Longest subsequence with different adjacent characters

Given string str. The task is to find the longest subsequence of str such that all the characters adjacent to each other in the subsequence are different.
Examples: 
 

Input: str = “ababa” 
Output:
Explanation: 
“ababa” is the subsequence satisfying the condition
Input: str = “xxxxy” 
Output:
Explanation: 
“xy” is the subsequence satisfying the condition 
 

 

Method 1: Greedy Approach 
It can be observed that choosing the first character which is not similar to the previously chosen character given the longest subsequence of the given string with different adjacent characters. 
The idea is to keep track of previously picked character while iterating through the string and if the current character is different than the previous character then count the current character to find the longest subsequence.

Below is the implementation of the above approach: 
 

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the longest Subsequence
// with different adjacent character
int longestSubsequence(string s)
{
    // Length of the string s
    int n = s.length();
    int answer = 0;
  
    // Previously picked character
    char prev = '-';
  
    for (int i = 0; i < n; i++) {
        // If the current character is
        // different from the previous
        // then include this character
        // and update previous character
        if (prev != s[i]) {
            prev = s[i];
            answer++;
        }
    }
  
    return answer;
}
  
// Driver Code
int main()
{
    string str = "ababa";
  
    // Function call
    cout << longestSubsequence(str);
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to find the longest subsequence
// with different adjacent character
static int longestSubsequence(String s)
{
      
    // Length of the String s
    int n = s.length();
    int answer = 0;
  
    // Previously picked character
    char prev = '-';
  
    for(int i = 0; i < n; i++)
    {
  
       // If the current character is
       // different from the previous
       // then include this character
       // and update previous character
       if (prev != s.charAt(i))
       {
           prev = s.charAt(i);
           answer++;
       }
    }
  
    return answer;
}
  
// Driver Code
public static void main(String[] args)
{
    String str = "ababa";
  
    // Function call
    System.out.print(longestSubsequence(str));
}
}
  
// This code is contributed by sapnasingh4991

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Python3

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# Python3 program for the above approach
  
# Function to find the longest Subsequence
# with different adjacent character
def longestSubsequence(s):
  
    # Length of the string s
    n = len(s);
    answer = 0;
  
    # Previously picked character
    prev = '-';
  
    for i in range(0, n):
          
        # If the current character is
        # different from the previous
        # then include this character
        # and update previous character
        if (prev != s[i]):
            prev = s[i];
            answer += 1;
          
    return answer;
  
# Driver Code
str = "ababa";
  
# Function call
print(longestSubsequence(str));
  
# This code is contributed by Code_Mech

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C#

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// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the longest subsequence
// with different adjacent character
static int longestSubsequence(String s)
{
      
    // Length of the String s
    int n = s.Length;
    int answer = 0;
  
    // Previously picked character
    char prev = '-';
  
    for(int i = 0; i < n; i++)
    {
         
       // If the current character is
       // different from the previous
       // then include this character
       // and update previous character
       if (prev != s[i])
       {
           prev = s[i];
           answer++;
       }
    }
    return answer;
}
  
// Driver Code
public static void Main(String[] args)
{
    String str = "ababa";
  
    // Function call
    Console.Write(longestSubsequence(str));
}
}
  
// This code is contributed by amal kumar choubey

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Output: 



5

 

Time Complexity: O(N), where N is the length of the given string.
Method 2: Dynamic Programming 
 

  1. For each characters in the given string str, do the following: 
    • Choose the current characters in the string for the resultant subsequence and recurr for the remaining string to find the next possible characters for the resultant subsequence.
    • Omit the current characters and recurr for the the remaining string to find the next possible characters for the resultant subsequence.
  2. The maximum value in the above recursive call will be the longest subsequence with different adjacent element.
  3. The recurrence relation is given by: 
     
    Let dp[pos][prev] be the length of longest subsequence 
    till index pos such that alphabet prev was picked previously.
    
    dp[pos][prev] = max(1 + function(pos+1, s[pos] - 'a' + 1, s), 
                        function(pos+1, prev, s));
    
    

Below is the implementation of the above approach: 
 

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// dp table
int dp[100005][27];
  
// A recursive function to find the
// update the dp[][] table
int calculate(int pos, int prev, string& s)
{
  
    // If we reach end of the string
    if (pos == s.length()) {
        return 0;
    }
  
    // If subproblem has been computed
    if (dp[pos][prev] != -1)
        return dp[pos][prev];
  
    // Initialise variable to find the
    // maximum length
    int val = 0;
  
    // Choose the current character
    if (s[pos] - 'a' + 1 != prev) {
        val = max(val,
                  1 + calculate(pos + 1,
                                s[pos] - 'a' + 1,
                                s));
    }
  
    // Omit the current character
    val = max(val, calculate(pos + 1, prev, s));
  
    // Return the store answer to the
    // current subproblem
    return dp[pos][prev] = val;
}
  
// Function to find the longest Subsequence
// with different adjacent character
int longestSubsequence(string s)
{
  
    // Length of the string s
    int n = s.length();
  
    // Initialise the memoisation table
    memset(dp, -1, sizeof(dp));
  
    // Return the final ans after every
    // recursive call
    return calculate(0, 0, s);
}
  
// Driver Code
int main()
{
    string str = "ababa";
  
    // Function call
    cout << longestSubsequence(str);
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
  
// dp table
static int dp[][] = new int[100005][27];
  
// A recursive function to find the
// update the dp[][] table
static int calculate(int pos, int prev, String s)
{
  
    // If we reach end of the String
    if (pos == s.length())
    {
        return 0;
    }
  
    // If subproblem has been computed
    if (dp[pos][prev] != -1)
        return dp[pos][prev];
  
    // Initialise variable to find the
    // maximum length
    int val = 0;
  
    // Choose the current character
    if (s.charAt(pos) - 'a' + 1 != prev)
    {
        val = Math.max(val, 1 + calculate(pos + 1,
                                s.charAt(pos) - 'a' + 1,
                                s));
    }
  
    // Omit the current character
    val = Math.max(val, calculate(pos + 1, prev, s));
  
    // Return the store answer to the
    // current subproblem
    return dp[pos][prev] = val;
}
  
// Function to find the longest Subsequence
// with different adjacent character
static int longestSubsequence(String s)
{
  
    // Length of the String s
    int n = s.length();
  
    // Initialise the memoisation table
    for(int i = 0; i < 100005; i++)
    {
        for (int j = 0; j < 27; j++) 
        {
            dp[i][j] = -1;
        }
    }
  
    // Return the final ans after every
    // recursive call
    return calculate(0, 0, s);
}
  
// Driver Code
public static void main(String[] args)
{
    String str = "ababa";
  
    // Function call
    System.out.print(longestSubsequence(str));
}
}
  
// This code is contributed by Rohit_ranjan

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Output: 

5

 

Time Complexity: O(N), where N is the length of the given string. 
Auxiliary Space: O(26*N) where N is the length of the given string.
 

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