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Bitwise and (or &) of a range
• Difficulty Level : Medium
• Last Updated : 31 Mar, 2021

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

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.#includeusing 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 3int 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 codeint 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 3    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 3def 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 codex, y = 10, 15print(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 3    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

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

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