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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 ``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

 `> (``\$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

 ``
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 ``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

 ``

## Javascript

 ``
Output:
`4`

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