Maximize count of 3-length palindromic subsequences with each index part of a single subsequence

Given a string, S, the task is to find the maximum number of distinct indexed palindromic subsequences of length 3 possible from the given string.

Examples:

Input: str = “geekforg”
Output: 2
Explanation:Possible palindromic subsequences of length 3 satisfying the conditions are “gkg” and “efe”. Therefore, the required output is 2.

Input: str = “geek” 
Output: 1
Explanation: Possible palindromic subsequences of length 3 satisfying the conditions are “ege” .

Approach: The idea is to count the frequency of every character of the string S, and count the frequency pairs such that pairs are of the same characters and count the number of subsequences of length 3 by dividing the string S by 3. Finally, print the minimum of frequency pairs as the number of subsequences. Follow the steps below to solve the problem:

• Initialize an array, say freq[], to store the frequencies of every character of the string S.
• Initialize a variable, say freqPair, to store the frequency pairs having pairs of the same characters.
• Initialize a variable, say len, to store the number of subsequences of length 3 of the string S.
• Iterate over the range [0, str.length() – 1]. For every i^th index of the string S increment the count of the character freq[S[i] – ‘a’] by 1.
• Iterate over the range [0, 26]. For every i^th index of the array freq[], count the frequency pairs by dividing the array element by 2.
• Finally, print the value of the minimum of freqPair and len.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count the maximum number
// oaf palindrome subsequences of length 3
// considering the same index only once
int maxNumPalindrome(string S)
{

    // Index of the string S
    int i = 0;

    // Stores the frequency of
    // every character
    int freq[26] = { 0 };

    // Stores the pair of frequency
    // containing same characters
    int freqPair = 0;

    // Number of subsequences
    // having length 3
    int len = S.length() / 3;

    // Counts the frequency
    while (i < S.length()) {

        freq[S[i] - 'a']++;
        i++;
    }

    // Counts the pair of frequency
    for (i = 0; i < 26; i++) {

        freqPair += (freq[i] / 2);
    }

    // Returns the minimum value
    return min(freqPair, len);
}

// Driver Code
int main()
{

    string S = "geeksforg";

    cout << maxNumPalindrome(S) << endl;

    return 0;
}

Java

// Java program to implement
// the above approach

import java.util.*;

class GFG {

    // Driver Code
    public static void main(String[] args)
    {

        String S = "geeksforg";

        System.out.println(maxNumPalindrome(S));
    }

    // Function to count the maximum number
    // of palindrome subsequences of length 3
    // considering the same index only once
    static int maxNumPalindrome(String S)
    {

        // Index of the string S
        int i = 0;

        // Stores the frequency of
        // every character
        int[] freq = new int[26];

        // Stores the pair of frequency
        // containing same characters
        int freqPair = 0;

        // Number of subsequences
        // having length 3
        int len = S.length() / 3;

        // Counts the frequency
        while (i < S.length()) {

            freq[S.charAt(i) - 'a']++;
            i++;
        }

        // Counts the pair of frequency
        for (i = 0; i < 26; i++) {

            freqPair += (freq[i] / 2);
        }

        // Returns the minimum value
        return Math.min(freqPair, len);
    }
}

Python3

# Python3 program to implement
# the above approach

# Function to count the maximum number
# of palindrome subsequences of length 3
# considering the same index only once
def maxNumPalindrome(S):

    # Index of the S
    i = 0

    # Stores the frequency of
    # every character
    freq = [0] * 26

    # Stores the pair of frequency
    # containing same characters
    freqPair = 0

    # Number of subsequences
    # having length 3
    ln = len(S) // 3

    # Counts the frequency
    while (i < len(S)):
        freq[ord(S[i]) - ord('a')] += 1
        i += 1

    # Counts the pair of frequency
    for i in range(26):
        freqPair += (freq[i] // 2)

    # Returns the minimum value
    return min(freqPair, ln)

# Driver Code
if __name__ == '__main__':

    S = "geeksforg"

    print(maxNumPalindrome(S))

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG
{

    // Driver Code
    public static void Main(String[] args)
    {
        string S = "geeksforg";
        Console.WriteLine(maxNumPalindrome(S));
    }

    // Function to count the maximum number
    // of palindrome subsequences of length 3
    // considering the same index only once
    static int maxNumPalindrome(string S)
    {

        // Index of the string S
        int i = 0;

        // Stores the frequency of
        // every character
        int[] freq = new int[26];

        // Stores the pair of frequency
        // containing same characters
        int freqPair = 0;

        // Number of subsequences
        // having length 3
        int len = S.Length / 3;

        // Counts the frequency
        while (i < S.Length)
        {
            freq[S[i] - 'a']++;
            i++;
        }

        // Counts the pair of frequency
        for (i = 0; i < 26; i++)
        {
            freqPair += (freq[i] / 2);
        }

        // Returns the minimum value
        return Math.Min(freqPair, len);
    }
}

// This code is contributed by susmitakundugoaldanga.

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to count the maximum number
// oaf palindrome subsequences of length 3
// considering the same index only once
function maxNumPalindrome(S)
{

    // Index of the string S
    let i = 0;

    // Stores the frequency of
    // every character
    let freq = new Array(26).fill(0);

    // Stores the pair of frequency
    // containing same characters
    let freqPair = 0;

    // Number of subsequences
    // having length 3
    let len = (S.length / 3);

    // Counts the frequency
    while (i < S.length)
    {
        freq[S[i].charCodeAt(0) -
              'a'.charCodeAt(0)]++;
        i++;
    }

    // Counts the pair of frequency
    for(i = 0; i < 26; i++)
    {
        freqPair += Math.floor(freq[i] / 2);
    }

    // Returns the minimum value
    return Math.min(freqPair, len);
}

// Driver Code
let S = "geeksforg";

document.write(maxNumPalindrome(S) + "<br>");

// This code is contributed by gfgking

</script>

Output:

2

Time Complexity: O(|S| + 26)
Auxiliary Space: O(26)