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 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.

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// 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);
  
    // find the maximum length
    // by traversing the array
    for (int i = 0; i < n; i++) {
        // if the current element is equal
        // to previous element +1
        if (i > 0 && arr[i] == arr[i - 1] + 1)
            count++;
        // reset the count
        else
            count = 1;
  
        // update the maximum
        ans = max(ans, count);
    }
    return ans;
}
  
// Driver program
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;
}

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Output:

Length of the Longest contiguous subsequence is 4

Complexity Analysis:



  • Time complexity: O(nLogn).
    Time to sort the array is O(nlogn).
  • Auxiliary space : O(1).
    As no extra space is needed.

Thanks to Hao.W for suggesting the above solution.

Efficient solution:
This problem can be solved in O(n) time using an Efficient Solution. The idea is to use Hashing. We first insert all elements in a Set. Then check all the possible starts of consecutive subsequences.

Algorithm:

  1. Create an empty hash.
  2. Insert all array elements to hash.
  3. Do following for every element arr[i]
  4. 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 a subsequence.
  5. 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.
  6. If the count is more than the previous longest subsequence found, then update this.

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

C/C++

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// 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++) {
        // if current element is the starting
        // element of a sequence
        if (S.find(arr[i] - 1) == S.end()) {
            // Then check for next elements
            // in 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 program
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;
}

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Java

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

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Python

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# Python program to find longest contiguous subsequence
  
from sets import Set
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 function 
if __name__=='__main__':
    n = 7
    arr = [1, 9, 3, 10, 4, 20, 2]
    print "Length of the Longest contiguous subsequence is "
    print  findLongestConseqSubseq(arr, n)
          
# Contributed by: Harshit Sidhwa

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C#

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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;
    }
  
    // Testing program
    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

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Output:

Length of the Longest contiguous subsequence is 4

Complexity Analysis:

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

Thanks to Gaurav Ahirwar for the above solution.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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Improved By : shrikanth13, andrew1234