查询在 [L, R] 范围内计算具有奇数频率的字符

给定一个长度为N的字符串S ，由小写字母组成，以及形式为[L, R] 的查询Q[][ ]，任务是计算在范围 [L] 中出现奇数次的字符数，R]。

例子 ：

Input: S = “geeksforgeeks”, Q[][] = {{2, 4}, {0, 3}, {0, 12}} 
Output: 3 2 3 
Explanation: 
Characters occurring odd number of times in [2, 4] : {‘e’, ‘k’, ‘s’}. 
Characters occurring odd number of times in [0, 3] : {‘g’, ‘k’}. 
Characters occurring odd number of times in [0, 12 ] : {‘f’, ‘o’, ‘r’}.

Input: S = “hello”, Q[][] = {{0, 4}} 
Output: 3 
Explanation: Characters occurring odd number of times in [0, 4] : {‘h’, ‘e’, ‘o’}.

方法 ：

请按照以下步骤解决问题：

• 每个字符都可以用唯一的 2 次幂（按升序）表示。例如， 2 ⁰表示‘a’ ， 2 ¹表示‘b’等等，最多2 ²⁵表示‘z’ 。
• 初始化大小为N的数组arr[] ，其中arr[i]是S[i]的对应整数值。
• 构造一个大小为N的前缀数组prefix[] ，其中prefix[i]是对从arr[0]到arr[i] 的所有数字执行的XOR运算的值。
• {arr[L], arr[L + 1], …, arr[R – 1], arr[R]}的 XOR 值中的设置位数给出给定范围[L, R]所需的答案.

下面是上述方法的实现：

C++

// C++ Program to implement
// the above problem
#include 
using namespace std;
 
// Function to print the number
// of characters having odd
// frequencies for each query
void queryResult(int prefix[],
                 pair Q)
{
    int l = Q.first;
    int r = Q.second;
 
    if (l == 0) {
        int xorval = prefix[r];
        cout << __builtin_popcount(xorval)
             << endl;
    }
    else {
        int xorval = prefix[r]
                     ^ prefix[l - 1];
        cout << __builtin_popcount(xorval)
             << endl;
    }
}
 
// A function to construct
// the arr[] and prefix[]
void calculateCount(string S,
                    pair Q[],
                    int m)
{
    // Stores array length
    int n = S.length();
 
    // Stores the unique powers of 2
    // associated to each character
    int arr[n];
    for (int i = 0; i < n; i++) {
        arr[i] = (1 << (S[i] - 'a'));
    }
 
    // Prefix array to store the
    // XOR values from array elements
    int prefix[n];
    int x = 0;
    for (int i = 0; i < n; i++) {
        x ^= arr[i];
        prefix[i] = x;
    }
 
    for (int i = 0; i < m; i++) {
        queryResult(prefix, Q[i]);
    }
}
 
// Driver Code
int main()
{
    string S = "geeksforgeeks";
    pair Q[] = { { 2, 4 },
                           { 0, 3 },
                           { 0, 12 } };
 
    calculateCount(S, Q, 3);
}

Java

// Java Program to implement
// the above problem
import java.util.*;
class GFG{
    static class pair
    {
        int first, second;
        public pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
    }
   
    // Function to print the number
    // of characters having odd
    // frequencies for each query
    static void queryResult(int prefix[], pair Q)
    {
        int l = Q.first;
        int r = Q.second;
        if (l == 0)
        {
            int xorval = prefix[r];
            System.out.print(Integer.bitCount(xorval) + "\n");
        }
        else
        {
            int xorval = prefix[r] ^ prefix[l - 1];
            System.out.print(Integer.bitCount(xorval) + "\n");
        }
    }
 
    // A function to construct
    // the arr[] and prefix[]
    static void calculateCount(String S, pair Q[], int m)
    {
       
        // Stores array length
        int n = S.length();
 
        // Stores the unique powers of 2
        // associated to each character
        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
        {
            arr[i] = (1 << (S.charAt(i) - 'a'));
        }
 
        // Prefix array to store the
        // XOR values from array elements
        int[] prefix = new int[n];
        int x = 0;
        for (int i = 0; i < n; i++)
        {
            x ^= arr[i];
            prefix[i] = x;
        }
 
        for (int i = 0; i < m; i++)
        {
            queryResult(prefix, Q[i]);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "geeksforgeeks";
        pair Q[] = {new pair(2, 4),
                    new pair(0, 3), new pair(0, 12)};
        calculateCount(S, Q, 3);
    }
}
// This code is contributed by shikhasingrajput

Python3

# Python3 program to implement
# the above approach
 
# Function to print the number
# of characters having odd
# frequencies for each query
def queryResult(prefix, Q):
 
    l = Q[0]
    r = Q[1]
 
    if(l == 0):
        xorval = prefix[r]
        print(bin(xorval).count('1'))
 
    else:
        xorval = prefix[r] ^ prefix[l - 1]
        print(bin(xorval).count('1'))
 
# A function to construct
# the arr[] and prefix[]
def calculateCount(S, Q, m):
 
    # Stores array length
    n = len(S)
 
    # Stores the unique powers of 2
    # associated to each character
    arr = [0] * n
    for i in range(n):
        arr[i] = (1 << (ord(S[i]) - ord('a')))
 
    # Prefix array to store the
    # XOR values from array elements
    prefix = [0] * n
    x = 0
     
    for i in range(n):
        x ^= arr[i]
        prefix[i] = x
 
    for i in range(m):
        queryResult(prefix, Q[i])
 
# Driver Code
if __name__ == '__main__':
 
    S = "geeksforgeeks"
 
    # Function call
    Q = [ [ 2, 4 ],
          [ 0, 3 ],
          [ 0, 12 ] ]
 
    calculateCount(S, Q, 3)
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above problem
using System;
 
class GFG{
     
class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to print the number
// of characters having odd
// frequencies for each query
static void queryResult(int []prefix, pair Q)
{
    int l = Q.first;
    int r = Q.second;
     
    if (l == 0)
    {
        int xorval = prefix[r];
        Console.Write(countSetBits(xorval) + "\n");
    }
    else
    {
        int xorval = prefix[r] ^ prefix[l - 1];
        Console.Write(countSetBits(xorval) + "\n");
    }
}
 
// A function to construct
// the []arr and prefix[]
static void calculateCount(String S, pair []Q,
                           int m)
{
 
    // Stores array length
    int n = S.Length;
 
    // Stores the unique powers of 2
    // associated to each character
    int[] arr = new int[n];
    for(int i = 0; i < n; i++)
    {
        arr[i] = (1 << (S[i] - 'a'));
    }
 
    // Prefix array to store the
    // XOR values from array elements
    int[] prefix = new int[n];
    int x = 0;
     
    for(int i = 0; i < n; i++)
    {
        x ^= arr[i];
        prefix[i] = x;
    }
 
    for(int i = 0; i < m; i++)
    {
        queryResult(prefix, Q[i]);
    }
}
 
static int countSetBits(long x)
{
    int setBits = 0;
     
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}
 
// Driver Code
public static void Main(String[] args)
{
    String S = "geeksforgeeks";
    pair []Q = { new pair(2, 4),
                 new pair(0, 3),
                 new pair(0, 12) };
                  
    calculateCount(S, Q, 3);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

    // Javascript program to implement the above problem
     
    // Function to print the number
    // of characters having odd
    // frequencies for each query
    function queryResult(prefix, Q)
    {
        let l = Q[0];
        let r = Q[1];
 
        if (l == 0)
        {
            let xorval = prefix[r];
            document.write(countSetBits(xorval) + "</br>");
        }
        else
        {
            let xorval = prefix[r] ^ prefix[l - 1];
            document.write(countSetBits(xorval) + "</br>");
        }
    }
 
    // A function to construct
    // the []arr and prefix[]
    function calculateCount(S, Q, m)
    {
 
        // Stores array length
        let n = S.length;
 
        // Stores the unique powers of 2
        // associated to each character
        let arr = new Array(n);
        for(let i = 0; i < n; i++)
        {
            arr[i] = (1 << (S[i].charCodeAt() - 'a'.charCodeAt()));
        }
 
        // Prefix array to store the
        // XOR values from array elements
        let prefix = new Array(n);
        let x = 0;
 
        for(let i = 0; i < n; i++)
        {
            x ^= arr[i];
            prefix[i] = x;
        }
 
        for(let i = 0; i < m; i++)
        {
            queryResult(prefix, Q[i]);
        }
    }
 
    function countSetBits(x)
    {
        let setBits = 0;
 
        while (x != 0)
        {
            x = x & (x - 1);
            setBits++;
        }
        return setBits;
    }
     
    let S = "geeksforgeeks";
    let Q = [ [2, 4], [0, 3], [0, 12] ];
                    
    calculateCount(S, Q, 3);
 

输出：

3
2
3

时间复杂度： O(N + Q)
辅助空间： O(N)

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