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
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
The subsequence 36, 35, 33, 34, 32 is the longest subsequence
of consecutive elements.
One Solution is to first sort the array and find the longest subarray with consecutive elements. Time complexity of this solution is O(nLogn). Thanks to Hao.W for suggesting this solution.
We can solve this problem in O(n) time using an Efficient Solution. The idea is to use Hashing. We first insert all elements in a Hash. Then check all the possible starts of consecutive subsequences. Below is complete algorithm.
1) Create an empty hash.
2) Insert all array elements to hash.
3) Do following for every element arr[i]
....a) Check if this element is the starting point of a
subsequence. To check this, we simply look for
arr[i] - 1 in hash, if not found, then this is
the first element a subsequence.
If this element is a first element, then count
number of elements in the consecutive starting
with this element.
If count is more than current res, then update
res.
Below is the implementation of above algorithm.
C/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++) { // 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; } |
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)) 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 |
Python
# Python program to find longest contiguous subsequence from sets import Setdef 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 |
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; } // 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 |
Output:
Length of the Longest contiguous subsequence is 4
Time Complexity: At first look, time complexity looks more than O(n). If we take a closer look, we can notice that it is O(n) under the assumption that hash insert and search take O(1) time. The function S.find() inside the while loop is called at most twice for every element. For example, consider the case when all array elements are consecutive. In this case, the outer find is called for every element, but we go inside the if condition only for the smallest element. Once we are inside the if condition, we call find() one more time for every other element.
Thanks to Gaurav Ahirwar for above solution.
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