Find largest sub-array formed by consecutive integers

Given an array of integers, find largest sub-array formed by consecutive integers. The sub-array should contain all distinct values.

 

For example,


Input:  { 2, 0, 2, 1, 4, 3, 1, 0 }
Output: The largest sub-array is { 0, 2, 1, 4, 3 }

 


 
The idea is to consider every sub-array and keep track of largest subarray found so far which is formed by consecutive integers. In order for an sub-array to contain consecutive integers,

  • The difference between maximum and minimum element in it should be exactly equal to length of the subarray minus one.
     
  • All elements in the array should be distinct (we can check this by inserting the elements in set or using a visited array).

C++

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Java

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Output:

The largest sub-array is [1, 5]

 
The time complexity of above solution is O(n3) and auxiliary space used by the program is O(n).
 

Exercise: Extend the solution to consider duplicates in the sub-array.

 
Thanks for reading.




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Dmitriy
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Dmitriy

I think a sliding-window approach can accomplish this in linear time.

Thank you for posting so many helpful practice problems!

Cliff Crosland
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Cliff Crosland

Cool – you can get O(n^2) time here by building the set as you go along in the inner loop that iterates through j. If max – min == j – i, and if all of the values between i and j are distinct (which you can tell from the set size), then you know immediately that the subsequence consists of the consecutive integers in the range [min, max].

You can also break out of the inner loop early if you ever run into a duplicate.

Will think about Dimitriy’s O(n) hint, cool…

Raj Pratim Bhattacharya
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Raj Pratim Bhattacharya

O(n*n) time and Omega(n) space –
http://ideone.com/87Hzjc

Sam
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I don’t understand the question. What are consecutive integers? To me consecutive integers are 2,3,4 or 77,78,79.

In the example provided at the beginning of this page, in what way are 0,2,1,4,3 consecutive integers?

Is the question really asking “find the largest subarray formed by unique integers” ? Or is it something else?

Akash kandpal
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Akash kandpal

A hashmap implementation can bring down the time complexity to O(n) keeping it in-place . Also an approach using merge sot can reduce the algorithm to O(nlogn) but with a space complexity of O(nlogn) . The above question was asked to me in my Facebook interview 🙂

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