使用单个子序列的每个索引部分最大化 3 长度回文子序列的计数

给定一个字符串S ，任务是从给定的字符串找到长度为 3 的不同索引回文子序列的最大数量。

例子：

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” .

方法：我们的想法是计数字符串S的每个字符的频率，和计数的频率对，使得对是相同的字符，并通过除以3字符串S计数长度为3的子序列的数目。最后，将频率对的最小值打印为子序列的数量。请按照以下步骤解决问题：

• 初始化一个数组，比如freq[] ，以存储字符串S的每个字符的频率。
• 初始化一个变量，比如freqPair ，以存储具有相同字符的频率对。
• 初始化一个变量，比如len ，以存储字符串S的长度为3的子序列的数量。
• 迭代范围[0, str.length() – 1] 。对于字符串S 的每个^第i^个索引，将字符freq[S[i] – ‘a’]的计数增加1 。
• 迭代范围[0, 26] 。对于数组freq[] 的每个^第i^个索引，通过将数组元素除以2 来计算频率对。
• 最后，打印freqPair和len的最小值。

下面是上述方法的实现：

C++

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

 
// 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
 

输出：

2

时间复杂度： O(|S| + 26)
辅助空间： O(26)

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