SIRIS

Longest consecutive subsequence

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 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}
Initialise countConsecutive with 0 which will increment when arr[i] == arr[i – 1] + 1 is true otherwise countConsecutive will re-initialise by 1.
Maintain a variable ans to store maximum count of consecutive elements so far.
At i = 0:
as i is 0 then re-initialise 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-initialise 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-initialise 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:

Initialise 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 above approach:

// 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

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:

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

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# 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.

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.

SIRIS

Longest consecutive subsequence

Longest Consecutive Subsequence using Hashing:

Longest Consecutive Subsequence using Priority Queue:

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