Longest Consecutive Subsequence

Given an array of integers, find the length of the longest sub-sequence such that elements in the subsequence are consecutive integers, the consecutive numbers can be in any order. 

Examples:  

Input: arr[] = {1, 9, 3, 10, 4, 20, 2}
Output: 4
Explanation: The subsequence 1, 3, 4, 2 is the longest subsequence of consecutive elements

Input: arr[] = {36, 41, 56, 35, 44, 33, 34, 92, 43, 32, 42}
Output: 5
Explanation: The subsequence 36, 35, 33, 34, 32 is the longest subsequence of consecutive elements.

Naive Approach:

The idea is to first sort the array and find the longest subarray with consecutive elements. After sorting the array and removing the multiple occurrences of elements, run a loop and keep a count and max (both initially zero). Run a loop from start to end and if the current element is not equal to the previous (element+1) then set the count to 1 else increase the count. Update max with a maximum of count and max. 

 Illustration:

Input: arr[] = {1, 9, 3, 10, 4, 20, 2}

First sort the array to arrange them in a consecutive fashion.
arr[] = {1, 2, 3, 4, 9, 10, 20}

Now, store the distinct elements from the sorted array.
dist[] = {1, 2, 3, 4, 9, 10, 20}

Initialize countConsecutive with 0 which will increment when arr[i] == arr[i – 1] + 1 is true otherwise countConsecutive will re-initialize by 1.

Maintain a variable ans to store the maximum count of consecutive elements so far.

At i = 0:

• as i is 0 then re-initialize countConsecutive by 1.
• ans = max(ans, countConsecutive) = max(0, 1) = 1

At i = 1:

• check if (dist[1] == dist[0] + 1) = (2 == 1 + 1) = true
• as the above condition is true, therefore increment countConsecutive by 1
  • countConsecutive = countConsecutive + 1 = 1 + 1 = 2
• ans = max(ans, countConsecutive) = max(1, 2) = 1

At i = 2:

• check if (dist[2] == dist[1] + 1) = (3 == 2 + 1) = true
• as the above condition is true, therefore increment countConsecutive by 1
  • countConsecutive = countConsecutive + 1 = 2 + 1 = 3
• ans = max(ans, countConsecutive) = max(2, 3) = 3

At i = 3:

• check if (dist[3] == dist[2] + 1) = (4 == 3 + 1) = true
• as the above condition is true, therefore increment countConsecutive by 1
  • countConsecutive = countConsecutive + 1 = 3 + 1 = 4
• ans = max(ans, countConsecutive) = max(3, 4) = 4

At i = 4:

• check if (dist[4] == dist[3] + 1) = (9 != 4 + 1) = false
• as the above condition is false, therefore re-initialize countConsecutive by 1
  • countConsecutive = 1
• ans = max(ans, countConsecutive) = max(4, 1) = 4

At i = 5:

• check if (dist[5] == dist[4] + 1) = (10 == 9 + 1) = true
• as the above condition is true, therefore increment countConsecutive by 1
  • countConsecutive = countConsecutive + 1 = 1 + 1 = 2
• ans = max(ans, countConsecutive) = max(4, 2) = 4

At i = 6:

• check if (dist[6] == dist[5] + 1) = (20 != 10 + 1) = false
• as the above condition is false, therefore re-initialize countConsecutive by 1
  • countConsecutive = 1
• ans = max(ans, countConsecutive) = max(4, 1) = 4

Therefore the longest consecutive subsequence is {1, 2, 3, 4}
Hence, ans is 4.

Follow the steps below to solve the problem:

• Initialize ans and countConsecutive with 0.
• Sort the arr[].
• Store the distinct elements in dist[] array by traversing over the arr[].
• Now, traverse on the dist[] array to find the count of consecutive elements.
• Simultaneously maintain the answer variable.

Below is the implementation of the above approach:

C++

// C++ program to find longest
// contiguous subsequence
#include <bits/stdc++.h>
using namespace std;

// Returns length of the longest
// contiguous subsequence
int findLongestConseqSubseq(int arr[], int n)
{
    int ans = 0, count = 0;

    // sort the array
    sort(arr, arr + n);

    vector<int> v;
    v.push_back(arr[0]);

    // insert repeated elements only once in the vector
    for (int i = 1; i < n; i++) {
        if (arr[i] != arr[i - 1])
            v.push_back(arr[i]);
    }
    // find the maximum length
    // by traversing the array
    for (int i = 0; i < v.size(); i++) {

        // Check if the current element is equal
        // to previous element +1
        if (i > 0 && v[i] == v[i - 1] + 1)
            count++;
        // reset the count
        else
            count = 1;

        // update the maximum
        ans = max(ans, count);
    }
    return ans;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 2, 3 };
    int n = sizeof arr / sizeof arr[0];
    cout << "Length of the Longest contiguous subsequence "
            "is "
         << findLongestConseqSubseq(arr, n);
    return 0;
}

Java

// Java program to find longest
// contiguous subsequence
import java.io.*;
import java.util.*;

class GFG {

    static int findLongestConseqSubseq(int arr[], int n)
    {

        // Sort the array
        Arrays.sort(arr);

        int ans = 0, count = 0;

        ArrayList<Integer> v = new ArrayList<Integer>();
        v.add(arr[0]);

        // Insert repeated elements
        // only once in the vector
        for (int i = 1; i < n; i++) {
            if (arr[i] != arr[i - 1])
                v.add(arr[i]);
        }

        // Find the maximum length
        // by traversing the array
        for (int i = 0; i < v.size(); i++) {

            // Check if the current element is
            // equal to previous element +1
            if (i > 0 && v.get(i) == v.get(i - 1) + 1)
                count++;
            else
                count = 1;

            // Update the maximum
            ans = Math.max(ans, count);
        }
        return ans;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
        int n = arr.length;

        System.out.println(
            "Length of the Longest "
            + "contiguous subsequence is "
            + findLongestConseqSubseq(arr, n));
    }
}

// This code is contributed by parascoding

Python3

# Python3 program to find longest
# contiguous subsequence

# Returns length of the longest
# contiguous subsequence


def findLongestConseqSubseq(arr, n):

    ans = 0
    count = 0

    # Sort the array
    arr.sort()

    v = []

    v.append(arr[0])

    # Insert repeated elements only
    # once in the vector
    for i in range(1, n):
        if (arr[i] != arr[i - 1]):
            v.append(arr[i])

    # Find the maximum length
    # by traversing the array
    for i in range(len(v)):

        # Check if the current element is
        # equal to previous element +1
        if (i > 0 and v[i] == v[i - 1] + 1):
            count += 1

        # Reset the count
        else:
            count = 1

        # Update the maximum
        ans = max(ans, count)

    return ans


# Driver code
arr = [1, 2, 2, 3]
n = len(arr)

print("Length of the Longest contiguous subsequence is",
      findLongestConseqSubseq(arr, n))

# This code is contributed by avanitrachhadiya2155

C#

// C# program to find longest
// contiguous subsequence
using System;
using System.Collections.Generic;

class GFG {

    static int findLongestConseqSubseq(int[] arr, int n)
    {

        // Sort the array
        Array.Sort(arr);

        int ans = 0, count = 0;

        List<int> v = new List<int>();
        v.Add(10);

        // Insert repeated elements
        // only once in the vector
        for (int i = 1; i < n; i++) {
            if (arr[i] != arr[i - 1])
                v.Add(arr[i]);
        }

        // Find the maximum length
        // by traversing the array
        for (int i = 0; i < v.Count; i++) {

            // Check if the current element is
            // equal to previous element +1
            if (i > 0 && v[i] == v[i - 1] + 1)
                count++;
            else
                count = 1;

            // Update the maximum
            ans = Math.Max(ans, count);
        }
        return ans;
    }

    // Driver code
    static void Main()
    {
        int[] arr = { 1, 9, 3, 10, 4, 20, 2 };
        int n = arr.Length;

        Console.WriteLine(
            "Length of the Longest "
            + "contiguous subsequence is "
            + findLongestConseqSubseq(arr, n));
    }
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

        // JavaScript program to find longest
        // contiguous subsequence

        // Returns length of the longest
        // contiguous subsequence
        function findLongestConseqSubseq(arr, n) {
            let ans = 0, count = 0;

            // sort the array
            arr.sort(function (a, b) { return a - b; })

            var v = [];
            v.push(arr[0]);

            //insert repeated elements only once in the vector
            for (let i = 1; i < n; i++) {
                if (arr[i] != arr[i - 1])
                    v.push(arr[i]);
            }
            // find the maximum length
            // by traversing the array
            for (let i = 0; i < v.length; i++) {

                // Check if the current element is equal
                // to previous element +1
                if (i > 0 && v[i] == v[i - 1] + 1)
                    count++;
                // reset the count
                else
                    count = 1;

                // update the maximum
                ans = Math.max(ans, count);
            }
            return ans;
        }

        // Driver code

        let arr = [1, 2, 2, 3];
        let n = arr.length;
        document.write(
        "Length of the Longest contiguous subsequence is "
        +findLongestConseqSubseq(arr, n)
        );

     // This code is contributed by Potta Lokesh

    </script>

Output

Length of the Longest contiguous subsequence is 3

Time complexity: O(Nlog(N)), Time to sort the array is O(Nlog(N)).
Auxiliary space: O(N). Extra space is needed for storing distinct elements.

Longest Consecutive Subsequence using Hashing:

The idea is to use Hashing. We first insert all elements in a Set. Then check all the possible starts of consecutive subsequences.

Illustration:

Below image is the dry run, for example, arr[] = {1, 9, 3, 10, 4, 20, 2}:

Follow the steps below to solve the problem:

• Create an empty hash.
• Insert all array elements to hash.
• Do the following for every element arr[i]
• Check if this element is the starting point of a subsequence. To check this, simply look for arr[i] – 1 in the hash, if not found, then this is the first element of a subsequence.
• If this element is the first element, then count the number of elements in the consecutive starting with this element. Iterate from arr[i] + 1 till the last element that can be found.
• If the count is more than the previous longest subsequence found, then update this.

Below is the implementation of the above approach: 

C++

// C++ program to find longest
// contiguous subsequence
#include <bits/stdc++.h>
using namespace std;

// Returns length of the longest
// contiguous subsequence
int findLongestConseqSubseq(int arr[], int n)
{
    unordered_set<int> S;
    int ans = 0;

    // Hash all the array elements
    for (int i = 0; i < n; i++)
        S.insert(arr[i]);

    // check each possible sequence from
    // the start then update optimal length
    for (int i = 0; i < n; i++) {
        // Check if the previous number (arr[i] - 1) is
        // present in the set,
        // if it is then that number (arr[i]) is not the
        // starting of the sequence.
        if (S.find(arr[i] - 1) != S.end()) {
            continue;
        }
        else {
            // If previous number is not present, that means
            // that number is the starting of the sequence.
            int j = arr[i];
            while (S.find(j) != S.end())
                j++;

            // update optimal length if
            // this length is more
            ans = max(ans, j - arr[i]);
        }
    }
    return ans;
}

// Driver code
int main()
{
    int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
    int n = sizeof arr / sizeof arr[0];
    cout << "Length of the Longest contiguous subsequence "
            "is "
         << findLongestConseqSubseq(arr, n);
    return 0;
}

Java

// Java program to find longest
// consecutive subsequence
import java.io.*;
import java.util.*;

class ArrayElements {
    // Returns length of the longest
    // consecutive subsequence
    static int findLongestConseqSubseq(int arr[], int n)
    {
        HashSet<Integer> S = new HashSet<Integer>();
        int ans = 0;

        // Hash all the array elements
        for (int i = 0; i < n; ++i)
            S.add(arr[i]);

        // check each possible sequence from the start
        // then update optimal length
        for (int i = 0; i < n; ++i) {
            // if current element is the starting
            // element of a sequence
            if (!S.contains(arr[i] - 1)) {
                // Then check for next elements
                // in the sequence
                int j = arr[i];
                while (S.contains(j)){
                      S.remove(Integer.valueOf(j));//this will improve runtime by avoiding the repetitive counts of elements
                    j++;
                    }

                // update  optimal length if this
                // length is more
                if (ans < j - arr[i])
                    ans = j - arr[i];
            }
        }
        return ans;
    }

    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
        int n = arr.length;
        System.out.println(
            "Length of the Longest consecutive subsequence is "
            + findLongestConseqSubseq(arr, n));
    }
}
// This code is contributed by Aakash Hasija

Python3

# Python program to find longest contiguous subsequence


def findLongestConseqSubseq(arr, n):

    s = set()
    ans = 0

    # Hash all the array elements
    for ele in arr:
        s.add(ele)

    # check each possible sequence from the start
    # then update optimal length
    for i in range(n):

         # if current element is the starting
        # element of a sequence
        if (arr[i]-1) not in s:

            # Then check for next elements in the
            # sequence
            j = arr[i]
            while(j in s):
                j += 1

            # update  optimal length if this length
            # is more
            ans = max(ans, j-arr[i])
    return ans


# Driver code
if __name__ == '__main__':
    n = 7
    arr = [1, 9, 3, 10, 4, 20, 2]
    print("Length of the Longest contiguous subsequence is ",
          findLongestConseqSubseq(arr, n))

# Contributed by: Harshit Sidhwa

C#

using System;
using System.Collections.Generic;

// C# program to find longest consecutive subsequence

public class ArrayElements {
    // Returns length of the
    // longest consecutive subsequence
    public static int findLongestConseqSubseq(int[] arr,
                                              int n)
    {
        HashSet<int> S = new HashSet<int>();
        int ans = 0;

        // Hash all the array elements
        for (int i = 0; i < n; ++i) {
            S.Add(arr[i]);
        }

        // check each possible sequence from the start
        // then update optimal length
        for (int i = 0; i < n; ++i) {
            // if current element is the starting
            // element of a sequence
            if (!S.Contains(arr[i] - 1)) {
                // Then check for next elements in the
                // sequence
                int j = arr[i];
                while (S.Contains(j)) {
                    j++;
                }

                // update  optimal length if this length
                // is more
                if (ans < j - arr[i]) {
                    ans = j - arr[i];
                }
            }
        }
        return ans;
    }

    // Driver code
    public static void Main(string[] args)
    {
        int[] arr = new int[] { 1, 9, 3, 10, 4, 20, 2 };
        int n = arr.Length;
        Console.WriteLine(
            "Length of the Longest consecutive subsequence is "
            + findLongestConseqSubseq(arr, n));
    }
}

// This code is contributed by Shrikant13

Javascript

<script>
// Javascript program to find longest
// contiguous subsequence


// Returns length of the longest
// contiguous subsequence
function findLongestConseqSubseq(arr, n) {
    let S = new Set();
    let ans = 0;

    // Hash all the array elements
    for (let i = 0; i < n; i++)
        S.add(arr[i]);

    // check each possible sequence from
    // the start then update optimal length
    for (let i = 0; i < n; i++)
    {

        // if current element is the starting
        // element of a sequence
        if (!S.has(arr[i] - 1))
        {

            // Then check for next elements
            // in the sequence
            let j = arr[i];
            while (S.has(j))
                j++;

            // update optimal length if
            // this length is more
            ans = Math.max(ans, j - arr[i]);
        }
    }
    return ans;
}

// Driver code
let arr = [1, 9, 3, 10, 4, 20, 2];
let n = arr.length;
document.write("Length of the Longest contiguous subsequence is "
    + findLongestConseqSubseq(arr, n));

    // This code is contributed by gfgking.
</script>

Output

Length of the Longest contiguous subsequence is 4

Time complexity: O(N), Only one traversal is needed and the time complexity is O(n) under the assumption that hash insert and search takes O(1) time.
Auxiliary space: O(N), To store every element in the hashmap O(n) space is needed

Longest Consecutive Subsequence using Priority Queue:

The Idea is to use Priority Queue. Using priority queue it will sort the elements and eventually it will help to find consecutive elements.

Illustration:

Input: arr[] = {1, 9, 3, 10, 4, 20, 2}

Insert all the elements in the Priority Queue:

1	2	3	4	9	10	20

Initialise variable prev with first element of priority queue, prev will contain last element has been picked and it will help to check whether the current element is contributing for consecutive sequence or not.

prev = 1, countConsecutive = 1, ans = 1

Run the algorithm till the priority queue becomes empty.

2	3	4	9	10	20

• current element is 2
  • prev + 1 == 2, therefore increment countConsecutive by 1
  • countConsecutive = countConsecutive + 1 = 1 + 1 = 2
  • update prev with current element, prev = 2
  • pop the current element
  • ans = max(ans, countConsecutive) = (1, 2) = 2

3	4	9	10	20

• current element is 3
  • prev + 1 == 3, therefore increment countConsecutive by 1
  • countConsecutive = countConsecutive + 1 = 2 + 1 = 3
  • update prev with current element, prev = 3
  • pop the current element
  • ans = max(ans, countConsecutive) = (2, 3) = 3

4	9	10	20

• current element is 4
  • prev + 1 == 4, therefore increment countConsecutive by 1
  • countConsecutive = countConsecutive + 1 = 3 + 1 = 4
  • update prev with current element, prev = 4
  • pop the current element
  • ans = max(ans, countConsecutive) = (3, 4) = 4

9	10	20

• current element is 9
  • prev + 1 != 9, therefore re-initialise countConsecutive by 1
  • countConsecutive = 1
  • update prev with current element, prev = 9
  • pop the current element
  • ans = max(ans, countConsecutive) = (4, 1) = 4

10	20

• current element is 10
  • prev + 1 == 10, therefore increment countConsecutive by 1
  • countConsecutive = countConsecutive + 1 = 1 + 1 = 2
  • update prev with current element, prev = 10
  • pop the current element
  • ans = max(ans, countConsecutive) = (4, 2) =4

20

• current element is 20
  • prev + 1 != 20, therefore re-initialise countConsecutive by 1
  • countConsecutive = 1
  • update prev with current element, prev = 20
  • pop the current element
  • ans = max(ans, countConsecutive) = (4, 1) = 4

Hence, the longest consecutive subsequence is 4.

Follow the steps below to solve the problem:

• Create a Priority Queue to store the element
• Store the first element in a variable
• Remove it from the Priority Queue
• Check the difference between this removed first element and the new peek element
• If the difference is equal to 1 increase the count by 1 and repeats step 2 and step 3
• If the difference is greater than 1 set counter to 1 and repeat step 2 and step 3
• if the difference is equal to 0 repeat step 2 and 3
• if counter greater than the previous maximum then store counter to maximum
• Continue step 4 to 7 until we reach the end of the Priority Queue
• Return the maximum value

Below is the implementation of the above approach: 

C++

// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;

int findLongestConseqSubseq(int arr[], int N)
{
    priority_queue<int, vector<int>, greater<int> > pq;
    for (int i = 0; i < N; i++) {

        // adding element from
        // array to PriorityQueue
        pq.push(arr[i]);
    }

    // Storing the first element
    // of the Priority Queue
    // This first element is also
    // the smallest element
    int prev = pq.top();
    pq.pop();

    // Taking a counter variable with value 1
    int c = 1;

    // Storing value of max as 1
    // as there will always be
    // one element
    int max = 1;
    while (!pq.empty()) {

        // check if current peek
        // element minus previous
        // element is greater than
        // 1 This is done because
        // if it's greater than 1
        // then the sequence
        // doesn't start or is broken here
        if (pq.top() - prev > 1) {

            // Store the value of counter to 1
            // As new sequence may begin
            c = 1;

            // Update the previous position with the
            // current peek And remove it
            prev = pq.top();
            pq.pop();
        }

        // Check if the previous
        //  element and peek are same
        else if (pq.top() - prev == 0) {

            // Update the previous position with the
            // current peek And remove it
            prev = pq.top();
            pq.pop();
        }

        // If the difference
        // between previous element and peek is 1
        else {

            // Update the counter
            // These are consecutive elements
            c++;

            // Update the previous position
            //  with the current peek And remove it
            prev = pq.top();
            pq.pop();
        }

        // Check if current longest
        // subsequence is the greatest
        if (max < c) {

            // Store the current subsequence count as
            // max
            max = c;
        }
    }
    return max;
}

// Driver Code
int main()
{
    int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
    int n = 7;

    cout << "Length of the Longest consecutive subsequence "
            "is "
         << findLongestConseqSubseq(arr, n);
    return 0;
}
// this code is contributed by Manu Pathria

Java

// Java Program to find longest consecutive
// subsequence This Program uses Priority Queue
import java.io.*;
import java.util.PriorityQueue;
public class Longset_Sub {
    // return the length of the longest
    // subsequence of consecutive integers
    static int findLongestConseqSubseq(int arr[], int N)
    {

        PriorityQueue<Integer> pq
            = new PriorityQueue<Integer>();
        for (int i = 0; i < N; i++) {
            // adding element from
            // array to PriorityQueue
            pq.add(arr[i]);
        }

        // Storing the first element
        // of the Priority Queue
        // This first element is also
        // the smallest element
        int prev = pq.poll();

        // Taking a counter variable with value 1
        int c = 1;

        // Storing value of max as 1
        // as there will always be
        // one element
        int max = 1;

        for (int i = 1; i < N; i++) {
            // check if current peek
            // element minus previous
            // element is greater than
            // 1 This is done because
            // if it's greater than 1
            // then the sequence
            // doesn't start or is broken here
            if (pq.peek() - prev > 1) {
                // Store the value of counter to 1
                // As new sequence may begin
                c = 1;

                // Update the previous position with the
                // current peek And remove it
                prev = pq.poll();
            }

            // Check if the previous
            //  element and peek are same
            else if (pq.peek() - prev == 0) {
                // Update the previous position with the
                // current peek And remove it
                prev = pq.poll();
            }
            // if the difference
            // between previous element and peek is 1
            else {
                // Update the counter
                // These are consecutive elements
                c++;

                // Update the previous position
                //  with the current peek And remove it
                prev = pq.poll();
            }

            // Check if current longest
            // subsequence is the greatest
            if (max < c) {
                // Store the current subsequence count as
                // max
                max = c;
            }
        }

        return max;
    }

    // Driver Code
    public static void main(String args[])
        throws IOException
    {
        int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
        int n = arr.length;
        System.out.println(
            "Length of the Longest consecutive subsequence is "
            + findLongestConseqSubseq(arr, n));
    }
}
// This code is contributed by Sudipa Sarkar

Python3

# Python program for the above approach
import bisect


def findLongestConseqSubseq(arr, N):
    pq = []
    for i in range(N):

        # adding element from
        # array to PriorityQueue
        bisect.insort(pq, arr[i])

    # Storing the first element
    # of the Priority Queue
    # This first element is also
    # the smallest element
    prev = pq[0]
    pq.pop(0)

    # Taking a counter variable with value 1
    c = 1

    # Storing value of max as 1
    # as there will always be
    # one element
    max = 1
    while(len(pq)):
        # check if current peek
        # element minus previous
        # element is greater than
        # 1 This is done because
        # if it's greater than 1
        # then the sequence
        # doesn't start or is broken here
        if(pq[0] - prev > 1):
            # Store the value of counter to 1
            # As new sequence may begin
            c = 1

            # Update the previous position with the
            # current peek And remove it
            prev = pq[0]
            pq.pop(0)

        # Check if the previous
        # element and peek are same
        elif(pq[0] - prev == 0):
            # Update the previous position with the
            # current peek And remove it
            prev = pq[0]
            pq.pop(0)

        # If the difference
        # between previous element and peek is 1
        else:
            # Update the counter
            # These are consecutive elements
            c = c + 1
            # Update the previous position
            # with the current peek And remove it
            prev = pq[0]
            pq.pop(0)

        # Check if current longest
        # subsequence is the greatest
        if(max < c):
            # Store the current subsequence count as
            # max
            max = c
    return max


# Driver Code
arr = [1, 9, 3, 10, 4, 20, 2]
n = 7
print("Length of the Longest consecutive subsequence is {}".format(
    findLongestConseqSubseq(arr, n)))


# This code is contributed by Pushpesh Raj

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;

class GFG {

    // return the length of the longest
    // subsequence of consecutive integers
    static int findLongestConseqSubseq(int[] arr, int N)
    {

        List<int> pq = new List<int>();
        for (int i = 0; i < N; i++) {

            // adding element from
            // array to PriorityQueue
            pq.Add(arr[i]);
            pq.Sort();
        }

        // Storing the first element
        // of the Priority Queue
        // This first element is also
        // the smallest element
        int prev = pq[0];

        // Taking a counter variable with value 1
        int c = 1;

        // Storing value of max as 1
        // as there will always be
        // one element
        int max = 1;

        for (int i = 1; i < N; i++) {

            // check if current peek
            // element minus previous
            // element is greater than
            // 1 This is done because
            // if it's greater than 1
            // then the sequence
            // doesn't start or is broken here
            if (pq[0] - prev > 1) {
                // Store the value of counter to 1
                // As new sequence may begin
                c = 1;

                // Update the previous position with the
                // current peek And remove it
                prev = pq[0];
                pq.RemoveAt(0);
            }

            // Check if the previous
            //  element and peek are same
            else if (pq[0] - prev == 0) {

                // Update the previous position with the
                // current peek And remove it
                prev = pq[0];
                pq.RemoveAt(0);
            }

            // if the difference
            // between previous element and peek is 1
            else {

                // Update the counter
                // These are consecutive elements
                c++;

                // Update the previous position
                //  with the current peek And remove it
                prev = pq[0];
                pq.RemoveAt(0);
            }

            // Check if current longest
            // subsequence is the greatest
            if (max < c) {

                // Store the current subsequence count as
                // max
                max = c;
            }
        }

        return max;
    }

    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 9, 3, 10, 4, 20, 2 };
        int n = arr.Length;
        Console.WriteLine(
            "Length of the Longest consecutive subsequence is "
            + findLongestConseqSubseq(arr, n));
    }
}

// This code is contributed by code_hunt.

Javascript

function findLongestConseqSubseq(arr, N) {
  let pq = new PriorityQueue();

  for (let i = 0; i < N; i++) {
    pq.push(arr[i]);
  }

  let prev = pq.poll();
  let c = 1;
  let max = 1;

  for (let i = 1; i < N; i++) {
    if (pq.peek() - prev > 1) {
      c = 1;
      prev = pq.poll();
    } else if (pq.peek() - prev == 0) {
      prev = pq.poll();
    } else {
      c++;
      prev = pq.poll();
    }

    if (max < c) {
      max = c;
    }
  }

  return max;
}

class PriorityQueue {
  constructor() {
    this.items = [];
  }

  push(element) {
    this.items.push(element);
    this.bubbleUp(this.items.length - 1);
  }

  peek() {
    if (this.items.length === 0) {
      return null;
    }

    return this.items[0];
  }

  poll() {
    if (this.items.length === 0) {
      return null;
    }

    const root = this.items[0];
    const last = this.items.pop();

    if (this.items.length > 0) {
      this.items[0] = last;
      this.bubbleDown(0);
    }

    return root;
  }

  bubbleUp(index) {
    while (index > 0) {
      const parentIndex = Math.floor((index - 1) / 2);

      if (this.items[parentIndex] <= this.items[index]) {
        break;
      }

      const temp = this.items[parentIndex];
      this.items[parentIndex] = this.items[index];
      this.items[index] = temp;

      index = parentIndex;
    }
  }

  bubbleDown(index) {
    while (true) {
      const leftChildIndex = 2 * index + 1;
      const rightChildIndex = 2 * index + 2;
      let smallestChildIndex = index;

      if (
        leftChildIndex < this.items.length &&
        this.items[leftChildIndex] < this.items[smallestChildIndex]
      ) {
        smallestChildIndex = leftChildIndex;
      }

      if (
        rightChildIndex < this.items.length &&
        this.items[rightChildIndex] < this.items[smallestChildIndex]
      ) {
        smallestChildIndex = rightChildIndex;
      }

      if (smallestChildIndex === index) {
        break;
      }

      const temp = this.items[smallestChildIndex];
      this.items[smallestChildIndex] = this.items[index];
      this.items[index] = temp;

      index = smallestChildIndex;
    }
  }
}

const arr = [1, 9, 3, 10, 4, 20, 2];
const n = arr.length;
console.log(
  "Length of the Longest consecutive subsequence is " +
    findLongestConseqSubseq(arr, n)
);

Output

Length of the Longest consecutive subsequence is 4

Time Complexity: O(N*log(N)), Time required to push and pop N elements is logN for each element.
Auxiliary Space: O(N), Space required by priority queue to store N elements.

Longest Consecutive Subsequence using Dynamic Programming:

For each number in the input array find the longest possible consecutive sequence starting at that number

For e.g. if input is [3, 2, 1], find the longest possible sequence starting with 3, 2, 1. Each time you find the longest possible sequence
memoize the sequence length for future use.

For e.g. once you find the longest possible sequence starting with 3, of length 1 (since there is no other number after 3) and you need to find the longest sequence starting at 2, you do not need to recalculate the length of sequence starting with 3 as you have solved this sub-problem before. So you can reuse the memoized value.

To solve this problem, we store the current consecutive count in dp[i].

When we have a sorted array: [1, 3, 4, 4, 5], the output is 3. So we want dp[i] to remains unchanged when sorted[i] equals to sorted[i-1], in this case dp is [1, 1, 2, 2, 3].

And we reset dp[i] to 1 if sorted[i] – sorted[i-1] is not equals to 0 or 1.

Sort the array to make sure the number is in ascending order.
Initialize the parameters
max = 0
dp[0] = 1
Loop through the sorted array
If sorted[i] – sorted[i-1] equals to 1 => dp[i] = dp[i-1] + 1 and update max if dp[i] > original max
else if sorted[i] – sorted[i-1] equals to 0 => dp[i] = 1
else dp[i] = 1
Return the maximum.

Java

// Java Program to find longest consecutive
// subsequence This Program uses dynamic programming and hashmap
import java.io.*;
import java.util.*;
public class Longset_Sub {
       public static int longestConsecutive(int[] nums) {
        //base case
        if (nums.length <= 1) {
            return nums.length;
        }
        //declare a max varaibale
        int max = 1;
        Map<Integer, Integer> longest = new HashMap<Integer, Integer>();
        for (int n : nums) {
            longest.put(n, null);
        }
        //Run a loop to store the maximum value in the max varibale
        for (int i = 0; i < nums.length; i++) {
            max = Math.max(max, dfs(nums[i], longest));
        }

        return max;
    }
    //Run a dfs across every consecutive values
    private static int dfs(int start, Map<Integer, Integer> longest) {
        if (!longest.containsKey(start)) {
            return 0;
        }
        //check if the values are present in the map before hand
        if (longest.get(start) != null) {
            return longest.get(start);
        }
        //store the values in the currentlongest variable
        int currentLongest = 1 + dfs(start + 1, longest);

        longest.put(start, currentLongest);

        return currentLongest;
    }

    // Driver Code
    public static void main(String args[])
        throws IOException
    {
      //declare a array
        int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
        int n = arr.length;
        System.out.println(
            "Length of the Longest consecutive subsequence is "
            + longestConsecutive(arr));
    }
}
// This code is contributed by Sudipa Sarkar

Output

Length of the Longest consecutive subsequence is 4

Time complexity: O(N), Only one traversal is needed and the time complexity is O(n) under the assumption that hash insert and search takes O(1) time.
Auxiliary space: O(N), To store every element in the hashmap O(n) space is needed.