# Longest Consecutive Subsequence

• Difficulty Level : Medium
• Last Updated : 31 Jan, 2023

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 using namespace std; // Returns length of the longest// contiguous subsequenceint findLongestConseqSubseq(int arr[], int n){    int ans = 0, count = 0;     // sort the array    sort(arr, arr + n);     vector 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 codeint 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 subsequenceimport 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 v = new ArrayList();        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 codearr = [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 subsequenceusing 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 v = new List();        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



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 using namespace std; // Returns length of the longest// contiguous subsequenceint findLongestConseqSubseq(int arr[], int n){    unordered_set 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++) {        // if current element is the starting        // element of a 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 codeint 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 subsequenceimport java.io.*;import java.util.*; class ArrayElements {    // Returns length of the longest    // consecutive subsequence    static int findLongestConseqSubseq(int arr[], int n)    {        HashSet S = new HashSet();        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 codeif __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 S = new HashSet();        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



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:

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.

• 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
• 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
• 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
• 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
• 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
• 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 using namespace std; int findLongestConseqSubseq(int arr[], int N){    priority_queue, greater > 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 Codeint 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 Queueimport 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 pq            = new PriorityQueue();        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 approachimport 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 Codearr = [1, 9, 3, 10, 4, 20, 2]n = 7print("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 approachusing 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 pq = new List();        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.

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.

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