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Longest subsequence with different adjacent characters
• Last Updated : 06 Sep, 2020

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 from the previous character then count the current character to find the longest subsequence.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``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;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`
Output:
```5

```

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

1. For each character 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 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 elements.
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.a 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++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// dp table``int` `dp;` `// 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;``}`

## Java

 `// 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`

## Python3

 `# Python3 program for the above approach``# dp table``dp ``=` `[[``-``1` `for` `i ``in` `range``(``27``)] ``for` `j ``in` `range``(``100005``)];` `# A recursive function to find the``# update the dp table``def` `calculate(pos, prev, s):``  ` `    ``# If we reach end of the String``    ``if` `(pos ``=``=` `len``(s)):``        ``return` `0``;` `    ``# If subproblem has been computed``    ``if` `(dp[pos][prev] !``=` `-``1``):``        ``return` `dp[pos][prev];` `    ``# Initialise variable to find the``    ``# maximum length``    ``val ``=` `0``;` `    ``# Choose the current character``    ``if` `(``ord``(s[pos]) ``-` `ord``(``'a'``) ``+` `1` `!``=` `prev):``        ``val ``=` `max``(val, ``1` `+` `calculate(pos ``+` `1``,``                                     ``ord``(s[pos]) ``-``                                     ``ord``(``'a'``) ``+` `1``, s));` `    ``# Omit the current character``    ``val ``=` `max``(val, calculate(pos ``+` `1``, prev, s));` `    ``# Return the store answer to``    ``# the current subproblem``    ``dp[pos][prev] ``=` `val;``    ``return` `dp[pos][prev];` `# Function to find the longest Subsequence``# with different adjacent character``def` `longestSubsequence(s):``  ` `    ``# Length of the String s``    ``n ``=` `len``(s);` `    ``# Return the final ans after every``    ``# recursive call``    ``return` `calculate(``0``, ``0``, s);` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``str` `=` `"ababa"``;` `    ``# Function call``    ``print``(longestSubsequence(``str``));` `# This code is contributed by shikhasingrajput`

## C#

 `// C# program for the above approach``using` `System;`` ` `public` `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[pos] - ``'a'` `+ 1 != prev)``    ``{``        ``val = Math.Max(val, 1 +``                       ``calculate(pos + 1,``                                 ``s[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 readonly ans after every``    ``// recursive call``    ``return` `calculate(0, 0, s);``}`` ` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``String str = ``"ababa"``;`` ` `    ``// Function call``    ``Console.Write(longestSubsequence(str));``}``}`` ` `// This code is contributed by shikhasingrajput`
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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