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# Longest subsequence with consecutive English alphabets

Given string S, the task is to find the length of the longest subsequence of the consecutive lowercase alphabets.

Examples:

Input: S = “acbdcfhg”
Output: 3
Explanation:
String “abc” is the longest subsequence of consecutive lowercase alphabets.
Therefore, print 3 as it is the length of the subsequence “abc”.

Input: S = “gabbsdcdggbe”
Output: 5

Naive Approach: The simplest approach is to generate all possible subsequences of the given string and if there exists any subsequence of the given string that has consecutive characters and is of maximum length then print that length.

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

Efficient Approach: The above approach can also be optimized by generating all possible consecutive subsequences of the given string starting from each lowercase alphabets and print the maximum among all the subsequence found. Follow the steps below to solve the problem:

• Initialize a variable, say ans, as 0 that stores the maximum length of the consecutive subsequence.
• For each character ch over the range [a, z] perform the following:
• Initialize a variable cnt as 0 that stores the length of a subsequence of consecutive characters starting from ch.
• Traverse the given string S and if the current character is ch then increment the cnt by 1 and update the current character ch to the next character by (ch + 1).
• Update ans = max(ans, cnt)
• After the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the length of subsequence` `// starting with character ch` `int` `findSubsequence(string S, ``char` `ch)` `{` `    ``// Length of the string` `    ``int` `N = S.length();`   `    ``// Stores the maximum length` `    ``int` `ans = 0;`   `    ``// Traverse the given string` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// If s[i] is required` `        ``// character ch` `        ``if` `(S[i] == ch) {`   `            ``// Increment  ans by 1` `            ``ans++;`   `            ``// Increment  character ch` `            ``ch++;` `        ``}` `    ``}`   `    ``// Return the current maximum` `    ``// length with character ch` `    ``return` `ans;` `}`   `// Function to find the maximum length` `// of subsequence of consecutive` `// characters` `int` `findMaxSubsequence(string S)` `{` `    ``// Stores the maximum length of` `    ``// consecutive characters` `    ``int` `ans = 0;`   `    ``for` `(``char` `ch = ``'a'``; ch <= ``'z'``; ch++) {`   `        ``// Update ans` `        ``ans = max(ans, findSubsequence(S, ch));` `    ``}`   `    ``// Return the maximum length of` `    ``// subsequence` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Input` `    ``string S = ``"abcabefghijk"``;`   `    ``// Function Call` `    ``cout << findMaxSubsequence(S);`   `    ``return` `0;` `}`

## Java

 `// C# program for the above approach` `import` `java.io.*;` `import` `java.util.*;` `import` `java.util.Arrays;`   `class` `GFG{`   `// Function to find the length of subsequence` `// starting with character ch` `static` `int` `findSubsequence(String S, ``char` `ch)` `{` `    `  `    ``// Length of the string` `    ``int` `N = S.length();`   `    ``// Stores the maximum length` `    ``int` `ans = ``0``;`   `    ``// Traverse the given string` `    ``for``(``int` `i = ``0``; i < N; i++)` `    ``{` `        `  `        ``// If s[i] is required` `        ``// character ch` `         ``if``(S.charAt(i) == ch)` `        ``{` `            `  `            ``// Increment  ans by 1` `            ``ans++;`   `            ``// Increment  character ch` `            ``ch++;` `        ``}` `    ``}`   `    ``// Return the current maximum` `    ``// length with character ch` `    ``return` `ans;` `}`   `// Function to find the maximum length` `// of subsequence of consecutive` `// characters` `static` `int` `findMaxSubsequence(String S)` `{` `    `  `    ``// Stores the maximum length of` `    ``// consecutive characters` `    ``int` `ans = ``0``;`   `    ``for``(``char` `ch = ``'a'``; ch <= ``'z'``; ch++)` `    ``{` `        `  `        ``// Update ans` `        ``ans = Math.max(ans, findSubsequence(S, ch));` `    ``}`   `    ``// Return the maximum length of` `    ``// subsequence` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args) ` `{` `    `  `    ``// Input` `    ``String S = ``"abcabefghijk"``;`   `    ``// Function Call` `    ``System.out.print(findMaxSubsequence(S));` `}` `}`   `// This code is contributed by shivanisinghss2110`

## Python3

 `# Python3 program for the above approach`   `# Function to find the length of subsequence` `# starting with character ch` `def` `findSubsequence(S, ch):` `    ``# Length of the string` `    ``N ``=` `len``(S)`   `    ``# Stores the maximum length` `    ``ans ``=` `0`   `    ``# Traverse the given string` `    ``for` `i ``in` `range``(N):` `      `  `        ``# If s[i] is required` `        ``# character ch` `        ``if` `(S[i] ``=``=` `ch):`   `            ``# Increment  ans by 1` `            ``ans ``+``=` `1`   `            ``# Increment  character ch` `            ``ch``=``chr``(``ord``(ch) ``+` `1``)`   `    ``# Return the current maximum` `    ``# length with character ch` `    ``return` `ans`   `# Function to find the maximum length` `# of subsequence of consecutive` `# characters` `def` `findMaxSubsequence(S):` `    ``#Stores the maximum length of` `    ``# consecutive characters` `    ``ans ``=` `0`   `    ``for` `ch ``in` `range``(``ord``(``'a'``),``ord``(``'z'``) ``+` `1``):` `        ``# Update ans` `        ``ans ``=` `max``(ans, findSubsequence(S, ``chr``(ch)))`   `    ``# Return the maximum length of` `    ``# subsequence` `    ``return` `ans`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``# Input` `    ``S ``=` `"abcabefghijk"`   `    ``# Function Call` `    ``print` `(findMaxSubsequence(S))`   `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to find the length of subsequence` `// starting with character ch` `static` `int` `findSubsequence(``string` `S, ``char` `ch)` `{` `    `  `    ``// Length of the string` `    ``int` `N = S.Length;`   `    ``// Stores the maximum length` `    ``int` `ans = 0;`   `    ``// Traverse the given string` `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        `  `        ``// If s[i] is required` `        ``// character ch` `        ``if` `(S[i] == ch) ` `        ``{` `            `  `            ``// Increment  ans by 1` `            ``ans++;`   `            ``// Increment  character ch` `            ``ch++;` `        ``}` `    ``}`   `    ``// Return the current maximum` `    ``// length with character ch` `    ``return` `ans;` `}`   `// Function to find the maximum length` `// of subsequence of consecutive` `// characters` `static` `int` `findMaxSubsequence(``string` `S)` `{` `    `  `    ``// Stores the maximum length of` `    ``// consecutive characters` `    ``int` `ans = 0;`   `    ``for``(``char` `ch = ``'a'``; ch <= ``'z'``; ch++)` `    ``{` `        `  `        ``// Update ans` `        ``ans = Math.Max(ans, findSubsequence(S, ch));` `    ``}`   `    ``// Return the maximum length of` `    ``// subsequence` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    `  `    ``// Input` `    ``string` `S = ``"abcabefghijk"``;`   `    ``// Function Call` `    ``Console.Write(findMaxSubsequence(S));` `}` `}`   `// This code is contributed by SURENDRA_GANGWAR`

## Javascript

 ``

Output

`7`

Time Complexity: O(26*N)
Auxiliary Space: O(1)

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