Queries to count characters having odd frequency in a range [L, R]

Given a string S of length N, consisting of lower case alphabets, and queries Q[][] of the form [L, R], the task is to count the number of characters appearing an odd number of times in the range [L, R].

Examples :

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:
Explanation: Characters occurring odd number of times in [0, 4] : {‘h’, ‘e’, ‘o’}.

Approach :



Follow the steps below to solve the problem:

  • Each character can be represented with a unique power of two (in ascending order). For example, 20 for ‘a’, 21 for ‘b’ and so on up to 225 for ‘z’.
  • Initialize an array arr[] of size N where arr[i] is the corresponding integer value of S[i].
  • Construct a prefix array prefix[] of size N where prefix[i] is the value of XOR operation performed on all the numbers from arr[0] to arr[i].
  • The number of set bits in the XOR value of {arr[L], arr[L + 1], …, arr[R – 1], arr[R]} gives the required answer for the given range [L, R].

Below is the implementation of the above approach :

C++14

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// C++ Program to implement
// the above problem
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the number
// of characters having odd
// frequencies for each query
void queryResult(int prefix[],
                 pair<int, int> 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<int, int> 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<int, int> Q[] = { { 2, 4 },
                           { 0, 3 },
                           { 0, 12 } };
  
    calculateCount(S, Q, 3);
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

3
2
3

Time Complexity: O(N + Q) 
Auxiliary Space: O(N)

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