Given a string S consisting of lowercase English alphabets, the task is to find the length of the longest subsequence from the given string such that the difference between the largest and smallest ASCII value is exactly 1.
Examples:
Input: S = “acbbebcg”
Output: 5
Explanation: The longest subsequence of required type is “cbbbc”, whose length is 5.
The difference between largest (‘c’) and smallest (‘b’) ASCII values is c – b = 99 – 98 = 1, which is minimum possible.Input: S = “abcd”
Output: 2
Explanation: The longest subsequence of the required type is “ab”, whose length is 2. Other possible subsequences are “bc” and “cd”.
The difference between largest(‘b’) and smallest(‘a’) ASCII values is b – a = 98 – 97 = 1.
Naive Approach: The simplest approach to solve the problem is to generate all possible subsequences of the given string S and print the length of the subsequence which is of maximum length and having a difference between ASCII values of the largest and smallest character is exactly equal to 1.
Time Complexity: O(N * 2N)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the main idea is to use a Map to optimize the above approach. Follow the steps below to solve the problem:
- Initialize a variable, say maxLength, that stores the maximum length of the resultant subsequence.
- Store the frequency of characters in a Map, say M.
- Traverse the string and for each character, say ch, check if there exists character c with ASCII value (ch – 1) in the map M or not. If found to be true, then update maxLength as the maximum of maxLength and (M[ch] + M[ch – 1]).
- After completing the above steps, print the value of maxLength as the result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum length of // subsequence having difference of ASCII // value of longest and smallest character as 1 int maximumLengthSubsequence(string str)
{ // Stores frequency of characters
unordered_map< char , int > mp;
// Iterate over characters
// of the string
for ( char ch : str) {
mp[ch]++;
}
// Stores the resultant
// length of subsequence
int ans = 0;
for ( char ch : str) {
// Check if there exists any
// elements with ASCII value
// one less than character ch
if (mp.count(ch - 1)) {
// Size of current subsequence
int curr_max = mp[ch] + mp[ch - 1];
// Update the value of ans
ans = max(ans, curr_max);
}
}
// Print the resultant count
cout << ans;
} // Driver Code int main()
{ string S = "acbbebcg" ;
maximumLengthSubsequence(S);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the maximum length of // subsequence having difference of ASCII // value of longest and smallest character as 1 static void maximumLengthSubsequence(String str)
{ // Stores frequency of characters
HashMap<Character, Integer> mp = new HashMap<>();
for ( char ch : str.toCharArray())
{
mp.put(ch, mp.getOrDefault(ch, 0 ) + 1 );
}
// Stores the resultant
// length of subsequence
int ans = 0 ;
for ( char ch : str.toCharArray())
{
// Check if there exists any
// elements with ASCII value
// one less than character ch
if (mp.containsKey(( char )(ch - 1 )))
{
// Size of current subsequence
int curr_max = mp.get(ch) +
mp.get(( char )(ch - 1 ));
// Update the value of ans
ans = Math.max(ans, curr_max);
}
}
// Print the resultant count
System.out.println(ans);
} // Driver Code public static void main(String[] args)
{ String S = "acbbebcg" ;
maximumLengthSubsequence(S);
} } // This code is contributed by aadityapburujwale |
# Python3 program for the above approach # Function to find the maximum length of # subsequence having difference of ASCII # value of longest and smallest character as 1 def maximumLengthSubsequence( str ):
# Stores frequency of characters
mp = {}
# Iterate over characters
# of the string
for ch in str :
if ch in mp.keys():
mp[ch] + = 1
else :
mp[ch] = 1
# Stores the resultant
# length of subsequence
ans = 0
for ch in str :
# Check if there exists any
# elements with ASCII value
# one less than character ch
if chr ( ord (ch) - 1 ) in mp.keys():
# Size of current subsequence
curr_max = mp[ch]
if chr ( ord (ch) - 1 ) in mp.keys():
curr_max + = mp[ chr ( ord (ch) - 1 )]
# Update the value of ans
ans = max (ans, curr_max)
# Print the resultant count
print (ans)
# Driver Code S = "acbbebcg"
maximumLengthSubsequence(S) # This code is contributed by Stream_Cipher |
// C# program for the above approach using System;
using System.Collections.Generic;
public class GFG
{ // Function to find the maximum length of // subsequence having difference of ASCII // value of longest and smallest character as 1 static void maximumLengthSubsequence(String str)
{ // Stores frequency of characters
Dictionary< char , int > mp = new Dictionary< char , int >();
foreach ( char ch in str.ToCharArray())
{
if (mp.ContainsKey(ch))
mp[ch] = mp[ch] + 1;
else
mp.Add(ch, 1);
}
// Stores the resultant
// length of subsequence
int ans = 0;
foreach ( char ch in str.ToCharArray())
{
// Check if there exists any
// elements with ASCII value
// one less than character ch
if (mp.ContainsKey(( char )(ch - 1)))
{
// Size of current subsequence
int curr_max = mp[ch] +
mp[( char )(ch - 1)];
// Update the value of ans
ans = Math.Max(ans, curr_max);
}
}
// Print the resultant count
Console.WriteLine(ans);
} // Driver Code public static void Main(String[] args)
{ String S = "acbbebcg" ;
maximumLengthSubsequence(S);
} } // This code is contributed by Amit Katiyar |
<script> // JavaScript program for the above approach
// Function to find the maximum length of
// subsequence having difference of ASCII
// value of longest and smallest character as 1
function maximumLengthSubsequence(str) {
// Stores frequency of characters
var mp = {};
var temp = str.split( "" );
for (const ch of temp) {
if (mp.hasOwnProperty(ch)) mp[ch] = mp[ch] + 1;
else mp[ch] = 1;
}
// Stores the resultant
// length of subsequence
var ans = 0;
for (const ch of temp) {
// Check if there exists any
// elements with ASCII value
// one less than character ch
if (mp.hasOwnProperty(String.fromCharCode(ch.charCodeAt(0) - 1))) {
// Size of current subsequence
var curr_max =
mp[ch] + mp[String.fromCharCode(ch.charCodeAt(0) - 1)];
// Update the value of ans
ans = Math.max(ans, curr_max);
}
}
// Print the resultant count
document.write(ans);
}
// Driver Code
var S = "acbbebcg" ;
maximumLengthSubsequence(S);
// This code is contributed by rdtank.
</script>
|
5
Time Complexity: O(N)
Auxiliary Space: O(1)