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Longest subsequence with consecutive English alphabets

  • Last Updated : 13 Aug, 2021

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

Examples:

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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 <bits/stdc++.h>
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




<script>
// Javascript program for the above approach
 
// Function to find the length of subsequence
// starting with character ch
function findSubsequence(S,ch)
{
    // Length of the string
    let N = S.length;
      
    // Stores the maximum length
    let ans = 0;
  
    // Traverse the given string
    for(let 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=String.fromCharCode(ch.charCodeAt(0)+1);
        }
    }
  
    // Return the current maximum
    // length with character ch
    return ans;
}
 
// Function to find the maximum length
// of subsequence of consecutive
// characters
function findMaxSubsequence(S)
{
    // Stores the maximum length of
    // consecutive characters
    let ans = 0;
  
    for(let ch = 'a'.charCodeAt(0); ch <= 'z'.charCodeAt(0); ch++)
    {
          
        // Update ans
        ans = Math.max(ans, findSubsequence(S, String.fromCharCode(ch)));
    }
  
    // Return the maximum length of
    // subsequence
    return ans;
}
 
// Driver Code
let S = "abcabefghijk";
  
// Function Call
document.write(findMaxSubsequence(S));
 
 
// This code is contributed by patel2127
</script>
Output
7

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

 




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