Queries on XOR of greatest odd divisor of the range

Given an array of N positive integers. There are Q queries, each include a range [L, R]. For each query output the xor of greatest odd divisor of each number in that range.

Examples:

Input : arr[] = { 3, 4, 5 }
query 1: [0, 2]
query 2: [1, 2]
Output : 7 4
Greatest odd divisor are: { 3, 1, 5 }
XOR of 3, 1, 5 is 7
XOR of 1, 5 is 4

Input : arr[] = { 2, 1, 2 }
query 1: [0, 2]
Output : 1

The idea is to precompute the greatest odd divisor of the array and store it in an array, say preXOR[]. Now, precompute and store prefix XOR of the array preXOR[]. To answer each query, return (preXOR[r] xor preXOR[l-1]).

Below is the implementation of this approach: 

C++

#include <bits/stdc++.h>
using namespace std;

// Precompute the prefix XOR of greatest
// odd divisor
void prefixXOR(int arr[], int preXOR[], int n)
{
    // Finding the Greatest Odd divisor
    for (int i = 0; i < n; i++) {
        while (arr[i] % 2 != 1)
            arr[i] /= 2;

        preXOR[i] = arr[i];
    }

    // Finding prefix XOR
    for (int i = 1; i < n; i++)
        preXOR[i] = preXOR[i - 1] ^ preXOR[i];
}

// Return XOR of the range
int query(int preXOR[], int l, int r)
{
    if (l == 0)
        return preXOR[r];
    else
        return preXOR[r] ^ preXOR[l - 1];
}

// Driven Program
int main()
{
    int arr[] = { 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);

    int preXOR[n];
    prefixXOR(arr, preXOR, n);

    cout << query(preXOR, 0, 2) << endl;
    cout << query(preXOR, 1, 2) << endl;

    return 0;
}

Java

// Java code Queries on XOR of
// greatest odd divisor of the range
import java.io.*;

class GFG
{
    // Precompute the prefix XOR of greatest
    // odd divisor
    static void prefixXOR(int arr[], int preXOR[], int n)
    {
        // Finding the Greatest Odd divisor
        for (int i = 0; i < n; i++)
        {
            while (arr[i] % 2 != 1)
                arr[i] /= 2;

            preXOR[i] = arr[i];
        }

        // Finding prefix XOR
        for (int i = 1; i < n; i++)
            preXOR[i] = preXOR[i - 1] ^ preXOR[i];
    }

    // Return XOR of the range
    static int query(int preXOR[], int l, int r)
    {
        if (l == 0)
            return preXOR[r];
        else
            return preXOR[r] ^ preXOR[l - 1];
    }

    // Driven Program
    public static void main (String[] args)
    {
        int arr[] = { 3, 4, 5 };
        int n = arr.length;

        int preXOR[] = new int[n];
        prefixXOR(arr, preXOR, n);

        System.out.println(query(preXOR, 0, 2)) ;
        System.out.println (query(preXOR, 1, 2));


    }
}

// This article is contributed by vt_m

Python3

# Precompute the prefix XOR of greatest
# odd divisor
def prefixXOR(arr, preXOR, n):

    # Finding the Greatest Odd divisor
    for i in range(0, n, 1):
        while (arr[i] % 2 != 1):
            arr[i] = int(arr[i] / 2)

        preXOR[i] = arr[i]

    # Finding prefix XOR
    for i in range(1, n, 1):
        preXOR[i] = preXOR[i - 1] ^ preXOR[i]

# Return XOR of the range
def query(preXOR, l, r):
    if (l == 0):
        return preXOR[r]
    else:
        return preXOR[r] ^ preXOR[l - 1]

# Driver Code
if __name__ == '__main__':
    arr = [3, 4, 5]
    n = len(arr)

    preXOR = [0 for i in range(n)]
    prefixXOR(arr, preXOR, n)

    print(query(preXOR, 0, 2))
    print(query(preXOR, 1, 2))

# This code is contributed by
# Sahil_shelangia

C#

// C# code Queries on XOR of
// greatest odd divisor of the range
using System;

class GFG
{
    // Precompute the prefix XOR of greatest
    // odd divisor
    static void prefixXOR(int []arr,
                    int []preXOR, int n)
    {
        // Finding the Greatest Odd divisor
        for (int i = 0; i < n; i++)
        {
            while (arr[i] % 2 != 1)
                arr[i] /= 2;

            preXOR[i] = arr[i];
        }

        // Finding prefix XOR
        for (int i = 1; i < n; i++)
            preXOR[i] = preXOR[i - 1] ^ preXOR[i];
    }

    // Return XOR of the range
    static int query(int [] preXOR, int l, int r)
    {
        if (l == 0)
            return preXOR[r];
        else
            return preXOR[r] ^ preXOR[l - 1];
    }

    // Driven Program
    public static void Main ()
    {
        int []arr = { 3, 4, 5 };
        int n = arr.Length;

        int []preXOR = new int[n];
        prefixXOR(arr, preXOR, n);

        Console.WriteLine(query(preXOR, 0, 2)) ;
        Console.WriteLine (query(preXOR, 1, 2));
    }
}

// This code is contributed by vt_m

Javascript

<script>

// Javascript code queries on XOR of
// greatest odd divisor of the range

// Precompute the prefix XOR of greatest
// odd divisor
function prefixXOR(arr, preXOR, n)
{

    // Finding the Greatest Odd divisor
    for(let i = 0; i < n; i++)
    {
        while (arr[i] % 2 != 1)
            arr[i] = parseInt(arr[i] / 2);

        preXOR[i] = arr[i];
    }

    // Finding prefix XOR
    for(let i = 1; i < n; i++)
        preXOR[i] = preXOR[i - 1] ^ preXOR[i];
}

// Return XOR of the range
function query(preXOR, l, r)
{
    if (l == 0)
        return preXOR[r];
    else
        return preXOR[r] ^ preXOR[l - 1];
}

// Driver code
let arr = [ 3, 4, 5 ];
let n = arr.length;
let preXOR = new Array(n);

prefixXOR(arr, preXOR, n);

document.write(query(preXOR, 0, 2) + "<br>");
document.write(query(preXOR, 1, 2) + "<br>");

// This code is contributed by subham348

</script>

Output

7
4

Time Complexity: O(n*log(n))
Auxiliary Space: O(n)

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.