log N + M\log M)$, followed by a linear merge $O(N+M)$. Concatenating and sorting once takes $O((N+M)\log(N+M))$. If $N$ and $M$ are similar in size, the individual sort + merge is often slightly more efficient due to better cache locality and lower constant factors in the merge step. However, if one array is significantly smaller, sorting the smaller one and inserting elements into the larger sorted array (if the larger is already sorted) might be preferable, but since both are unsorted, the standard approach is to sort both and merge, or simply concatenate and sort if implementation simplicity is prioritized.
Q2In a distributed system, two nodes send unsorted telemetry data. You need to produce a globally sorted view. If the data volumes are massive (terabytes), how does the choice of sorting algorithm impact memory usage and I/O operations?
For massive data, in-memory sorting is impossible. You would use External Merge Sort. You split the data into chunks that fit in memory, sort each chunk, write to disk, and then perform a k-way merge. The key is that the merge phase is linear $O(N)$, but the initial sorting of chunks dominates. The choice of algorithm for the in-memory chunks (e.g., Timsort) is critical for performance. The final merge requires reading from multiple sorted files, which is I/O bound. The theoretical complexity remains $O(N\log N)$, but the constant factor and I/O patterns are what matter in practice.
Q3You are given two unsorted arrays. Can you merge them into a sorted array in $O(N+M)$ time without using extra space for a new array?
No, not if the arrays are unsorted. To achieve $O(N+M)$, the inputs must be pre-sorted. If they are unsorted, you must incur at least $O(N\log N + M\log M)$ time to sort them. If the question implies the arrays are already sorted, then yes, you can use a two-pointer technique to merge them in-place if one array has sufficient extra space, or in $O(N+M)$ space otherwise. For unsorted inputs, the lower bound is determined by the comparison-based sorting lower bound.
Examples
Input
streamA=[5,1,3] streamB=[4,2,6]
Output
[1,2,3,4,5,6]
Explanation: Sort streamA to [1,3,5] and streamB to [2,4,6]. Merge the two sorted lists: 1,2,3,4,5,6.
Input
streamA=[10,-1,10] streamB=[0,10]
Output
[-1,0,10,10,10]
Explanation: Sorted streamA: [-1,10,10]; sorted streamB: [0,10]. Merging yields -1,0,10,10,10.
Input
streamA=[7] streamB=[7,7,7]
Output
[7,7,7,7]
Explanation: streamA sorted is [7]; streamB sorted is [7,7,7]. Merging keeps all four 7s in order.
Constraints
- 1 <= streamA.length, streamB.length <= 100000
- -1000000000 <= streamA[i], streamB[i] <= 1000000000
- Total number of elements across both streams does not exceed 200000
Optimal Approach & Strategy
Sort each input array individually using an efficient sorting algorithm, then merge the two sorted arrays using a two-pointer technique to produce the final sorted result. This separates the sorting and merging concerns, allowing for potential parallelization of the sorting steps.
Brute Force Approach
Concatenate the two arrays into a single array and apply a standard sorting algorithm like QuickSort or MergeSort to the entire combined array. This approach is simple to implement but does not leverage any structure in the input.
Verified Code Solutions
function mergeStreams(streamA, streamB) {
const merged = streamA.concat(streamB);
merged.sort((a, b) => a - b);
return merged;
}
// Example usage:
console.log(mergeStreams([5,1,3],[4,2,6]));#include <bits/stdc++.h>
using namespace std;
vector<int> mergeStreams(const vector<int>& streamA, const vector<int>& streamB) {
vector<int> result;
result.reserve(streamA.size() + streamB.size());
result.insert(result.end(), streamA.begin(), streamA.end());
result.insert(result.end(), streamB.begin(), streamB.end());
sort(result.begin(), result.end());
return result;
}
int main(){
vector<int> streamA = {5,1,3};
vector<int> streamB = {4,2,6};
vector<int> merged = mergeStreams(streamA, streamB);
for(size_t i=0;i<merged.size();++i){
if(i) cout << ",";
cout << merged[i];
}
cout << endl;
return 0;
}import java.util.*;
public class Solution {
public int[] mergeStreams(int[] streamA, int[] streamB) {
int total = streamA.length + streamB.length;
int[] merged = new int[total];
System.arraycopy(streamA, 0, merged, 0, streamA.length);
System.arraycopy(streamB, 0, merged, streamA.length, streamB.length);
Arrays.sort(merged);
return merged;
}
public static void main(String[] args) {
int[] streamA = {5,1,3};
int[] streamB = {4,2,6};
int[] result = new Solution().mergeStreams(streamA, streamB);
System.out.println(Arrays.toString(result));
}
}def merge_streams(streamA, streamB):
merged = streamA + streamB
merged.sort()
return merged
# Example usage:
print(merge_streams([5,1,3],[4,2,6]))function mergeStreams(streamA, streamB) {
const merged = streamA.concat(streamB);
merged.sort((a, b) => a - b);
return merged;
}
// Example usage:
console.log(mergeStreams([5,1,3],[4,2,6]));Asked in Top Tech Interviews
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