Galactic Transmission Sequence — Problem Statement & Solution Guide
Problem Description
Given an integer array that represents a circular buffer, determine the maximum number of consecutive elements that can be read without encountering a duplicate value. The reading may start at any index and proceeds forward, wrapping to the beginning after the last element, but it must stop before a value repeats. Return the length of the longest such duplicate‑free segment.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Transmission Sequence"
WHY DOES IT MATTER?
Detecting the longest duplicate‑free stretch in a circular buffer is a classic example of the “longest subarray with distinct elements” pattern, which appears in cache eviction, network packet sequencing, and sliding‑window analytics. Mastery of this pattern sharpens a candidate’s ability to reason about wrap‑around data structures and constant‑time membership checks.
OPTIMIZATION CHALLENGE
The key insight is to linearize the circular nature by virtually concatenating the array to itself, then enforce a window size ≤ n. This avoids O(n²) re‑scanning and keeps each element’s entry/exit cost constant.
REAL-WORLD CONNECTION
Think of a rotating log buffer in a distributed system: you can read logs sequentially until a log entry repeats (e.g., a heartbeat ID), after which you must stop to avoid re‑processing. The algorithm ensures you read the maximal fresh segment before a duplicate appears.
When coding, keep the window length check (right‑left < n) right after you increment the right pointer; forgetting this subtle bound is the most common source of WA on circular variants.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem asks for the longest contiguous segment of a circular array that contains no duplicate values. A naïve solution would examine every possible start index and expand until a repeat is found, yielding O(n²) time – unacceptable for n up to 10⁵ or more. The optimal paradigm treats the circular buffer as a linear sequence of length 2n by concatenating the array to itself; this allows a standard sliding‑window (two‑pointer) technique to traverse the “wrapped” region while maintaining a hash map of element frequencies. The window is always kept ≤ n elements long, guaranteeing that any segment we consider corresponds to a valid wrap‑around segment in the original circle. This yields a linear‑time solution because each element enters and leaves the window at most once, and the hash map provides O(1) duplicate detection.
Interview Questions on This Problem
Q1How would you adapt the sliding‑window solution if the array could contain negative numbers and the range of values is unbounded?
Use an unordered_map (or HashMap) to store frequencies instead of a fixed‑size array; the map works for any integer range and still offers O(1) average updates.
Q2What changes are needed if the buffer is read‑only and you cannot duplicate the array in memory?
Maintain two pointers that wrap modulo n, and keep a hash set of current window elements; when the right pointer reaches the end, continue from index 0 while ensuring the window size never exceeds n.
Q3At a fintech firm you must process a stream of transaction IDs in a circular buffer; how would you guarantee O(n) processing while also reporting the start index of the maximal duplicate‑free segment?
Run the same sliding‑window over the duplicated view, tracking the maximum length and its left‑pointer position; the start index in the original array is left % n.
Examples
Input
[2,5,1,2,3,5,1]
Output
4
Explanation: Starting at index 1 yields the segment [5,1,2,3]; the next element (5) repeats a value already seen, so the segment length is 4. No other start position produces a longer duplicate‑free segment.
Input
[7,7,7,7]
Output
1
Explanation: Every element is identical, so the longest segment without a repeat consists of a single element.
Input
[9,1,2,3,4,5,6,7,8]
Output
9
Explanation: All values are distinct, therefore the entire array can be traversed once without repeats, giving a length equal to the array size.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The algorithm should run in O(n) time and O(min(n, k)) extra space, where k is the number of distinct values.
Optimal Approach & Strategy
Duplicate the array to length 2n, then use a sliding window with a hash map, limiting the window size to n – O(n) time, O(n) space.
Brute Force Approach
For each index, expand forward (wrapping as needed) until a duplicate is hit, tracking the longest length – O(n²) time.
Verified Code Solutions
const fs = require('fs');
function maxUniqueCircular(arr) {
const n = arr.length;
if(n===0) return 0;
const extended = arr.concat(arr);
const lastPos = new Map();
let left = 0, best = 0;
for(let right=0; right<2*n; ++right) {
const val = extended[right];
if(lastPos.has(val) && lastPos.get(val) >= left) {
left = lastPos.get(val) + 1;
}
lastPos.set(val, right);
if(right - left + 1 > n) left++;
best = Math.max(best, right - left + 1);
}
return best;
}
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0){process.exit(0);}
let idx=0; const n=data[idx++]; const arr=data.slice(idx, idx+n);
console.log(maxUniqueCircular(arr));#include <bits/stdc++.h>
using namespace std;
int maxUniqueCircular(const vector<int>& arr) {
int n = arr.size();
if(n==0) return 0;
vector<int> extended(arr);
extended.insert(extended.end(), arr.begin(), arr.end()); // size 2n
unordered_map<int,int> lastPos; // value -> last index in extended
int left = 0, best = 0;
for(int right=0; right<2*n; ++right) {
int val = extended[right];
if(lastPos.find(val)!=lastPos.end() && lastPos[val]>=left) {
left = lastPos[val] + 1;
}
lastPos[val] = right;
// window size cannot exceed n (cannot use more than original length)
if(right - left + 1 > n) {
++left;
}
best = max(best, right - left + 1);
}
return best;
}
int main(){ios::sync_with_stdio(false);cin.tie(nullptr);
int n; if(!(cin>>n)) return 0; vector<int>a(n); for(int i=0;i<n;++i)cin>>a[i];
cout<<maxUniqueCircular(a);
return 0;}
import java.io.*;
import java.util.*;
public class Main {
public static int maxUniqueCircular(int[] arr) {
int n = arr.length;
if(n==0) return 0;
int[] extended = new int[2*n];
System.arraycopy(arr,0,extended,0,n);
System.arraycopy(arr,0,extended,n,n);
Map<Integer,Integer> lastPos = new HashMap<>();
int left = 0, best = 0;
for(int right=0; right<2*n; ++right) {
int val = extended[right];
Integer prev = lastPos.get(val);
if(prev != null && prev >= left) {
left = prev + 1;
}
lastPos.put(val, right);
if(right - left + 1 > n) left++;
best = Math.max(best, right - left + 1);
}
return best;
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
if(line==null || line.isEmpty()) return;
int n = Integer.parseInt(line.trim());
int[] arr = new int[n];
StringTokenizer st = new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) arr[i]=Integer.parseInt(st.nextToken());
System.out.println(maxUniqueCircular(arr));
}
}import sys
def max_unique_circular(arr):
n = len(arr)
if n == 0:
return 0
extended = arr + arr
last = {}
left = 0
best = 0
for right, val in enumerate(extended):
if val in last and last[val] >= left:
left = last[val] + 1
last[val] = right
if right - left + 1 > n:
left += 1
best = max(best, right - left + 1)
return best
if __name__ == "__main__":
data = list(map(int, sys.stdin.read().strip().split()))
if not data:
sys.exit()
n = data[0]
arr = data[1:1+n]
print(max_unique_circular(arr))const fs = require('fs');
function maxUniqueCircular(arr) {
const n = arr.length;
if(n===0) return 0;
const extended = arr.concat(arr);
const lastPos = new Map();
let left = 0, best = 0;
for(let right=0; right<2*n; ++right) {
const val = extended[right];
if(lastPos.has(val) && lastPos.get(val) >= left) {
left = lastPos.get(val) + 1;
}
lastPos.set(val, right);
if(right - left + 1 > n) left++;
best = Math.max(best, right - left + 1);
}
return best;
}
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0){process.exit(0);}
let idx=0; const n=data[idx++]; const arr=data.slice(idx, idx+n);
console.log(maxUniqueCircular(arr));Asked in Top Tech Interviews
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