 Open in App
Not now

# Bitwise and (or &) of a range

• Difficulty Level : Medium
• Last Updated : 13 Jan, 2023

Given two non-negative long integers, x and y given x <= y, the task is to find bit-wise and of all integers from x and y, i.e., we need to compute value of x & (x+1) & … & (y-1) & y.7
Examples:

```Input  : x = 12, y = 15
Output : 12
12 & 13 & 14 & 15 = 12

Input  : x = 10, y = 20
Output : 0 ```
Recommended Practice

A simple solution is to traverse all numbers from x to y and do bit-wise and of all numbers in range.
An efficient solution is to follow following steps.
1) Find position of Most Significant Bit (MSB) in both numbers.
2) If positions of MSB are different, then result is 0.
3) If positions are same. Let positions be msb_p.
……a) We add 2msb_p to result.
……b) We subtract 2msb_p from x and y,
……c) Repeat steps 1, 2 and 3 for new values of x and y.

```Example 1 :
x = 10, y = 20
Result is initially 0.
Position of MSB in x = 3
Position of MSB in y = 4
Since positions are different, return result.

Example 2 :
x = 17, y = 19
Result is initially 0.
Position of MSB in x = 4
Position of MSB in y = 4
Since positions are same, we compute 24.

Result becomes 16.

We subtract this value from x and y.
New value of x  = x - 24  = 17 - 16 = 1
New value of y  = y - 24  = 19 - 16 = 3

Position of MSB in new x = 1
Position of MSB in new y = 2
Since positions are different, we return result.```

## C++

 `// An efficient C++ program to find bit-wise & of all``// numbers from x to y.``#include``using` `namespace` `std;``typedef` `long` `long` `int` `ll;` `// Find position of MSB in n. For example if n = 17,``// then position of MSB is 4. If n = 7, value of MSB``// is 2``int` `msbPos(ll n)``{``    ``int` `msb_p = -1;``    ``while` `(n)``    ``{``        ``n = n>>1;``        ``msb_p++;``    ``}``    ``return` `msb_p;``}` `// Function to find Bit-wise & of all numbers from x``// to y.``ll andOperator(ll x, ll y)``{``    ``ll res = 0; ``// Initialize result` `    ``while` `(x && y)``    ``{``        ``// Find positions of MSB in x and y``        ``int` `msb_p1 = msbPos(x);``        ``int` `msb_p2 = msbPos(y);` `        ``// If positions are not same, return``        ``if` `(msb_p1 != msb_p2)``            ``break``;` `        ``// Add 2^msb_p1 to result``        ``ll msb_val =  (1 << msb_p1);``        ``res = res + msb_val;` `        ``// subtract 2^msb_p1 from x and y.``        ``x = x - msb_val;``        ``y = y - msb_val;``    ``}` `    ``return` `res;``}` `// Driver code``int` `main()``{``    ``ll x = 10, y = 15;``    ``cout << andOperator(x, y);``    ``return` `0;``}`

## Java

 `// An efficient Java program to find bit-wise``// & of all numbers from x to y.``class` `GFG {``    ` `    ``// Find position of MSB in n. For example``    ``// if n = 17, then position of MSB is 4.``    ``// If n = 7, value of MSB is 2``    ``static` `int` `msbPos(``long` `n)``    ``{``        ` `        ``int` `msb_p = -``1``;``        ``while` `(n > ``0``) {``            ``n = n >> ``1``;``            ``msb_p++;``        ``}``        ` `        ``return` `msb_p;``    ``}` `    ``// Function to find Bit-wise & of all``    ``// numbers from x to y.``    ``static` `long` `andOperator(``long` `x, ``long` `y)``    ``{``        ` `        ``long` `res = ``0``; ``// Initialize result` `        ``while` `(x > ``0` `&& y > ``0``) {``            ` `            ``// Find positions of MSB in x and y``            ``int` `msb_p1 = msbPos(x);``            ``int` `msb_p2 = msbPos(y);` `            ``// If positions are not same, return``            ``if` `(msb_p1 != msb_p2)``                ``break``;` `            ``// Add 2^msb_p1 to result``            ``long` `msb_val = (``1` `<< msb_p1);``            ``res = res + msb_val;` `            ``// subtract 2^msb_p1 from x and y.``            ``x = x - msb_val;``            ``y = y - msb_val;``        ``}` `        ``return` `res;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ` `        ``long` `x = ``10``, y = ``15``;``        ` `        ``System.out.print(andOperator(x, y));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# An efficient Python program to find``# bit-wise & of all numbers from x to y.` `# Find position of MSB in n. For example``# if n = 17, then position of MSB is 4.``# If n = 7, value of MSB is 2``def` `msbPos(n):` `    ``msb_p ``=` `-``1``    ``while` `(n > ``0``):``    ` `        ``n ``=` `n >> ``1``        ``msb_p ``+``=` `1``    ` `    ``return` `msb_p` `# Function to find Bit-wise & of``# all numbers from x to y.``def` `andOperator(x, y):` `    ``res ``=` `0` `# Initialize result` `    ``while` `(x > ``0` `and` `y > ``0``):``    ` `        ``# Find positions of MSB in x and y``        ``msb_p1 ``=` `msbPos(x)``        ``msb_p2 ``=` `msbPos(y)` `        ``# If positions are not same, return``        ``if` `(msb_p1 !``=` `msb_p2):``            ``break` `        ``# Add 2^msb_p1 to result``        ``msb_val ``=` `(``1` `<< msb_p1)``        ``res ``=` `res ``+` `msb_val` `        ``# subtract 2^msb_p1 from x and y.``        ``x ``=` `x ``-` `msb_val``        ``y ``=` `y ``-` `msb_val` `    ``return` `res``    ` `# Driver code``x, y ``=` `10``, ``15``print``(andOperator(x, y))` `# This code is contributed by Anant Agarwal.`

## C#

 `// An efficient C# program to find bit-wise & of all``// numbers from x to y.``using` `System;` `class` `GFG``{``    ``// Find position of MSB in n.``    ``// For example if n = 17,``    ``// then position of MSB is 4.``    ``// If n = 7, value of MSB``    ``// is 2``    ``static` `int` `msbPos(``long` `n)``    ``{``        ``int` `msb_p = -1;``        ``while` `(n > 0)``        ``{``            ``n = n >> 1;``            ``msb_p++;``        ``}``        ``return` `msb_p;``    ``}``    ` `    ``// Function to find Bit-wise``    ``// & of all numbers from x``    ``// to y.``    ``static` `long` `andOperator(``long` `x, ``long` `y)``    ``{``        ``// Initialize result``        ``long` `res = 0;``    ` `        ``while` `(x > 0 && y > 0)``        ``{``            ``// Find positions of MSB in x and y``            ``int` `msb_p1 = msbPos(x);``            ``int` `msb_p2 = msbPos(y);``    ` `            ``// If positions are not same, return``            ``if` `(msb_p1 != msb_p2)``                ``break``;``    ` `            ``// Add 2^msb_p1 to result``            ``long` `msb_val = (1 << msb_p1);``            ``res = res + msb_val;``    ` `            ``// subtract 2^msb_p1 from x and y.``            ``x = x - msb_val;``            ``y = y - msb_val;``        ``}``    ` `        ``return` `res;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``long` `x = 10, y = 15;``        ``Console.WriteLine(andOperator(x, y));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## PHP

 ` 0)``    ``{``        ``\$n` `= ``\$n` `>> 1;``        ``\$msb_p``++;``    ``}``    ``return` `\$msb_p``;``}` `// Function to find Bit-wise &``// of all numbers from x to y.``function` `andOperator(``\$x``, ``\$y``)``{``    ``\$res` `= 0; ``// Initialize result` `    ``while` `(``\$x` `> 0 && ``\$y` `> 0)``    ``{``        ``// Find positions of``        ``// MSB in x and y``        ``\$msb_p1` `= msbPos(``\$x``);``        ``\$msb_p2` `= msbPos(``\$y``);` `        ``// If positions are not``        ``// same, return``        ``if` `(``\$msb_p1` `!= ``\$msb_p2``)``            ``break``;` `        ``// Add 2^msb_p1 to result``        ``\$msb_val` `= (1 << ``\$msb_p1``);``        ``\$res` `= ``\$res` `+ ``\$msb_val``;` `        ``// subtract 2^msb_p1``        ``// from x and y.``        ``\$x` `= ``\$x` `- ``\$msb_val``;``        ``\$y` `= ``\$y` `- ``\$msb_val``;``    ``}` `    ``return` `\$res``;``}` `// Driver code``\$x` `= 10;``\$y` `= 15;``echo` `andOperator(``\$x``, ``\$y``);` `// This code is contributed``// by ihritik``?>`

## Javascript

 ``

Output

`8`

Time Complexity: O(log(max(x, y)))
Auxiliary Space: O(1)

More efficient solution

1. Flip the LSB of b.
2. And check if the new number is in range(a < number < b) or not
• if the number greater than ‘a’ again flip lsb
• if it is not then that’s the answer

## C++

 `// An efficient C++ program to find bit-wise & of all``// numbers from x to y.``#include``using` `namespace` `std;``typedef` `long` `long` `int` `ll;`  `// Function to find Bit-wise & of all numbers from x``// to y.``ll andOperator(ll x, ll y)``{``    ``// Iterate over all bits of y, starting from the lsb, if it's equal to 1, flip it``    ``for``(``int` `i=0; i<(``int``)log2(y)+1;i++)``    ``{``        ``//repeat till x >= y, otherwise return the answer.``        ``if` `(y <= x) {``            ``return` `y;``        ``}``        ``if` `(y & (1 << i)) {``            ``y &= ~(1UL << i);``        ``}``    ``}``    ``return` `y;``}` `// Driver code``int` `main()``{``    ``ll x = 10, y = 15;``    ``cout << andOperator(x, y);``    ``return` `0;``}`

## Java

 `// An efficient Java program to find bit-wise & of all``// numbers from x to y.``import` `java.util.*;` `class` `GFG {` `  ``// Function to find Bit-wise & of all numbers from x``  ``// to y.``  ``static` `int` `andOperator(``int` `x, ``int` `y)``  ``{` `    ``// Iterate over all bits of y, starting from the``    ``// lsb, if it's equal to 1, flip it``    ``for` `(``int` `i = ``0``; i < (Math.log(y) / Math.log(``2``)) + ``1``;``         ``i++) {``      ``// repeat till x >= y, otherwise return the``      ``// answer.``      ``if` `(y <= x) {``        ``return` `y;``      ``}``      ``if` `((y & (``1` `<< i)) != ``0``) {``        ``y &= ~(``1` `<< i);``      ``}``    ``}``    ``return` `y;``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `x = ``10``;``    ``int` `y = ``15``;``    ``System.out.print(andOperator(x, y));``  ``}``}` `// This code is contributed by phasing17`

## Python3

 `# An efficient Python program to find bit-wise & of``# all numbers from x to y.` `# Importing math module for using logarithm``import` `math` `# Function to find Bit-wise & of all numbers from x``# to y.``def` `andOperator(x, y):``  ` `    ``# Iterate over all bits of y, starting from the lsb, if it's equal to 1, flip it``    ``for` `i ``in` `range``(``int``(math.log2(y) ``+` `1``)):``      ` `        ``# repeat till x >= y, otherwise return the answer``        ``if``(y <``=` `x):``            ``return` `y``        ``if``(y & ``1` `<< i):``            ``y ``=` `y & (~(``1``<

## C#

 `// An efficient C# program to find bit-wise & of all``// numbers from x to y.` `using` `System;` `class` `GFG {` `    ``// Function to find Bit-wise & of all numbers from x``    ``// to y.``    ``static` `int` `andOperator(``int` `x, ``int` `y)``    ``{` `        ``// Iterate over all bits of y, starting from the``        ``// lsb, if it's equal to 1, flip it``        ``for` `(``int` `i = 0; i < (Math.Log(y) / Math.Log(2)) + 1;``             ``i++) {``            ``// repeat till x >= y, otherwise return the``            ``// answer.``            ``if` `(y <= x) {``                ``return` `y;``            ``}``            ``if` `((y & (1 << i)) != 0) {``                ``y &= ~(1 << i);``            ``}``        ``}``        ``return` `y;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int` `x = 10;``        ``int` `y = 15;``        ``Console.Write(andOperator(x, y));``    ``}``}` `// This code is contributed by phasing17`

## Javascript

 `// An efficient JavaScript program to find bit-wise & of all``// numbers from x to y.`  `// Function to find Bit-wise & of all numbers from x``// to y.``function` `andOperator(x, y)``{``    ``// Iterate over all bits of y, starting from the lsb, if it's equal to 1, flip it``    ``for``(``var` `i=0; i= y, otherwise return the answer.``        ``if` `(y <= x) {``            ``return` `y;``        ``}``        ``if` `(y & (1 << i)) {``            ``y &= ~(1 << i);``        ``}``    ``}``    ``return` `y;``}` `// Driver code``var` `x = 10, y = 15;``console.log(andOperator(x, y));`  `//This code is contributed by phasing17`

Output

`8`

Time Complexity: O(log(y))
Auxiliary Space: O(1)

Another Approach

We know that if a number num is a power of 2 then (num &(num – 1)) is equal to 0. So if a is less than 2^k  and b is greater than or equal to 2^k, then the & of all values in between a and b should be zero as (2^k & (2^k – 1)) is equal to 0. So, if both a and b lies within the same number of bits then only answer wont be zero. Now, in every case last bit is bound to be zero because even if a and b are 2 side by side numbers last bit will be different. Similarly 2nd last bit will be zero if difference between a and b is greater than 2 and this goes on for every bit. Now, take example a = 1100(12) and b = 1111(15), then last bit should be zero of the answer. For 2nd last bit we need to check whether a/2 == b/2 because if they are equal then we know that b – a <= 2. So if a/2 and b/2 is not equal then we proceed. Now, 3rd last bit should have a difference of 4 which can be checked by a/ 4 != b/4. Hence we check every bit from last until a!=b and in every step we modify a/=2(a >> 1) and b/=2(b >> 1) to reduce a bit from end.

1. Run a while loop as long as a != b and a > 0
2. Right shift a by 1 and right shift b by 1
3. increment shiftcount
4. after while loop return left * 2^(shiftcount)

## C++

 `// An efficient C++ program to find bit-wise & of all``// numbers from x to y.``#include``using` `namespace` `std;``#define int long long int` `// Function to find Bit-wise & of all numbers from x``// to y.``int` `andOperator(``int` `a, ``int` `b) {``      ``// ShiftCount variables counts till which bit every value will convert to 0``      ``int` `shiftcount = 0;``    ``//Iterate through every bit of a and b simultaneously``      ``//If a == b then we know that beyond that the and value will remain constant``      ``while``(a != b and a > 0) {``          ``shiftcount++;``          ``a = a >> 1;``          ``b = b >> 1;``    ``}``      ``return` `int64_t(a << shiftcount);``}` `// Driver code``int32_t main() {``    ``int` `a = 10, b = 15;``    ``cout << andOperator(a, b);``    ``return` `0;``}`

## Java

 `// An efficient Java program to find bit-wise & of all``// numbers from x to y.``import` `java.util.*;` `class` `GFG {` `  ``// Function to find Bit-wise & of all numbers from x``  ``// to y.``  ``static` `long` `andOperator(``int` `a, ``int` `b)``  ``{` `    ``// ShiftCount variables counts till``    ``// which bit every value will convert to 0``    ``int` `shiftcount = ``0``;` `    ``// Iterate through every bit of a and b``    ``// simultaneously If a == b then we know that beyond``    ``// that the and value will remain constant``    ``while` `(a != b && a > ``0``) {``      ``shiftcount++;``      ``a = a >> ``1``;``      ``b = b >> ``1``;``    ``}``    ``return` `(``long``)(a << shiftcount);``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `a = ``10``, b = ``15``;``    ``System.out.println(andOperator(a, b));``  ``}``}` `// This code is contributed by phasing17`

## Python3

 `# An efficient Python program to find bit-wise & of``# all numbers from x to y.` `# Function to find Bit-wise & of all numbers from x``# to y.``def` `andOperator(a,b):``    ` `    ``# ShiftCount variables counts till which bit every value will convert to 0``    ``shiftcount``=``0``    ` `    ``# Iterate through every bit of a and b simultaneously``    ``# If a == b then we know that beyond that the and value will remain constant``    ``while``(a!``=``b ``and` `a>``0``):``        ``shiftcount``=``shiftcount``+``1``        ``a``=``a>>``1``        ``b``=``b>>``1``        ` `    ``return` `a<

## C#

 `// An efficient C# program to find bit-wise & of all``// numbers from x to y.``using` `System;` `class` `GFG``{``  ` `  ``// Function to find Bit-wise & of all numbers from x``  ``// to y.``  ``static` `Int64 andOperator(``int` `a, ``int` `b)``  ``{` `    ``// ShiftCount variables counts till``    ``// which bit every value will convert to 0``    ``int` `shiftcount = 0;` `    ``// Iterate through every bit of a and b simultaneously``    ``// If a == b then we know that beyond that``    ``// the and value will remain constant``    ``while``(a != b && a > 0)``    ``{``      ``shiftcount++;``      ``a = a >> 1;``      ``b = b >> 1;``    ``}``    ``return` `(Int64)(a << shiftcount);``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main(``string``[] args) {``    ``int` `a = 10, b = 15;``    ``Console.WriteLine(andOperator(a, b));``  ``}``}` `// This code is contributed by phasing17`

## Javascript

 `// An efficient JavaScript program to find bit-wise & of all``// numbers from x to y.` `// Function to find Bit-wise & of all numbers from x``// to y.``function` `andOperator(a, b)``{` `      ``// ShiftCount variables counts till which bit every value will convert to 0``      ``let shiftcount = 0;``      ` `    ``// Iterate through every bit of a and b simultaneously``      ``//If a == b then we know that beyond that the and value will remain constant``      ``while``(a != b && a > 0) {``          ``shiftcount++;``          ``a = a >> 1;``          ``b = b >> 1;``    ``}``      ``return` `(a << shiftcount);``}` `// Driver code``let a = 10, b = 15;``console.log(andOperator(a, b));` `// This code is contributed by phasing17`

Output

`8`

Time Complexity: O(log(max(x, y)))
Auxiliary Space: O(1)

My Personal Notes arrow_drop_up