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++
// 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
// 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 |
5
Time Complexity: O(N)
Auxiliary Space: O(1)
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