计算给定字符串(其字谜是回文)的子字符串

给定长度为N的字符串S ，其中仅包含小写字母，任务是打印给定字符串(其字谜是回文)的子字符串的计数。

例子：

Input: S = “aaaa”
Output: 10
Explanation:
Possible substrings are {“a”, “a”, “a”, “a”, “aa”, “aa”, “aa”, “aaa”, “aaa”, “aaaa”}. Since all substrings are have palindromic anagrams, the required answer is 10.

Input: S = “abc”
Output: 3

天真的方法：想法是生成给定字符串的所有子字符串，并为每个子字符串检查其字谜是否是回文。不断增加发现上述条件成立的子字符串的数量。最后，打印所有这些子字符串的计数。
时间复杂度： O(N ³ )
辅助空间： O(N)

位掩码方法：这个想法是生成26位数字的掩码，因为有26个字符。另外，请注意，如果某个字符串的字谜是回文，则除最多一个字符外，其他字符的频率必须是偶数。
请按照以下步骤解决问题：

• 遍历字符串并在每个索引处初始化变量X = 0 。
• 从每个i^的索引，遍历字符串在一定范围的指数的[I，N – 1]，并且对于每个字符S [j]时，X的计算按位XOR和^{2(S [j]的- ‘A’)，}其中^第0位代表是否的频率是奇数，1^个位表示如果b的频率为奇数，等等。
• 然后，检查X＆(X – 1)是否等于0 。如果发现为真，则将计数增加1 。

Illustration: 
Suppose, X = (0001000)₂ 
=> (X – 1) will be represented as (0000111)₂. 
Therefore, X & (X – 1) = 0

• 完成上述步骤后，请打印获得的计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to print count of substrings
// whose anagrams are palindromic
void countSubstring(string s)
{
    // Stores the answer
    int res = 0;
 
    // Iterate over the string
    for (int i = 0;
         i < s.length(); i++) {
 
        int x = 0;
 
        for (int j = i;
             j < s.length(); j++) {
 
            // Set the current character
            int temp = 1 << s[j] - 'a';
 
            // Parity update
            x ^= temp;
            if ((x & (x - 1)) == 0)
                res++;
        }
    }
 
    // Print the final count
    cout << res;
}
 
// Driver Code
int main()
{
    string str = "aaa";
 
    // Function Call
    countSubstring(str);
 
    return 0;
}

Java

// Java program for
// the above approach
class GFG{
 
// Function to print count of subStrings
// whose anagrams are palindromic
static void countSubString(String s)
{
  // Stores the answer
  int res = 0;
 
  // Iterate over the String
  for (int i = 0; i < s.length(); i++)
  {
    int x = 0;
    for (int j = i; j < s.length(); j++)
    {
      // Set the current character
      int temp = 1 << s.charAt(j) - 'a';
 
      // Parity update
      x ^= temp;
      if ((x & (x - 1)) == 0)
        res++;
    }
  }
 
  // Print the final count
  System.out.print(res);
}
 
// Driver Code
public static void main(String[] args)
{
  String str = "aaa";
 
  // Function Call
  countSubString(str);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for
# the above approach
 
# Function to prcount of subStrings
# whose anagrams are palindromic
def countSubString(s):
   
    # Stores the answer
    res = 0;
 
    # Iterate over the String
    for i in range(len(s)):
        x = 0;
        for j in range(i, len(s)):
           
            # Set the current character
            temp = 1 << ord(s[j]) - ord('a');
 
            # Parity update
            x ^= temp;
            if ((x & (x - 1)) == 0):
                res += 1;
 
    # Print final count
    print(res);
 
# Driver Code
if __name__ == '__main__':
    str = "aaa";
 
    # Function Call
    countSubString(str);
 
# This code is contributed by 29AjayKumar

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function to print count of subStrings
// whose anagrams are palindromic
static void countSubString(String s)
{
  // Stores the answer
  int res = 0;
 
  // Iterate over the String
  for (int i = 0; i < s.Length; i++)
  {
    int x = 0;
 
    for (int j = i; j < s.Length; j++)
    {
      // Set the current character
      int temp = 1 << s[j] - 'a';
 
      // Parity update
      x ^= temp;
      if ((x & (x - 1)) == 0)
        res++;
    }
  }
 
  // Print the readonly count
  Console.Write(res);
}
 
// Driver Code
public static void Main(String[] args)
{
  String str = "aaa";
 
  // Function Call
  countSubString(str);
}
}
 
// This code is contributed by shikhasingrajput

输出

6

时间复杂度： O(N ² )
辅助空间： O(N)

高效方法：为了优化上述位屏蔽技术，我们的想法是使用映射。请按照以下步骤解决问题：

1. 初始化地图以存储蒙版的频率。初始化变量X = 0 。
2. 遍历字符串，对于每个^第i^个索引，计算X和2 ^{(S [j]} –’a ^‘)的按位异或，并通过在Map中添加X的当前值的频率来更新答案，因为如果有任何子字符串从0开始到j具有与X相同的掩码， X是从0到i的子字符串的掩码，然后子字符串S [j + 1，i]的偶数频率为j 。
3. 还要加上掩码X xor 2 ^k的频率，其中0≤k <26 。之后，将X的频率增加1 。
4. 对给定字符串的每个索引重复上述步骤。
5. 遍历给定的字符串，打印所需的答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to get the count of substrings
// whose anagrams are palindromic
void countSubstring(string s)
{
    // Store the answer
    int answer = 0;
 
    // Map to store the freq of masks
    unordered_map m;
 
    // Set frequency for mask
    // 00...00 to 1
    m[0] = 1;
 
    // Store mask in x from 0 to i
    int x = 0;
    for (int j = 0; j < s.length(); j++) {
        x ^= 1 << (s[j] - 'a');
 
        // Update answer
        answer += m[x];
        for (int i = 0; i < 26; ++i) {
            answer += m[x ^ (1 << i)];
        }
 
        // Update frequency
        m[x] += 1;
    }
 
    // Print the answer
    cout << answer;
}
 
// Driver Code
int main()
{
    string str = "abab";
 
    // Function Call
    countSubstring(str);
    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG {
 
// Function to get the count of subStrings
// whose anagrams are palindromic
static void countSubString(String s)
{
  // Store the answer
  int answer = 0;
 
  // Map to store the freq of masks
  HashMap m = new HashMap();
   
  // Set frequency for mask
  // 00...00 to 1
  m.put(0, 1);
 
  // Store mask in x from 0 to i
  int x = 0;
  for (int j = 0; j < s.length(); j++)
  {
    x ^= 1 << (s.charAt(j) - 'a');
 
    // Update answer
    answer += m.containsKey(x) ? m.get(x) : 0;
    for (int i = 0; i < 26; ++i)
    {
      answer += m.containsKey(x ^ (1 << i)) ?
                m.get(x ^ (1 << i)) : 0;
    }
 
    // Update frequency
    if (m.containsKey(x))
      m.put(x, m.get(x) + 1);
    else
      m.put(x, 1);
  }
 
  // Print the answer
  System.out.print(answer);
}
 
// Driver Code
public static void main(String[] args)
{
  String str = "abab";
 
  // Function Call
  countSubString(str);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
from collections import defaultdict
 
# Function to get the count of substrings
# whose anagrams are palindromic
def countSubstring(s):
 
    # Store the answer
    answer = 0
 
    # Map to store the freq of masks
    m = defaultdict(lambda : 0)
 
    # Set frequency for mask
    # 00...00 to 1
    m[0] = 1
 
    # Store mask in x from 0 to i
    x = 0
     
    for j in range(len(s)):
        x ^= 1 << (ord(s[j]) - ord('a'))
 
        # Update answer
        answer += m[x]
        for i in range(26):
            answer += m[x ^ (1 << i)]
 
        # Update frequency
        m[x] += 1
 
    # Print the answer
    print(answer)
 
# Driver Code
str = "abab"
 
# Function call
countSubstring(str)
 
# This code is contributed by Shivam Singh

C＃

// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to get the count of
// subStrings whose anagrams
// are palindromic
static void countSubString(String s)
{
  // Store the answer
  int answer = 0;
 
  // Map to store the freq of masks
  Dictionary m = new Dictionary();
   
  // Set frequency for mask
  // 00...00 to 1
  m.Add(0, 1);
 
  // Store mask in x from 0 to i
  int x = 0;
  for (int j = 0; j < s.Length; j++)
  {
    x ^= 1 << (s[j] - 'a');
 
    // Update answer
    answer += m.ContainsKey(x) ? m[x] : 0;
    for (int i = 0; i < 26; ++i)
    {
      answer += m.ContainsKey(x ^ (1 << i)) ?
                m[x ^ (1 << i)] : 0;
    }
 
    // Update frequency
    if (m.ContainsKey(x))
      m[x] = m[x] + 1;
    else
      m.Add(x, 1);
  }
 
  // Print the answer
  Console.Write(answer);
}
 
// Driver Code
public static void Main(String[] args)
{
  String str = "abab";
 
  // Function Call
  countSubString(str);
}
}
 
// This code is contributed by shikhasingrajput

输出

7

时间复杂度： O(N)
辅助空间： O(N)