Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

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

  • Difficulty Level : Hard
  • Last Updated : 25 Oct, 2021

Given a string S, the task is to count the maximum occurrence of subsequences in the given string such that the indices of the characters of the subsequence are 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 collective 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 every two integers will always form an arithmetic progression. Below is an 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 lengths 1 and 2.

Below is the implementation of the above approach:

C++




// C++ implementation to find the
// maximum occurrence 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 occurrence 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;
}

Java




// Java implementation to find the
// maximum occurrence of the subsequence
// such that the indices of characters
// are in arithmetic progression
import java.util.*;
 
class GFG
{
    // Function to find the
    // maximum occurrence 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

Python3




# Python3 implementation to find the
# maximum occurrence of the subsequence
# such that the indices of characters
# are in arithmetic progression
 
# Function to find the
# maximum occurrence 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

C#




// C# implementation to find the
// maximum occurrence 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 occurrence 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

Javascript




<script>
 
// Javascript implementation to find the
// maximum occurrence of the subsequence
// such that the indices of characters
// are in arithmetic progression
 
// Function to find the
// maximum occurrence of the subsequence
// such that the indices of characters
// are in arithmetic progression
function maximumOccurrence(s)
{
    var n = s.length;
 
    // Frequencies of subsequence
    var freq = new Map();
 
    // Loop to find the frequencies
    // of subsequence of length 1
    for (var i = 0; i < n; i++) {
        var temp = "";
        temp += s[i];
        if(freq.has(temp))
                freq.set(temp, freq.get(temp)+1)
            else
                freq.set(temp, 1)
    }
     
    // Loop to find the frequencies
    // subsequence of length 2
    for (var i = 0; i < n; i++) {
        for (var j = i + 1; j < n; j++) {
            var temp = "";
            temp += s[i];
            temp += s[j];
 
            if(freq.has(temp))
                freq.set(temp, freq.get(temp)+1)
            else
                freq.set(temp, 1)
        }
    }
 
    var answer = -1000000000;
 
    // Finding maximum frequency
    freq.forEach((value, key) => {
        answer = Math.max(answer, value);
    });
    return answer;
}
 
// Driver Code
var s = "xxxyy";
document.write( maximumOccurrence(s));
 
 
</script>
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 lengths 1 and 2 in the string. Below is an 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++




// C++ implementation to find the
// maximum occurrence 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 occurrence 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 occurrence
    // 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 occurrence of subsequence
    // of length 1 in given string
    for (int i = 0; i < 26; i++)
        answer = max(answer, freq[i]);
         
    // Maximum occurrence 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;
}

Java




// Java implementation to find the
// maximum occurrence of the subsequence
// such that the indices of characters
// are in arithmetic progression
 
 
class GFG{
  
// Function to find the
// maximum occurrence 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 occurrence
    // 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 occurrence of subsequence
    // of length 1 in given String
    for (int i = 0; i < 26; i++)
        answer = Math.max(answer, freq[i]);
          
    // Maximum occurrence 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

Python3




# Python3 implementation to find the
# maximum occurrence of the subsequence
# such that the indices of characters
# are in arithmetic progression
import sys
 
# Function to find the maximum occurrence
# 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 occurrence
    # 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 occurrence of subsequence
    # of length 1 in given String
    for i in range(26):
        answer = max(answer, freq[i])
 
    # Maximum occurrence 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

C#




// C# implementation to find the
// maximum occurrence of the subsequence
// such that the indices of characters
// are in arithmetic progression
using System;
class GFG{
  
// Function to find the maximum
// occurrence 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 occurrence
    // 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 occurrence of subsequence
    // of length 1 in given String
    for(int i = 0; i < 26; i++)
       answer = Math.Max(answer, freq[i]);
              
    // Maximum occurrence 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

Javascript




<script>
 
// javascript implementation to find the
// maximum occurrence of the subsequence
// such that the indices of characters
// are in arithmetic progression
    // Function to find the
    // maximum occurrence of the subsequence
    // such that the indices of characters
    // are in arithmetic progression
    function maximumOccurrence(s) {
        var n = s.length;
 
        // Frequency for characters
        var freq = Array(26).fill(0);
        var dp = Array(26).fill().map(()=>Array(26).fill(0));
 
        // Loop to count the occurrence
        // of ith character before jth
        // character in the given String
        for (var i = 0; i < n; i++) {
            var c = (s.charCodeAt(i) - 'a'.charCodeAt(0));
 
            for (var j = 0; j < 26; j++)
                dp[j] += freq[j];
 
            // Increase the frequency
            // of s[i] or c of String
            freq++;
        }
 
        var answer = Number.MIN_VALUE;
 
        // Maximum occurrence of subsequence
        // of length 1 in given String
        for (var i = 0; i < 26; i++)
            answer = Math.max(answer, freq[i]);
 
        // Maximum occurrence of subsequence
        // of length 2 in given String
        for (var i = 0; i < 26; i++) {
            for (var j = 0; j < 26; j++) {
                answer = Math.max(answer, dp[i][j]);
            }
        }
 
        return answer;
    }
 
    // Driver Code
     
        var s = "xxxyy";
 
        document.write(maximumOccurrence(s));
 
// This code contributed by Princi Singh
</script>
Output: 
6

 

Time complexity: O(26 * N)
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :