Longest sequence of numbers

Question

I was recently asked this question in an interview for which i could give an O(nlogn) solution, but couldn't find a logic for O(n) . Can someone help me with O(n) solution?

In an array find the length of longest sequence of numbers

Example : Input : 2 4 6 7 3 1 Output: 4 (because 1,2,3,4 is a sequence even though they are not in consecutive positions)

The solution should also be realistic in terms of space consumed . i.e the solution should be realistic even with an array of 1 billion numbers

Does it have to be simply increasing? Or it needs to be increasing with a step of 1? — Tunaki, Feb 06 '16 at 17:54
Already answered over on CodeReview - https://codereview.stackexchange.com/questions/71578/print-length-of-longest-sequence-of-consecutive-numbers — OneCricketeer, Feb 06 '16 at 17:55
@cricket_007 That is for consecutive numbers, not non-consecutive numbers. — Peter Lawrey, Feb 06 '16 at 17:57
it believe it is a O(n^2) solution . I had given this solution . however the recruiter wanted a more optimized solution — Amogh Huilgol, Feb 06 '16 at 17:57
@cricket_007, probably, though we don't even know the exact question yet. The example the OP gave is not the same as the one in the link — Gavriel, Feb 06 '16 at 17:57
i'll be interested to in an answer. honestly, i think it's not possible in O(N) with non consecutive numbers — Jérémie B, Feb 06 '16 at 18:01
@PeterLawrey - Huh? That question says "prints the length of the longest sequence of consecutive natural successors" the answer there is "5 6 7", the answer here is "1 2 3 4". It is the same question — OneCricketeer, Feb 06 '16 at 18:01
@cricket_007 the answer should be 4 . I gave "1 2 3 4" as an explaination for why the answer is 4 — Amogh Huilgol, Feb 06 '16 at 18:03
I know the output is 4. If you can get the numbers, then you can count them just as easily, the interviewer just asked for the length, though — OneCricketeer, Feb 06 '16 at 18:05
I did get an hint about using hash set . But I was confused about maintaining the order of elements to check sequence — Amogh Huilgol, Feb 06 '16 at 18:05
@AmoghHuilgol HashSet is unordered so I don't see how it helps. — Peter Lawrey, Feb 06 '16 at 18:14
@AmoghHuilgol Hash set in Java is effectively a hash map whose value is boolean. You can easily modify my solution to use a hash set, when processing the neighbors just remove them from the hash set. — Miljen Mikic, Feb 06 '16 at 18:24

Peter Lawrey · Answer 1 · 2016-02-06T18:03:40.023

3

For non-consecutive numbers you needs a means of sorting them in O(n). In this case you can use BitSet.

int[] ints = {2, 4, 6, 7, 3, 1};
BitSet bs = new BitSet();
IntStream.of(ints).forEach(bs::set); 

// you can search for the longer consecutive sequence.
int last = 0, max = 0;
do {
    int set = bs.nextSetBit(last);
    int clear = bs.nextClearBit(set + 1);
    int len = clear - set;
    if (len > max)
        max = len;
    last = clear;
} while (last > 0);
System.out.println(max);

edited Feb 06 '16 at 18:03

answered Feb 06 '16 at 18:00

Peter Lawrey

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I did suggest this solution . However the recruiter suggested that if there were 1 billion records in array , then the solution would not be feasible – Amogh Huilgol Feb 06 '16 at 18:02
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@AmoghHuilgol N billion records of up to 4 billion possible values uses 512 MB of memory. This would work on a mobile phone. – Peter Lawrey Feb 06 '16 at 18:04
@AmoghHuilgol If there is a space requirement, please add it to the question. – Déjà vu Feb 06 '16 at 18:10
@Peter Thanks for the solution . – Amogh Huilgol Feb 06 '16 at 18:11

Miljen Mikic · Answer 2 · 2016-02-06T19:24:52.513

1

Traverse the array once and build the hash map whose key is a number from the input array and value is a boolean variable indicating whether the element has been processed or not (initially all are false). Traverse once more and do the following: when you check number a, put value true for that element in the hash map and immediately check the hash map for the existence of the elements a-1 and a+1. If found, denote their values in the hash map as true and proceed checking their neighbors, incrementing the length of the current contigous subsequence. Stop when there are no neighbors, and update longest length. Move forward in the array and continue checking unprocessed elements. It is not obvious at the first glance that this solution is O(n), but there are only two array traversals and hash map ensures that every element of the input is processed only once.

Main lesson - if you have to reduce time complexity, it is often neccesary to use additional space.

edited Feb 06 '16 at 19:24

answered Feb 06 '16 at 18:03

Miljen Mikic

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Note: Java's `boolean[]` typically uses 1 byte per boolean. BitSet or similar would be more compact. – Peter Lawrey Feb 06 '16 at 18:07
i didn't do the math, but it's not O(N) – Jérémie B Feb 06 '16 at 18:23
@JérémieB Although it doesn't seem O(n) at the first glance, it is. You process each element from the input only once, hash map ensures that. When you stumble upon the processed element in the second traversal of the input array, just skip it. – Miljen Mikic Feb 06 '16 at 18:29
yes, but a hashmap is not free. it's O(1) at best, not on average/worst. – Jérémie B Feb 06 '16 at 18:32
@NiklasB Of course not. Hash code of an integer is the value of number itself, but the hash map implementation is still prone to collisions and rehashing, so yes, it is not O(1) always - that was incorrect. However, when evaluating the complexity of hash map operations, one usually assumes O(1), at interviews as well. – Miljen Mikic Feb 06 '16 at 19:24

Longest sequence of numbers

2 Answers2