Continuous Sequence Validator — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a validator that identifies the maximum length of a contiguous integer sequence present within a given collection of data points. A contiguous sequence is defined as a set of integers where each subsequent element is exactly one greater than the previous element (e.g., 4, 5, 6). The input is provided as an unsorted array of integers, which may contain duplicates.
Your objective is to determine the length of the longest such sequence. If no valid sequence of length greater than 1 exists, or if the array is empty, the result should be 0. The solution must efficiently handle large datasets by leveraging hash-based lookups to avoid the O(n log n) complexity of sorting, aiming for an average-case O(n) time complexity.
The function should accept a single parameter: an array of integers. It must return a single integer representing the length of the longest consecutive sequence found in the input array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Continuous Sequence Validator"
WHY DOES IT MATTER?
Detecting longest consecutive sequences appears in many domains such as time‑series gap analysis, version control diffing, and gaming leaderboards; mastering this pattern teaches you to replace sorting with hash‑based linear scans, a core optimization skill.
OPTIMIZATION CHALLENGE
The key insight is to start expanding a sequence only from numbers that have no predecessor in the set. This prevents re‑scanning the same run from every interior element, collapsing what could be O(n^2) work into O(n).
REAL-WORLD CONNECTION
Think of a distributed log system where each log entry has a sequence number. Finding the longest uninterrupted block of logs is analogous to our contiguous integer sequence, and the hash‑set method mirrors how systems track seen IDs to quickly detect gaps.
During an interview, first insert all numbers into a set, then loop over the set and for each element check if (num-1) is absent; if so, walk forward while (num+1) exists. This two‑step check instantly signals a start point and guarantees linear work.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem asks for the longest contiguous integer sequence that can be formed from an unsorted array possibly containing duplicates. A naive solution would sort the array (O(n log n)) or use nested loops to check every possible start point, both of which become prohibitive for large n (up to 10^5 or more). The optimal paradigm leverages a hash set to achieve O(n) time by allowing constant‑time membership checks. By inserting every unique element into a set, we can then iterate over the set and only start expanding a sequence from numbers that are the beginning of a potential run (i.e., numbers whose predecessor is not present). This eliminates redundant work because each element is visited at most twice – once when checking if it is a start, and once while walking forward through its consecutive neighbors. The overall space cost is O(n) for the hash set, which is acceptable given the linear time gain.
Interview Questions on This Problem
Q1How would you modify the solution if the array could contain negative integers and the required sequence length must be at least k?
The same hash‑set approach works for negatives because membership is value‑agnostic. After building the set, when expanding a sequence you simply count its length and only update the answer if length >= k. If no such sequence exists, return 0 or -1 as per spec.
Q2What is the time‑space trade‑off if you are constrained to O(1) extra space?
Without extra space you cannot achieve linear time; you would need to sort in‑place (O(n log n) time, O(1) extra space) and then scan for consecutive runs, which is the best you can do under the strict space limit.
Q3Explain how you could parallelize the computation for a massive distributed dataset stored across multiple nodes.
Each node can compute the longest run within its local partition and also expose its minimum and maximum values. A coordinator then merges overlapping boundary runs by checking if max of one node +1 equals min of another, stitching runs across partitions. The overall complexity remains linear in total data size, with communication overhead proportional to the number of partitions.
Examples
Input
nums = [100, 4, 200, 1, 3, 2]
Output
4
Explanation: The array contains the sequence 1, 2, 3, 4. The number 100 and 200 are isolated. The longest consecutive sequence is [1, 2, 3, 4], which has a length of 4.
Input
nums = [0, 3, 7, 2, 5, 8, 4, 6]
Output
4
Explanation: The integers can be grouped into sequences: [0], [2, 3, 4], [5, 6, 7, 8]. The sequence [5, 6, 7, 8] is the longest, with a length of 4.
Input
nums = [1, 2, 0, 3]
Output
4
Explanation: The entire array forms a single consecutive sequence: 0, 1, 2, 3. The length is 4.
Input
nums = [1, 1, 1, 1]
Output
1
Explanation: All elements are identical. A consecutive sequence requires distinct integers increasing by 1. Therefore, the longest sequence consists of a single element, length 1.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The input array may contain duplicate values.
- The input array is not guaranteed to be sorted.
Optimal Approach & Strategy
Insert all unique numbers into a hash set, then for each number that lacks a predecessor, expand forward while consecutive numbers exist, tracking the maximum length.
Brute Force Approach
Sort the array and then scan for consecutive runs, or use nested loops to test every possible start, both costing O(n log n) or O(n^2) time respectively.
Verified Code Solutions
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx=0;
const n = input[idx++]||0;
const nums = input.slice(idx, idx+n);
function longestConsecutive(nums) {
const set = new Set(nums);
let best = 0;
for (const x of set) {
if (!set.has(x-1)) {
let y = x;
while (set.has(y)) y++;
best = Math.max(best, y - x);
}
}
return best;
}
console.log(longestConsecutive(nums));#include <bits/stdc++.h>
using namespace std;
int longestConsecutive(const vector<int>& nums) {
unordered_set<int> s(nums.begin(), nums.end());
int best = 0;
for(int x: s) {
if(!s.count(x-1)) {
int y = x;
while(s.count(y)) ++y;
best = max(best, y - x);
}
}
return best;
}
int main(){ios::sync_with_stdio(false);cin.tie(nullptr);int n; if(!(cin>>n)) return 0; vector<int> nums(n); for(int i=0;i<n;++i)cin>>nums[i]; cout<<longestConsecutive(nums); return 0;}import java.util.*;
public class Main {
public static int longestConsecutive(int[] nums) {
Set<Integer> set = new HashSet<>();
for(int num: nums) set.add(num);
int best = 0;
for(int x: set) {
if(!set.contains(x-1)) {
int y = x;
while(set.contains(y)) y++;
best = Math.max(best, y - x);
}
}
return best;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
if(!sc.hasNextInt()) return;
int n = sc.nextInt();
int[] nums = new int[n];
for(int i=0;i<n;i++) nums[i]=sc.nextInt();
System.out.print(longestConsecutive(nums));
}
}import sys
def longest_consecutive(nums):
s = set(nums)
best = 0
for x in s:
if x-1 not in s:
y = x
while y in s:
y += 1
best = max(best, y - x)
return best
if __name__ == "__main__":
data = sys.stdin.read().strip().split()
if not data:
sys.exit()
n = int(data[0])
nums = list(map(int, data[1:1+n]))
print(longest_consecutive(nums))const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx=0;
const n = input[idx++]||0;
const nums = input.slice(idx, idx+n);
function longestConsecutive(nums) {
const set = new Set(nums);
let best = 0;
for (const x of set) {
if (!set.has(x-1)) {
let y = x;
while (set.has(y)) y++;
best = Math.max(best, y - x);
}
}
return best;
}
console.log(longestConsecutive(nums));Asked in Top Tech Interviews
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