以排序形式查找字符串的回文子字符串的计数

给定一个由小写英文字母组成的字符串str ，任务是找出以str的排序形式存在的回文子字符串的总数。
例子：

Input: str = “acbbd”
Output: 6
All palindromic sub-string in it’s sorted form (“abbcd”) are “a”, “b”, “b”, “bb”, “c” and “d”.
Input: str = “abbabdbd”
Output: 16

编程需要懂一点英语

幼稚的方法：一种方法是对给定的字符串进行排序，然后计算存在的回文子字符串的总数。为了找到回文子串的数量，可以使用这种方法，其时间复杂度为 O(n^2)。
优化方法：一种有效的方法是计算每个字符的频率，然后对于每个频率，回文数的总数将为(n*(n+1))/2 ，因为已排序字符串将由相同的字符。
例如，字符串“aabbbcd”的回文子字符串将是“a”、“aa”、...、“bbb”、“c”、...等。这种方法的时间复杂度将为 O(n)。

创建一个哈希表来存储字符串str的每个字符的频率。
遍历哈希表并为每个非零频率添加(hash[i] * (hash[i]+1)) / 2到sum 。
最后打印总和。

下面是上述方法的实现：

C++

// CPP program to find the count of palindromic sub-string
// of a string in it's ascending form
#include 
using namespace std;
 
const int MAX_CHAR = 26;
 
// function to return count of palindromic sub-string
int countPalindrome(string str)
{
    int n = str.size();
    int sum = 0;
 
    // calculate frequency
    int hashTable[MAX_CHAR];
    for (int i = 0; i < n; i++)
        hashTable[str[i] - 'a']++;
 
    // calculate count of palindromic sub-string
    for (int i = 0; i < 26; i++) {
        if (hashTable[i])
            sum += (hashTable[i] * (hashTable[i] + 1) / 2);
    }
 
    // return result
    return sum;
}
 
// driver program
int main()
{
    string str = "ananananddd";
 
    cout << countPalindrome(str);
    return 0;
}

Java

// Java program to find the count of palindromic sub-string
// of a string in it's ascending form
 
class GFG {
 
    final static int MAX_CHAR = 26;
 
// function to return count of palindromic sub-string
    static int countPalindrome(String str) {
        int n = str.length();
        int sum = 0;
 
        // calculate frequency
        int hashTable[] = new int[MAX_CHAR];
        for (int i = 0; i < n; i++) {
            hashTable[str.charAt(i) - 'a']++;
        }
 
        // calculate count of palindromic sub-string
        for (int i = 0; i < 26; i++) {
            if (hashTable[i] != 0) {
                sum += (hashTable[i] * (hashTable[i] + 1) / 2);
            }
        }
 
        // return result
        return sum;
    }
 
// driver program
    public static void main(String[] args) {
        String str = "ananananddd";
 
        System.out.println(countPalindrome(str));
 
    }
}

Python3

# Python3 program to find the count of
# palindromic sub-string of a string
# in it's ascending form
MAX_CHAR = 26
 
# function to return count of
# palindromic sub-string
def countPalindrome(str):
 
    n = len (str)
    sum = 0
 
    # calculate frequency
    hashTable = [0] * MAX_CHAR
    for i in range(n):
        hashTable[ord(str[i]) -
                  ord('a')] += 1
 
    # calculate count of palindromic
    # sub-string
    for i in range(26) :
        if (hashTable[i]):
            sum += (hashTable[i] *
                   (hashTable[i] + 1) // 2)
 
    # return result
    return sum
 
# Driver Code
if __name__ == "__main__":
 
    str = "ananananddd"
 
    print (countPalindrome(str))
 
# This code is contributed by ita_c

C#

// C# program to find the count of palindromic sub-string
// of a string in it's ascending form
using System;
                     
public class GFG{
  
    readonly static int MAX_CHAR = 26;
  
// function to return count of palindromic sub-string
    static int countPalindrome(String str) {
        int n = str.Length;
        int sum = 0;
  
        // calculate frequency
        int []hashTable = new int[MAX_CHAR];
        for (int i = 0; i < n; i++) {
            hashTable[str[i] - 'a']++;
        }
  
        // calculate count of palindromic sub-string
        for (int i = 0; i < 26; i++) {
            if (hashTable[i] != 0) {
                sum += (hashTable[i] * (hashTable[i] + 1) / 2);
            }
        }
  
        // return result
        return sum;
    }
  
// driver program
    public static void Main() {
        String str = "ananananddd";
  
        Console.Write(countPalindrome(str));
  
    }
}
 
// This code is contributed by Rajput-Ji

PHP

Javascript

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