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XOR of all the elements in the given range [L, R]
  • Difficulty Level : Medium
  • Last Updated : 03 May, 2021

Given a range [L, R], the task is to find the XOR of all the integers in the given range i.e. (L) ^ (L + 1) ^ (L + 2) ^ … ^ (R)
Examples: 
 

Input: L = 1, R = 4 
Output:
1 ^ 2 ^ 3 ^ 4 = 4
Input: L = 3, R = 9 
Output:
 

 

A simple solution is to find XOR of all the numbers iteratively from L to R. This will take linear time.
A better solution is to first find the most significant bit in the integer R. Our answer cannot have it’s a most significant bit larger than that of ‘R’. For each bit ‘i’ between 0 and MSB inclusive, we will try to determine the parity of count of a number of integers between L and R inclusive such that there ‘ith‘ bit is set. If the count is odd, then the ‘ith‘ bit of our final answer will also be set. 
Now the real question is, for ith bit, how do we determine the parity of the count? 
For an idea, let’s look at the binary representation of the first 16 integers. 
 

0: 0000 
1: 0001 
2: 0010 
3: 0011 
4: 0100 
5: 0101 
6: 0110 
7: 0111 
8: 1000 
9: 1001 
10: 1010 
11: 1011 
12: 1100 
13: 1101 
14: 1110 
15: 1111 
 



It’s easily noticeable that the state of ith bit changes after every 2i number. We will use this idea to predict the count of a number of integers with ith bit set in the range L to R inclusive.
We have two cases here: 
 

  1. Case 1(i != 0): We try to determine whether the ith bit of L is set or not. If set, we try to find the parity of the count of numbers between L and L + 2i inclusive, such that their ith bit is set. If ith bit of L is set and L is odd, then this count will be odd else even.
    Similarly for R, we try to determine the parity of the count of a number of elements between R – 2i and R, such that their ith bit is set. If ith bit of L is set and L is even, then this count will be odd else even.
    We ignore all the other integers in between because they will have even number of integers with ith bit set.
  2. Case 2(i = 0): Here, we have the following cases: 
    • If L and R, both are odd, the count of the number of integers with 0th bit set will be (R – L)/2 + 1
    • In any other case, the count will be floor((R – L + 1)/2).

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the
// most significant bit
int msb(int x)
{
    int ret = 0;
    while ((x >> (ret + 1)) != 0)
        ret++;
    return ret;
}
 
// Function to return the required XOR
int xorRange(int l, int r)
{
 
    // Finding the MSB
    int max_bit = msb(r);
 
    // Value of the current bit to be added
    int mul = 2;
 
    // To store the final answer
    int ans = 0;
 
    // Loop for case 1
    for (int i = 1; i <= max_bit; i++) {
 
        // Edge case when both the integers
        // lie in the same segment of continuous
        // 1s
        if ((l / mul) * mul == (r / mul) * mul) {
            if (((l & (1 << i)) != 0) && (r - l + 1) % 2 == 1)
                ans += mul;
            mul *= 2;
            continue;
        }
 
        // To store whether parity of count is odd
        bool odd_c = 0;
 
        if (((l & (1 << i)) != 0) && l % 2 == 1)
            odd_c = (odd_c ^ 1);
        if (((r & (1 << i)) != 0) && r % 2 == 0)
            odd_c = (odd_c ^ 1);
 
        // Updating the answer if parity is odd
        if (odd_c)
            ans += mul;
 
        // Updating the number to be added
        mul *= 2;
    }
 
    // Case 2
    int zero_bit_cnt = zero_bit_cnt = (r - l + 1) / 2;
 
    if (l % 2 == 1 && r % 2 == 1)
        zero_bit_cnt++;
 
    if (zero_bit_cnt % 2 == 1)
        ans++;
 
    return ans;
}
 
// Driver code
int main()
{
    int l = 1, r = 4;
 
    // Final answer
    cout << xorRange(l, r);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the
// most significant bit
static int msb(int x)
{
    int ret = 0;
    while ((x >> (ret + 1)) != 0)
        ret++;
    return ret;
}
 
// Function to return the required XOR
static int xorRange(int l, int r)
{
 
    // Finding the MSB
    int max_bit = msb(r);
 
    // Value of the current bit to be added
    int mul = 2;
 
    // To store the final answer
    int ans = 0;
 
    // Loop for case 1
    for (int i = 1; i <= max_bit; i++)
    {
 
        // Edge case when both the integers
        // lie in the same segment of continuous
        // 1s
        if ((l / mul) * mul == (r / mul) * mul)
        {
            if (((l & (1 << i)) != 0) && (r - l + 1) % 2 == 1)
                ans += mul;
            mul *= 2;
            continue;
        }
 
        // To store whether parity of count is odd
        int odd_c = 0;
 
        if (((l & (1 << i)) != 0) && l % 2 == 1)
            odd_c = (odd_c ^ 1);
        if (((r & (1 << i)) != 0) && r % 2 == 0)
            odd_c = (odd_c ^ 1);
 
        // Updating the answer if parity is odd
        if (odd_c!=0)
            ans += mul;
 
        // Updating the number to be added
        mul *= 2;
    }
 
    // Case 2
    int zero_bit_cnt = zero_bit_cnt = (r - l + 1) / 2;
 
    if (l % 2 == 1 && r % 2 == 1)
        zero_bit_cnt++;
 
    if (zero_bit_cnt % 2 == 1)
        ans++;
 
    return ans;
}
 
// Driver code
public static void main(String args[])
{
    int l = 1, r = 4;
 
    // Final answer
    System.out.print(xorRange(l, r));
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 implementation of the approach
 
# Function to return the most significant bit
def msb(x) :
 
    ret = 0
    while ((x >> (ret + 1)) != 0) :
        ret = ret + 1
    return ret
 
# Function to return the required XOR
def xorRange(l, r) :
 
    # Finding the MSB
    max_bit = msb(r)
 
    # Value of the current bit to be added
    mul = 2
 
    # To store the final answer
    ans = 0
 
    # Loop for case 1
    for i in range (1, max_bit + 1) :
 
        # Edge case when both the integers
        # lie in the same segment of continuous
        # 1s
        if ((l // mul) * mul == (r // mul) * mul) :
            if ((((l & (1 << i)) != 0) and
                 (r - l + 1) % 2 == 1)) :
                ans = ans + mul
            mul = mul * 2
            continue
         
        # To store whether parity of count is odd
        odd_c = 0
 
        if (((l & (1 << i)) != 0) and l % 2 == 1) :
            odd_c = (odd_c ^ 1)
        if (((r & (1 << i)) != 0) and r % 2 == 0) :
            odd_c = (odd_c ^ 1)
 
        # Updating the answer if parity is odd
        if (odd_c) :
            ans = ans + mul
 
        # Updating the number to be added
        mul = mul * 2
     
    # Case 2
    zero_bit_cnt = (r - l + 1) // 2
 
    if ((l % 2 == 1 ) and (r % 2 == 1)) :
        zero_bit_cnt = zero_bit_cnt + 1
 
    if (zero_bit_cnt % 2 == 1):
        ans = ans + 1
 
    return ans
 
# Driver code
l = 1
r = 4
 
# Final answer
print(xorRange(l, r))
 
# This code is contributed by ihritik

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the
// most significant bit
static int msb(int x)
{
    int ret = 0;
    while ((x >> (ret + 1)) != 0)
        ret++;
    return ret;
}
 
// Function to return the required XOR
static int xorRange(int l, int r)
{
 
    // Finding the MSB
    int max_bit = msb(r);
 
    // Value of the current bit to be added
    int mul = 2;
 
    // To store the final answer
    int ans = 0;
 
    // Loop for case 1
    for (int i = 1; i <= max_bit; i++)
    {
 
        // Edge case when both the integers
        // lie in the same segment of continuous
        // 1s
        if ((l / mul) * mul == (r / mul) * mul)
        {
            if (((l & (1 << i)) != 0) && (r - l + 1) % 2 == 1)
                ans += mul;
            mul *= 2;
            continue;
        }
 
        // To store whether parity of count is odd
        int odd_c = 0;
 
        if (((l & (1 << i)) != 0) && l % 2 == 1)
            odd_c = (odd_c ^ 1);
        if (((r & (1 << i)) != 0) && r % 2 == 0)
            odd_c = (odd_c ^ 1);
 
        // Updating the answer if parity is odd
        if (odd_c!=0)
            ans += mul;
 
        // Updating the number to be added
        mul *= 2;
    }
 
    // Case 2
    int zero_bit_cnt = zero_bit_cnt = (r - l + 1) / 2;
 
    if (l % 2 == 1 && r % 2 == 1)
        zero_bit_cnt++;
 
    if (zero_bit_cnt % 2 == 1)
        ans++;
 
    return ans;
}
 
// Driver code
public static void Main(String []args)
{
    int l = 1, r = 4;
 
    // Final answer
    Console.Write(xorRange(l, r));
}
}
 
// This code contributed by Rajput-Ji

PHP




<?php
// PHP implementation of the approach
 
// Function to return the
// most significant bit
function msb($x)
{
    $ret = 0;
    while (($x >> ($ret + 1)) != 0)
        $ret++;
    return $ret;
}
 
// Function to return the required XOR
function xorRange($l, $r)
{
 
    // Finding the MSB
    $max_bit = msb($r);
 
    // Value of the current bit to be added
    $mul = 2;
 
    // To store the final answer
    $ans = 0;
 
    // Loop for case 1
    for ($i = 1; $i <= $max_bit; $i++)
    {
 
        // Edge case when both the integers
        // lie in the same segment of continuous
        // 1s
        if ((int)(($l / $mul) * $mul) ==
            (int)(($r / $mul) * $mul))
        {
            if ((($l & (1 << $i)) != 0) &&
                 ($r - $l + 1) % 2 == 1)
                $ans += $mul;
            $mul *= 2;
            continue;
        }
 
        // To store whether parity of count is odd
        $odd_c = 0;
 
        if ((($l & (1 << $i)) != 0) && $l % 2 == 1)
            $odd_c = ($odd_c ^ 1);
        if ((($r & (1 << $i)) != 0) && $r % 2 == 0)
            $odd_c = ($odd_c ^ 1);
 
        // Updating the answer if parity is odd
        if ($odd_c)
            $ans += $mul;
 
        // Updating the number to be added
        $mul *= 2;
    }
 
    // Case 2
    $zero_bit_cnt = (int)(($r - $l + 1) / 2);
 
    if ($l % 2 == 1 && $r % 2 == 1)
        $zero_bit_cnt++;
 
    if ($zero_bit_cnt % 2 == 1)
        $ans++;
 
    return $ans;
}
 
// Driver code
$l = 1;
$r = 4;
 
// Final answer
echo xorRange($l, $r);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the
// most significant bit
function msb(x)
{
    let ret = 0;
    while ((x >> (ret + 1)) != 0)
        ret++;
    return ret;
}
 
// Function to return the required XOR
function xorRange(l, r)
{
 
    // Finding the MSB
    let max_bit = msb(r);
 
    // Value of the current bit to be added
    let mul = 2;
 
    // To store the final answer
    let ans = 0;
 
    // Loop for case 1
    for (let i = 1; i <= max_bit; i++) {
 
        // Edge case when both the integers
        // lie in the same segment of continuous
        // 1s
        if ((parseInt(l / mul) * mul) ==
        (parseInt(r / mul) * mul))
        {
            if (((l & (1 << i)) != 0) &&
            (r - l + 1) % 2 == 1)
            ans += mul;
            mul *= 2;
            continue;
        }
 
        // To store whether parity of count is odd
        let odd_c = 0;
 
        if (((l & (1 << i)) != 0) && l % 2 == 1)
            odd_c = (odd_c ^ 1);
        if (((r & (1 << i)) != 0) && r % 2 == 0)
            odd_c = (odd_c ^ 1);
 
        // Updating the answer if parity is odd
        if (odd_c)
            ans += mul;
 
        // Updating the number to be added
        mul *= 2;
    }
 
    // Case 2
    let zero_bit_cnt = parseInt((r - l + 1) / 2);
 
    if (l % 2 == 1 && r % 2 == 1)
        zero_bit_cnt++;
 
    if (zero_bit_cnt % 2 == 1)
        ans++;
 
    return ans;
}
 
// Driver code
    let l = 1, r = 4;
 
    // Final answer
    document.write(xorRange(l, r));
 
</script>
Output: 
4

 

Time Complexity: O(log2(R))
Efficient Approach: Let F(N) be a function that computes XOR of all the natural numbers less than or equal to N. Thus, for range (L-R), the answer will be F(R) ^ F(L-1)
Finding the value of this function for any given number is possible in O(1) as discussed in this article.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the required XOR
long computeXOR(const int n)
{
    // Modulus operator are expensive
    // on most of the computers.
    // n & 3 will be equivalent to n % 4
    // n % 4
    switch (n & 3) {
 
    // If n is a multiple of 4
    case 0:
        return n;
 
    // If n % 4 gives remainder 1
    case 1:
        return 1;
 
    // If n % 4 gives remainder 2
    case 2:
        return n + 1;
 
    // If n % 4 gives remainder 3
    case 3:
        return 0;
    }
}
 
// Driver code
int main()
{
    int l = 1, r = 4;
    cout << (computeXOR(r) ^ computeXOR(l - 1));
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the required XOR
static long computeXOR(int n)
{
    // Modulus operator are expensive
    // on most of the computers.
    // n & 3 will be equivalent to n % 4
    // n % 4
    int x = n & 3;
    switch (x)
    {
 
        // If n is a multiple of 4
        case 0:
            return n;
     
        // If n % 4 gives remainder 1
        case 1:
            return 1;
     
        // If n % 4 gives remainder 2
        case 2:
            return n + 1;
     
        // If n % 4 gives remainder 3
        case 3:
            return 0;
    }
    return 0;
}
 
// Driver code
public static void main(String args[])
{
    int l = 1, r = 4;
    System.out.println(computeXOR(r) ^
                       computeXOR(l - 1));
}
}
 
// This code is contributed by Ryuga

Python3




# Python3 implementation of the approach
 
# Function to return the required XOR
def computeXOR(n) :
 
    # Modulus operator are expensive
    # on most of the computers.
    # n & 3 will be equivalent to n % 4
    # n % 4
    switch = {
 
        # If n is a multiple of 4
        0 : n,
 
        # If n % 4 gives remainder 1
        1 : 1,
 
        # If n % 4 gives remainder 2
        2: n + 1,
 
        # If n % 4 gives remainder 3
        3 : 0,
    }
    return switch.get( n & 3, "")
 
# Driver code
l = 1
r = 4
print(computeXOR(r) ^ computeXOR(l - 1))
 
# This code is contributed by ihritik

C#




// C# implementation of the approach
using System;
class GFG
{
     
// Function to return the required XOR
static long computeXOR(int n)
{
    // Modulus operator are expensive
    // on most of the computers.
    // n & 3 will be equivalent to n % 4
    // n % 4
    int x=n&3;
    switch (x)
    {
 
    // If n is a multiple of 4
    case 0:
        return n;
 
    // If n % 4 gives remainder 1
    case 1:
        return 1;
 
    // If n % 4 gives remainder 2
    case 2:
        return n + 1;
 
    // If n % 4 gives remainder 3
    case 3:
        return 0;
    }
    return 0;
}
 
// Driver code
static void Main()
{
    int l = 1, r = 4;
    Console.WriteLine(computeXOR(r) ^ computeXOR(l - 1));
}
}
 
// This code is contributed by mits

PHP




<?php
// PHP implementation of the approach
 
// Function to return the required XOR
function computeXOR($n)
{
    // Modulus operator are expensive
    // on most of the computers.
    // n & 3 will be equivalent to n % 4
    // n % 4
    $x = $n & 3;
    switch ($x)
    {
 
        // If n is a multiple of 4
        case 0:
            return $n;
     
        // If n % 4 gives remainder 1
        case 1:
            return 1;
     
        // If n % 4 gives remainder 2
        case 2:
            return $n + 1;
     
        // If n % 4 gives remainder 3
        case 3:
            return 0;
    }
    return 0;
}
 
// Driver code
$l = 1; $r = 4;
echo(computeXOR($r) ^ computeXOR($l - 1));
 
// This code is contributed by Code_Mech
?>

Javascript




<script>
// Javascript implementation of the approach
 
// Function to return the required XOR
function computeXOR(n)
{
    // Modulus operator are expensive
    // on most of the computers.
    // n & 3 will be equivalent to n % 4
    // n % 4
    switch (n & 3) {
 
    // If n is a multiple of 4
    case 0:
        return n;
 
    // If n % 4 gives remainder 1
    case 1:
        return 1;
 
    // If n % 4 gives remainder 2
    case 2:
        return n + 1;
 
    // If n % 4 gives remainder 3
    case 3:
        return 0;
    }
}
 
// Driver code
    let l = 1, r = 4;
    document.write(computeXOR(r) ^ computeXOR(l - 1));
 
// This code is contributed by subhammahato348.
</script>
Output: 
4

 

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