Count maximum occurrence of subsequence in string such that indices in subsequence is in A.P.

Given a string S, the task is to count the maximum occurrence of subsequence in the given string such that indices of the characters of the subsequence are in Arithmetic Progression.

Examples:

Input: S = “xxxyy”
Output: 6
Explanation:
There is a subsequence “xy”, where indices of each character of the subsequence are in A.P.
The indices of the different characters that form the subsequence “xy” –
{(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

Input: S = “pop”
Output: 2
Explanation:
There is a subsequence “p”, where indices of each character of the subsequence are in A.P.
The indices of the different characters that form the subsequence “p” –
{(1), (2)}

Approach: The key observation in the problem is if there are two characters in string whose collectively occurrence is greater than the occurrence of any single character, then these characters will form the maximum occurrence subsequence in the string with the character in Arithmetic progression because of every two integers will always form an arithmetic progression. Below is the illustration of the steps:



  • Iterate over the string and count the frequency of the characters of the string. That is considering the subsequences of length 1.
  • Iterate over the string and choose every two possible characters of the string and increment the frequency of the subsequence of the string.
  • Finally, find the maximum frequency of the subsequence from length 1 and 2.

Below is the implementation of the above approach:

C++

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// C++ implementation to find the 
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
#include <bits/stdc++.h>
  
using namespace std;
  
// Function to find the 
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
int maximumOccurrence(string s)
{
    int n = s.length();
  
    // Frequencies of subsequence
    map<string, int> freq;
  
    // Loop to find the frequencies 
    // of subsequence of length 1
    for (int i = 0; i < n; i++) {
        string temp = "";
        temp += s[i];
        freq[temp]++;
    }
      
    // Loop to find the frequencies 
    // subsequence of length 2 
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            string temp = "";
            temp += s[i];
            temp += s[j];
            freq[temp]++;
        }
    }
  
    int answer = INT_MIN;
  
    // Finding maximum frequency
    for (auto it : freq)
        answer = max(answer, it.second);
    return answer;
}
  
// Driver Code
int main()
{
    string s = "xxxyy";
  
    cout << maximumOccurrence(s);
    return 0;
}

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Java

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// Java implementation to find the 
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
import java.util.*;
  
class GFG 
{
    // Function to find the 
    // maximum occurence of the subsequence
    // such that the indices of characters
    // are in arithmetic progression
    static int maximumOccurrence(String s)
    {
        int n = s.length();
       
        // Frequencies of subsequence
        HashMap<String, Integer> freq = new HashMap<String,Integer>();
        int i, j;
  
        // Loop to find the frequencies 
        // of subsequence of length 1
        for ( i = 0; i < n; i++) {
            String temp = "";
            temp += s.charAt(i);
            if (freq.containsKey(temp)){
                freq.put(temp,freq.get(temp)+1); 
            }
            else{
                freq.put(temp, 1); 
            }
        }
           
        // Loop to find the frequencies 
        // subsequence of length 2 
        for (i = 0; i < n; i++) {
            for (j = i + 1; j < n; j++) {
                String temp = "";
                temp += s.charAt(i);
                temp += s.charAt(j);
                if(freq.containsKey(temp))
                    freq.put(temp,freq.get(temp)+1);
                else 
                    freq.put(temp,1);
            }
        }
        int answer = Integer.MIN_VALUE;
       
        // Finding maximum frequency
        for (int it : freq.values())
            answer = Math.max(answer, it);
        return answer;
    }
       
    // Driver Code
    public static void main(String []args)
    {
        String s = "xxxyy";
       
        System.out.print(maximumOccurrence(s));
    }
}
  
// This code is contributed by chitranayal

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Python3

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# Python3 implementation to find the
# maximum occurence of the subsequence
# such that the indices of characters
# are in arithmetic progression
  
# Function to find the
# maximum occurence of the subsequence
# such that the indices of characters
# are in arithmetic progression
def maximumOccurrence(s):
    n = len(s)
  
    # Frequencies of subsequence
    freq = {}
  
    # Loop to find the frequencies
    # of subsequence of length 1
    for i in s:
        temp = ""
        temp += i
        freq[temp] = freq.get(temp, 0) + 1
  
    # Loop to find the frequencies
    # subsequence of length 2
    for i in range(n):
        for j in range(i + 1, n):
            temp = ""
            temp += s[i]
            temp += s[j]
            freq[temp] = freq.get(temp, 0) + 1
  
    answer = -10**9
  
    # Finding maximum frequency
    for it in freq:
        answer = max(answer, freq[it])
    return answer
  
# Driver Code
if __name__ == '__main__':
    s = "xxxyy"
  
    print(maximumOccurrence(s))
  
# This code is contributed by mohit kumar 29

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Output:

6

Time Complexity: O(N2)

Efficient Approach: The idea is to use the dynamic programming paradigm to compute the frequency of the subsequences of the length 1 and 2 in the string. Below is the illustration of the steps:

  • Compute the frequency of the characters of the string in a frequency array.
  • For subsequences of the string of length 2, the DP state will be
    dp[i][j] = Total number of times ith
      character occured before jth character.
    

Below is the implementation of the above approach:

C++

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// C++ implementation to find the 
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
  
#include <bits/stdc++.h>
  
using namespace std;
  
// Function to find the 
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
int maximumOccurrence(string s)
{
    int n = s.length();
  
    // Frequency for characters
    int freq[26] = { 0 };
    int dp[26][26] = { 0 };
      
    // Loop to count the occurence
    // of ith character before jth
    // character in the given string
    for (int i = 0; i < n; i++) {
        int c = (s[i] - 'a');
  
        for (int j = 0; j < 26; j++)
            dp[j] += freq[j];
  
        // Increase the frequency 
        // of s[i] or c of string
        freq++;
    }
  
    int answer = INT_MIN;
      
    // Maximum occurence of subsequence
    // of length 1 in given string
    for (int i = 0; i < 26; i++)
        answer = max(answer, freq[i]);
          
    // Maximum occurence of subsequence
    // of length 2 in given string
    for (int i = 0; i < 26; i++) {
        for (int j = 0; j < 26; j++) {
            answer = max(answer, dp[i][j]);
        }
    }
  
    return answer;
}
  
// Driver Code
int main()
{
    string s = "xxxyy";
  
    cout << maximumOccurrence(s);
    return 0;
}

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Java

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// Java implementation to find the 
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
  
  
class GFG{
   
// Function to find the 
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
static int maximumOccurrence(String s)
{
    int n = s.length();
   
    // Frequency for characters
    int freq[] = new int[26];
    int dp[][] = new int[26][26];
       
    // Loop to count the occurence
    // of ith character before jth
    // character in the given String
    for (int i = 0; i < n; i++) {
        int c = (s.charAt(i) - 'a');
   
        for (int j = 0; j < 26; j++)
            dp[j] += freq[j];
   
        // Increase the frequency 
        // of s[i] or c of String
        freq++;
    }
   
    int answer = Integer.MIN_VALUE;
       
    // Maximum occurence of subsequence
    // of length 1 in given String
    for (int i = 0; i < 26; i++)
        answer = Math.max(answer, freq[i]);
           
    // Maximum occurence of subsequence
    // of length 2 in given String
    for (int i = 0; i < 26; i++) {
        for (int j = 0; j < 26; j++) {
            answer = Math.max(answer, dp[i][j]);
        }
    }
   
    return answer;
}
   
// Driver Code
public static void main(String[] args)
{
    String s = "xxxyy";
   
    System.out.print(maximumOccurrence(s));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

6

Time complexity: O(26 * N)

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