每个字符出现偶数次的子串数

给定一个由N个小写字母组成的字符串S ，任务是计算每个字符频率为偶数的子字符串的数量。

例子：

Input: S = “abbaa”
Output: 4
Explanation:
The substrings having frequency of each character is even are {“abba”, “aa”, “bb”, “bbaa”}.
Therefore, the count is 4.

Input: S = “geeksforgeeks”
Output: 2

朴素方法：解决给定问题的最简单方法是生成给定字符串的所有可能子字符串 并计算每个字符具有偶数频率的那些子串。检查所有子字符串后，打印获得的总数。

下面是上述方法的实现：

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG
{
 
    // Function to count substrings having
    // even frequency of each character
    static int subString(String s, int n)
    {
 
        // Stores the total
        // count of substrings
        int count = 0;
 
        // Traverse the range [0, N]:
        for (int i = 0; i < n; i++) {
 
            // Traverse the range [i + 1, N]
            for (int len = i + 1; len <= n; len++) {
 
                // Stores the substring over
                // the range of indices [i, len]
                String test_str = s.substring(i, len);
 
                // Stores the frequency of characters
                HashMap res
                    = new HashMap<>();
 
                // Count frequency of each character
                for (char keys : test_str.toCharArray()) {
                    res.put(keys,
                            res.getOrDefault(keys, 0) + 1);
                }
 
                int flag = 0;
 
                // Traverse the dictionary
                for (char keys : res.keySet()) {
 
                    // If any of the keys
                    // have odd count
                    if (res.get(keys) % 2 != 0) {
 
                        flag = 1;
                        break;
                    }
                }
               
                // Otherwise
                if (flag == 0)
                    count += 1;
            }
        }
 
        // Return count
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "abbaa";
        int N = S.length();
        System.out.println(subString(S, N));
    }
}
 
// This code is contributed by Kingash.

Python3

# Python program for the above approach
 
# Function to count substrings having
# even frequency of each character
def subString(s, n):
 
    # Stores the total
    # count of substrings
    count = 0
 
    # Traverse the range [0, N]:
    for i in range(n):
       
        # Traverse the range [i + 1, N]
        for len in range(i + 1, n + 1):
           
            # Stores the substring over
            # the range of indices [i, len]
            test_str = (s[i: len])
 
            # Stores the frequency of characters
            res = {}
 
            # Count frequency of each character
            for keys in test_str:
                res[keys] = res.get(keys, 0) + 1
                 
            flag = 0
             
            # Traverse the dictionary
            for keys in res:
               
                # If any of the keys
                # have odd count
                if res[keys] % 2 != 0:
 
                    flag = 1
                    break
                     
            # Otherwise
            if flag == 0:
                count += 1
 
    # Return count
    return count
 
 
# Driver Code
 
S = "abbaa"
N = len(S)
print(subString(S, N))

Javascript

 
// JavaScript program for the above approach
// Function to count substrings having
// even frequency of each character
function subString(s, n)
{
     
    // Stores the total
    // count of substrings
    var count = 0;
     
    // Traverse the range [0, N]:
    for(var i = 0; i < n; i++)
    {
         
        // Traverse the range [i + 1, N]
        for(var len = i + 1; len <= n; len++)
        {
             
            // Stores the substring over
            // the range of indices [i, len]
            var test_str = s.substring(i, len);
             
            // Stores the frequency of characters
            var res = {};
             
            // Count frequency of each character
             
            var temp = test_str.split("");
             
            for(const keys of temp)
            {
                res[keys] = (res[keys] ? res[keys] : 0) + 1;
            }
             
            var flag = 0;
             
            // Traverse the dictionary
            for(const [key, value] of Object.entries(res))
            {
                 
                // If any of the keys
                // have odd count
                if (res[key] % 2 != 0)
                {
                    flag = 1;
                    break;
                }
            }
             
            // Otherwise
            if (flag == 0) count += 1;
        }
    }
     
    // Return count
    return count;
}
 
// Driver Code
var S = "abbaa";
var N = S.length;
 
document.write(subString(S, N));
 
// This code is contributed by rdtank
 

输出：

4

时间复杂度： O(N ² * 26)
辅助空间： O(N)

高效的方法：上述方法可以通过使用Bitmasking的概念进行优化 和 字典。请按照以下步骤解决问题：

• 初始化一个字典，比如hash来存储一个字符的计数。
• 初始化两个变量，比如count为0和pre为0以存储每个字符的偶数计数的子字符串的总数，并存储包含在子字符串中的字符的掩码。
• 遍历给定的字符串并执行以下步骤：
  • 翻转变量pre中的^第(S[i] – ‘a’)位。
  • 通过递增散列[预]的计数和预先在哈希计数。
• 完成以上步骤后，打印count的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count substrings having
// even frequency of each character
int subString(string s, int n)
{
 
    // Stores the count of a character
    map hash;
    hash[0] = 1;
 
    // Stores bitmask
    int pre = 0;
 
    // Stores the count of substrings
    // with even count of each character
    int count = 0;
 
    // Traverse the string S
    for (int i = 0; i < n; i++) {
 
        // Flip the ord(i)-97 bits in pre
        pre ^= (1 << int(s[i]) - 97);
 
        // Increment the count by hash[pre]
        count += hash[pre];
 
        // Increment count of pre in hash
        hash[pre] = hash[pre] + 1;
    }
 
    // Return the total count obtained
    return count;
}
 
// Driver Code
int main()
{
    string S = "abbaa";
    int N = S.length();
    cout << (subString(S, N));
}
 
// THIS CODE IS CONTRIBUTED BY UKASP.

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
     
// Function to count substrings having
// even frequency of each character
static int subString(String s, int n)
{
     
    // Stores the count of a character
    Map hash = new HashMap<>();
    hash.put(0, 1);
 
    // Stores bitmask
    int pre = 0;
 
    // Stores the count of substrings
    // with even count of each character
    int count = 0;
 
    // Traverse the string S
    for(int i = 0; i < n; i++)
    {
         
        // Flip the ord(i)-97 bits in pre
        pre ^= (1 << (int)(s.charAt(i) - 97));
 
        // Increment the count by hash[pre]
        count += hash.getOrDefault(pre, 0);
 
        // Increment count of pre in hash
        hash.put(pre, hash.getOrDefault(pre, 0) + 1);
    }
 
    // Return the total count obtained
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    String S = "abbaa";
    int N = S.length();
     
    System.out.print(subString(S, N));
}
}
 
// This code is contributed by offbeat

蟒蛇3

# Python program for the above approach
 
# Function to count substrings having
# even frequency of each character
def subString(s, n):
 
    # Stores the count of a character
    hash = {0: 1}
 
    # Stores bitmask
    pre = 0
 
    # Stores the count of substrings
    # with even count of each character
    count = 0
 
    # Traverse the string S
    for i in s:
       
        # Flip the ord(i)-97 bits in pre
        pre ^= (1 << ord(i) - 97)
 
        # Increment the count by hash[pre]
        count += hash.get(pre, 0)
 
        # Increment count of pre in hash
        hash[pre] = hash.get(pre, 0) + 1
         
    # Return the total count obtained
    return count
 
 
# Driver Code
 
S = "abbaa"
N = len(S)
print(subString(S, N))

C＃

// C# program for the above approach
using System.IO;
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count substrings having
// even frequency of each character
static int subString(string s, int n)
{
     
    // Stores the count of a character
    Dictionary hash = new Dictionary();
 
    hash[0] = 1;
 
    // Stores bitmask
    int pre = 0;
 
    // Stores the count of substrings
    // with even count of each character
    int count = 0;
 
    // Traverse the string S
    for(int i = 0; i < n; i++)
    {
         
        // Flip the ord(i)-97 bits in pre
        pre ^= (1 << (int)(s[i]) - 97);
 
        // Increment the count by hash[pre]
        if (hash.ContainsKey(pre))
            count += hash[pre];
        else
            count += 0;
 
        // Increment count of pre in hash
        if (hash.ContainsKey(pre))
            hash[pre] = hash[pre] + 1;
        else
            hash.Add(pre, 1);
    }
 
    // Return the total count obtained
    return count;
}
 
// Driver code
static void Main()
{
    String S = "abbaa";
    int N = S.Length;
     
    Console.WriteLine(subString(S, N));
}
}
 
// This code is contributed by sk944795

输出：

4

时间复杂度： O(N)
辅助空间： O(N)

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