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:
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:
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 a 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: 



Below is the implementation of the above approach:

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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 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 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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// C# implementation to find the
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
using System;
using System.Collections.Generic;
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
  Dictionary<string,
             int> freq = new Dictionary<string,
                                        int>();
  int i, j;
 
  // Loop to find the frequencies
  // of subsequence of length 1
  for ( i = 0; i < n; i++)
  {
    string temp = "";
    temp += s[i];
    if (freq.ContainsKey(temp))
    {
      freq[temp]++;
    }
    else
    {
      freq[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[i];
      temp += s[j];
      if(freq.ContainsKey(temp))
        freq[temp]++;
      else
        freq[temp] = 1;
    }
  }
  int answer =int.MinValue;
 
  // Finding maximum frequency
  foreach(KeyValuePair<string,
                       int> it in freq)
    answer = Math.Max(answer, it.Value);
  return answer;
}
       
// Driver Code
public static void Main(string []args)
{
  string s = "xxxyy";
  Console.Write(maximumOccurrence(s));
}
}
 
// This code is contributed by Rutvik_56
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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 length 1 and 2 in the string. Below is the illustration of the steps: 

dp[i][j] = Total number of times ith
  character occured before jth character.


Below is the implementation of the above approach:

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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 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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# Python3 implementation to find the
# maximum occurence of the subsequence
# such that the indices of characters
# are in arithmetic progression
import sys
 
# 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)
 
    # Frequency for characters
    freq = [0] * (26)
 
    dp = [[0 for i in range(26)]
             for j in range(26)]
 
    # Loop to count the occurence
    # of ith character before jth
    # character in the given String
    for i in range(n):
        c = (ord(s[i]) - ord('a'))
 
        for j in range(26):
            dp[j] += freq[j]
 
        # Increase the frequency
        # of s[i] or c of String
        freq += 1
 
    answer = -sys.maxsize
 
    # Maximum occurence of subsequence
    # of length 1 in given String
    for i in range(26):
        answer = max(answer, freq[i])
 
    # Maximum occurence of subsequence
    # of length 2 in given String
    for i in range(26):
        for j in range(26):
            answer = max(answer, dp[i][j])
 
    return answer
 
# Driver Code
if __name__ == '__main__':
     
    s = "xxxyy"
 
    print(maximumOccurrence(s))
 
# This code is contributed by Princi Singh
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// C# implementation to find the
// maximum occurence of the subsequence
// such that the indices of characters
// are in arithmetic progression
using System;
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 x = (s[i] - 'a');
        
       for(int j = 0; j < 26; j++)
          dp[x, j] += freq[j];
           
       // Increase the frequency
       // of s[i] or c of String
       freq[x]++;
    }
      
    int answer = int.MinValue;
          
    // 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";
      
    Console.Write(maximumOccurrence(s));
}
}
 
// This code is contributed by Yash_R
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Output: 
6




 

Time complexity: O(26 * N)
 

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