最多一个字符的频率为奇数的子串计数

给定一个包含N个字符的字符串S ，任务是计算非空子字符串的总数，使得最多一个字符出现奇数次。

示例：

Input: S = “aba”
Output: 4
Explanation: The valid substrings are “a”, “b”, “a”, and “aba”. Therefore, the total number of required substrings are 4.

Input: “aabb”
Output: 9
Explanation: The valid substrings are “a”, “aa”, “aab”, “aabb”, “a”, “abb”, “b”, “bb”, and “b”.

编程需要懂一点英语

方法：上述问题可以在Bit Masking using HashMaps的帮助下解决。请按照以下步骤解决问题：

每个字符频率的奇偶校验可以存储在位掩码掩码中，其中第i^个字符由2 ⁱ表示。最初mask = 0的值。
创建一个无序的 map seen ，它存储每个位掩码的出现频率。最初， seen[0] = 1的值。
创建一个变量cnt ，它存储有效子字符串的计数。最初， cnt = 0的值。
迭代 [0, N) 范围内的每个i并将掩码的值与表示字符串的第 i^个字符的整数按位异或，并通过seen[mask]增加cnt的值。
对于每个有效的i ，遍历[a, z]范围内的所有字符并通过翻转当前掩码中的第 j^个设置位来增加其频率，并在翻转第j^个位掩码后将cnt的值增加位掩码的频率设置位。
存储在cnt中的值是所需的答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the count of substrings
// such that at most one character occurs
// odd number of times
int uniqueSubstrings(string S)
{
    // Stores the frequency of the bitmasks
    unordered_map seen;
 
    // Initial Condition
    seen[0] = 1;
 
    // Store the current value of the bitmask
    int mask = 0;
 
    // Stores the total count of the
    // valid substrings
    int cnt = 0;
 
    for (int i = 0; i < S.length(); ++i) {
 
        // XOR the mask with current character
        mask ^= (1 << (S[i] - 'a'));
 
        // Increment the count by mask count
        // of strings with all even frequencies
        cnt += seen[mask];
 
        for (int j = 0; j < 26; ++j) {
            // Increment count by mask count
            // of strings if exist with the
            // jth character having odd frequency
            cnt += seen[mask ^ (1 << j)];
        }
        seen[mask]++;
    }
 
    // Return Answer
    return cnt;
}
 
// Driver Code
int main()
{
    string word = "aabb";
    cout << uniqueSubstrings(word);
 
    return 0;
}

Python3

# Python program for the above approach
 
# Function to find the count of substrings
# such that at most one character occurs
# odd number of times
def uniqueSubstrings(S):
 
    # Stores the frequency of the bitmasks
    seen = {}
 
    # Initial Condition
    seen[0] = 1
 
    # Store the current value of the bitmask
    mask = 0
 
    # Stores the total count of the
    # valid substrings
    cnt = 0
 
    for i in range(len(S)):
 
        # XOR the mask with current character
        mask ^= (1 << (ord(S[i]) - ord('a')))
 
        # Increment the count by mask count
        # of strings with all even frequencies
        if mask in seen:
            cnt += seen[mask]
        else:
            cnt += 0
 
        for j in range(26):
            # Increment count by mask count
            # of strings if exist with the
            # jth character having odd frequency
            if mask ^ (1 << j) in seen:
                cnt += seen[mask ^ (1 << j)]
            else:
                cnt += 0
        if mask in seen:
            seen[mask] += 1
        else:
            seen[mask] = 1
 
    # Return Answer
    return cnt
 
# Driver Code
word = "aabb"
print(uniqueSubstrings(word))
 
# This code is contributed by rj13to.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the count of substrings
// such that at most one character occurs
// odd number of times
static int uniqueSubstrings(string S)
{
     
    // Stores the frequency of the bitmasks
    Dictionary seen = new Dictionary();
 
    // Initial Condition
    seen[0] = 1;
 
    // Store the current value of the bitmask
    int mask = 0;
 
    // Stores the total count of the
    // valid substrings
    int cnt = 0;
 
    for(int i = 0; i < S.Length; ++i)
    {
         
        // XOR the mask with current character
        mask ^= (1 << (S[i] - 'a'));
 
        // Increment the count by mask count
        // of strings with all even frequencies
        if (seen.ContainsKey(mask))
            cnt += seen[mask];
 
        for(int j = 0; j < 26; ++j)
        {
             
            // Increment count by mask count
            // of strings if exist with the
            // jth character having odd frequency
            if (seen.ContainsKey(mask ^ (1 << j)))
                cnt += seen[mask ^ (1 << j)];
        }
        if (seen.ContainsKey(mask))
            seen[mask]++;
        else
            seen[mask] = 1;
    }
 
    // Return Answer
    return cnt;
}
 
// Driver Code
public static void Main()
{
    string word = "aabb";
     
    Console.WriteLine(uniqueSubstrings(word));
}
}
 
// This code is contributed by ukasp

输出：

时间复杂度：O(N)
辅助空间：O(N)