Longest Consecutive Sequence
You are given an unsorted array/list 'ARR' of 'N' integers. Your task is to return the length of the longest consecutive sequence.

The consecutive sequence is in the form ['NUM', 'NUM' + 1, 'NUM' + 2, ..., 'NUM' + L] where 'NUM' is the starting integer of the sequence and 'L' + 1 is the length of the sequence.

Note:

If there are any duplicates in the given array we will count only one of them in the consecutive sequence.

For example-

For the given 'ARR' [9,5,4,9,10,10,6]. Output = 3 The longest consecutive sequence is [4,5,6].

Follow Up:

Can you solve this in O(N) time and O(N) space complexity?

Input format :

The first line of input contains a single integer 'T', representing the number of test cases or queries to be run. Then the 'T' test cases follow. The first line of each test case contains integer 'N' denoting the size of the array. The second line of each test case contains 'N' single space-separated integers, elements of the array.

Output format :

For each test case, print an integer in a single line that represents the length of the longest consecutive sequence.

Note :

You are not required to print the expected output; it has already been taken care of. Just implement the function.

Constraints :

1

Question

Longest Consecutive Sequence

You are given an unsorted array/list 'ARR' of 'N' integers. Your task is to return the length of the longest consecutive sequence.

The consecutive sequence is in the form ['NUM', 'NUM' + 1, 'NUM' + 2, ..., 'NUM' + L] where 'NUM' is the starting integer of the sequence and 'L' + 1 is the length of the sequence.

Note:

If there are any duplicates in the given array we will count only one of them in the consecutive sequence.

For example-

For the given 'ARR' [9,5,4,9,10,10,6].

Output = 3
The longest consecutive sequence is [4,5,6].

Follow Up:

Can you solve this in O(N) time and O(N) space complexity?

Input format :

The first line of input contains a single integer 'T', representing the number of test cases or queries to be run. Then the 'T' test cases follow.

The first line of each test case contains integer 'N' denoting the size of the array.

The second line of each test case contains 'N' single space-separated integers, elements of the array.

Output format :

For each test case, print an integer in a single line that represents the length of the longest consecutive sequence.

Note :

You are not required to print the expected output; it has already been taken care of. Just implement the function.

Constraints :

1 <= T <= 10
1 <= N <= 10^5
-10^9 <= ARR[i] <= 10^9

Time Limit: 1 sec

Accepted Answer

I thought of two points on a number line a and b and formulated a consecutive sequence [a....b], (b*(b+1))/2-(a*(a+1))/2=n. Now we have a clearly quadratic solution by brute-forcing a and b but if we simplify it further we have (b-a+1)*(b+a) = 2*n. So now we can easily find all divisors of 2*n and satisfy the equation for which we get a>=0 and b>=0 and thus we get the answer.

Accepted Answer

Brute Force

As we only need the consecutive elements in the form ['NUM', 'NUM' + 1, 'NUM' + 2,...,'NUM' + 'L']. The brute force approach is to traverse each element in the array ('NUM' = ‘ARR[i]’) and then keep finding ('NUM' + 1) in the array until we don't find the next consecutive element in the array.

Keep a track of the current length of the sequence. If the current length of the consecutive subsequence is greater than the longest length of consecutive subsequence then update it.

Space Complexity: O(1)Explanation:

O(1)

Constant extra space is required.

Time Complexity: O(1)Explanation:

O(N^3), where ‘N’ is the size of the array.

In the worst case, The outer loop will take O(N) time to traverse through the array, and then for each element, the inner while loop can also take up to O(N) time for selecting the next consecutive element.

And to check each next consecutive element is present in the array or not takes again O(N). So there are 3 nested loops therefore, the overall time complexity will be O(N^3).

Accepted Answer

Sorting

The idea is to sort the array, then iterate through the array and find the longest subarray containing consecutive elements.
We first initialize the variable ‘COUNT’ = 0 which stores the length of the consecutive sequence and ‘MX’ = 0 which stores the longest length of consecutive subsequence.
Now run a loop and check if ‘ARR[i - 1]’ + 1 equal to ‘ARR[i]’ then it will include in the current consecutive subsequence by increment ‘COUNT’ by 1.
If ‘ARR[i - 1]’ is equal to ‘ARR[i]’ then it means it shouldn’t be considered in consecutive sequence because the consecutive sequence is of the form ['NUM', 'NUM' + 1, 'NUM' + 2,...,'NUM' + L].
Else If ‘ARR[i - 1]’ + 1 is not equal to ‘ARR[i]’ then we set to ‘COUNT’ to 1. For finding the longest length we update ‘MX’ with a maximum of ‘COUNT’ and ‘MX’.

Space Complexity: O(1)Explanation:

O(1)

Constant extra space is required.

Time Complexity: O(1)Explanation:

O(N * log(N)) where ‘N’ is the size of an array.

Since we are using the sort function that takes O(N * log(N)) time and a loop till ‘N’ and so, the total time complexity is O(N * log(N) + N) = O(N * log(N)).

Accepted Answer

Using Hash Table

We can improve our time complexity of searching the next consecutive element in the array by using a Hash Table which can check the presence of an element in O(1).

The steps are as follows:

Store all the elements in the Hash table first.
For every element check, if it is a starting element of sequence or not. by simply checking if ‘ARR[i]’ - 1 is present in a hashtable or not. If it is present so it means ‘ARR[i]’ can’t be the first element of the sequence.
If the element is the first element then count the total number of elements that can occur in sequence by incrementing the element by 1 in each iteration of the while loop.
If the count is more than the longest consecutive sequence then we update it.

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the size of the array.

Since we are using a HashSet to store the elements and so, the total space complexity is O(N).

Time Complexity: O(1)Explanation:

O(N), where ‘N’ is the size of an array.

A solution seems to be quadratic as while loop is in for loop but if we look closer then in the worst case, while loop will only iterate if the current element is the first of sequence and so the while loop will only take O(N) throughout the whole algorithm and nested loop actually run in O(N+N) i.e O(N). Hence the total time complexity is O(N).

Accepted Answer

create an array of size maximum element then mark all by 1, in new array now find longest subarray which has maximum 1 in new array...for give eg create array of size 10, 0 1 2 3 4 5 6 7 8 9 10, now element at index 4 5 6 is mark as 1 now in the array it is the maximum consecutive

You are given an unsorted array/list 'ARR' of 'N' integers. Your task is to return the length of the longest consecutive sequence.

The consecutive sequence is in the form ['NUM', 'NUM' + 1, 'NUM' + 2, ..., 'NUM' + L] where 'NUM' is the starting integer of the sequence and 'L' + 1 is the length of the sequence.

Note:

For example-

Follow Up:

Input format :

Output format :

Note :

Constraints :

Top Paytm Software Engineer interview questions & answers

Popular interview questions of Software Engineer

Top HR questions asked in Paytm Software Engineer