Length of longest subsequence consisting of distinct adjacent elements

Given an array arr[], the task is to find the length of the longest subsequence of the array arr[] such that all adjacent elements in the subsequence are different.

Examples:

Input: arr[] = {4, 2, 3, 4, 3}
Output: 5
Explanation:
The longest subsequence where no two adjacent elements are equal is {4, 2, 3, 4, 3}. Length of the subsequence is 5.

Input: arr[] = {7, 8, 1, 2, 2, 5, 5, 1}
Output: 6
Explanation: Longest subsequence where no two adjacent elements are equal is {7, 8, 1, 2, 5, 1}. Length of the subsequence is 5.

Naive Approach: The simplest approach is to generate all possible subsequence of the given array and print the maximum length of that subsequence having all adjacent elements different. 

Time Complexity: O(2^N)
Auxiliary Space: O(1)

Efficient Approach: Follow the steps below to solve the problem:

• Initialize count to 1 to store the length of the longest subsequence.
• Traverse the array over the indices [1, N - 1] and for each element, check if the current element is equal to the previous element or not. If found to be not equal, then increment count by 1.
• After completing the above steps, print the value of count as the maximum possible length of subsequence.

Below is the implementation of the above approach:

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

// Function that finds the length of
// longest subsequence having different
// adjacent elements
void longestSubsequence(int arr[], int N)
{
    // Stores the length of the
    // longest subsequence
    int count = 1;

    // Traverse the array
    for (int i = 1; i < N; i++) {

        // If previous and current
        // element are not same
        if (arr[i] != arr[i - 1]) {

            // Increment the count
            count++;
        }
    }

    // Print the maximum length
    cout << count << endl;
}

// Driver Code
int main()
{
    int arr[] = { 7, 8, 1, 2, 2, 5, 5, 1 };

    // Size of Array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    longestSubsequence(arr, N);

    return 0;
}

// Java program for the 
// above approach
import java.util.*;

class GFG{
   
// Function that finds the length of
// longest subsequence having different
// adjacent elements
static void longestSubsequence(int arr[], 
                               int N)
{
  // Stores the length of the
  // longest subsequence
  int count = 1;

  // Traverse the array
  for (int i = 1; i < N; i++) 
  {
    // If previous and current
    // element are not same
    if (arr[i] != arr[i - 1]) 
    {
      // Increment the count
      count++;
    }
  }

  // Print the maximum length
  System.out.println(count);
}

// Driver Code
public static void main(String args[])
{
  int arr[] = {7, 8, 1, 2, 
               2, 5, 5, 1};

  // Size of Array
  int N = arr.length;

  // Function Call
  longestSubsequence(arr, N);
}
}

// This code is contributed by bgangwar59

# Python3 program for the above approach

# Function that finds the length of
# longest subsequence having different
# adjacent elements
def longestSubsequence(arr, N):
    
    # Stores the length of the
    # longest subsequence
    count = 1

    # Traverse the array
    for i in range(1, N, 1):
        
        # If previous and current
        # element are not same
        if (arr[i] != arr[i - 1]):
            
            # Increment the count
            count += 1

    # Print the maximum length
    print(count)

# Driver Code
if __name__ == '__main__':
    
    arr = [ 7, 8, 1, 2, 2, 5, 5, 1 ]
    
    # Size of Array
    N = len(arr)
    
    # Function Call
    longestSubsequence(arr, N)

# This code is contributed by ipg2016107

// C# program for the 
// above approach
using System;
 
class GFG{
    
// Function that finds the length of
// longest subsequence having different
// adjacent elements
static void longestSubsequence(int[] arr, 
                               int N)
{
  
  // Stores the length of the
  // longest subsequence
  int count = 1;
 
  // Traverse the array
  for(int i = 1; i < N; i++) 
  {
    
    // If previous and current
    // element are not same
    if (arr[i] != arr[i - 1]) 
    {
      
      // Increment the count
      count++;
    }
  }
 
  // Print the maximum length
  Console.WriteLine(count);
}
 
// Driver Code
public static void Main()
{
  int[] arr = { 7, 8, 1, 2, 
                2, 5, 5, 1 };
  
  // Size of Array
  int N = arr.Length;
  
  // Function Call
  longestSubsequence(arr, N);
}
}

// This code is contributed by susmitakundugoaldanga

<script>

// JavaScript program to implement
// the above approach

// Function that finds the length of
// longest subsequence having different
// adjacent elements
function longestSubsequence(arr, N)
{
  // Stores the length of the
  // longest subsequence
  let count = 1;
 
  // Traverse the array
  for (let i = 1; i < N; i++)
  {
    // If previous and current
    // element are not same
    if (arr[i] != arr[i - 1])
    {
      // Increment the count
      count++;
    }
  }
 
  // Print the maximum length
  document.write(count);
}

// Driver Code

    let arr = [7, 8, 1, 2,
               2, 5, 5, 1];
 
  // Size of Array
  let N = arr.length;
 
  // Function Call
  longestSubsequence(arr, N);
          
</script>

6

Time Complexity: O(N)
Auxiliary Space: O(1)

supratik_mitra

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Article Tags :

• Greedy
• School Programming
• DSA
• Arrays
• subsequence
• frequency-counting

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