Bitwise and (or &) of a range
Last Updated :
23 Jul, 2025
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.
Steps to implement-
- Initialize a variable res with the first number x
- Run a loop from x+1 to y
- In that loop store res & element of the loop into res
- Finally return/print res
Code-
C++
// C++ program to find bit-wise & of all
// numbers from x to y.
#include <bits/stdc++.h>
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)
{
// Initialize result as the first number
ll res = x;
// Traverse from x+1 to y
for (int i = x + 1; i <= y; i++) {
res = res & i;
}
return res;
}
// Driver code
int main()
{
ll x = 10, y = 15;
cout << andOperator(x, y);
return 0;
}
// Java program to find bit-wise & of all
// numbers from x to y.
import java.io.*;
class GFG {
// Function to find Bit-wise & of along numbers from x
// to y.
static long andOperator(long x, long y)
{
// Initialize result as the first number
long res = x;
// Traverse from x+1 to y
for (long i = x + 1; i <= y; i++) {
res = res & i;
}
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 Utkarsh Kumar
# Function to find bitwise AND of all numbers from x to y
def and_operator(x, y):
# Initialize the result as the first number
res = x
# Traverse from x+1 to y
for i in range(x + 1, y + 1):
res = res & i
return res
# Driver code
if __name__ == "__main__":
x = 10
y = 15
print(and_operator(x, y))
using System;
class GFG
{
// Function to find Bit-wise AND of all numbers from x to y.
static long AndOperator(long x, long y)
{
// Initialize result as the first number
long res = x;
// Traverse from x+1 to y
for (long i = x + 1; i <= y; i++)
{
res = res & i;
}
return res;
}
// Driver code
public static void Main(string[] args)
{
long x = 10, y = 15;
Console.Write(AndOperator(x, y));
}
}
//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)
{
// Initialize result as the first number
let res = x;
//Traverse from x+1 to y
for(let i=x+1;i<=y;i++){
res=res&i;
}
return res;
}
// Driver code
let x = 10, y = 15;
console.log(andOperator(x, y));
//This code is contributed by Tushar.
Time Complexity: O(y-x-1), because we have loop from x+1 to y
Auxiliary Space: O(1), because no extra space has been used
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.
We add 24 to result.
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<bits/stdc++.h>
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;
}
// 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.
# 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.
// 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.
<script>
// Javascript 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
function msbPos(n)
{
let 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.
function andOperator(x, y)
{
let res = 0; // Initialize result
while (x > 0 && y > 0) {
// Find positions of MSB in x and y
let msb_p1 = msbPos(x);
let msb_p2 = msbPos(y);
// If positions are not same, return
if (msb_p1 != msb_p2)
break;
// Add 2^msb_p1 to result
let 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
let x = 10, y = 15;
document.write(andOperator(x, y));
// This code is contributed by avijitmondal1998.
</script>
<?php
// An efficient C++ 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
function msbPos($n)
{
$msb_p = -1;
while ($n > 0)
{
$n = $n >> 1;
$msb_p++;
}
return $msb_p;
}
// Function to find Bit-wise &
// of all numbers from x to y.
function andOperator($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
?>
Time Complexity: O(log(max(x, y)))
Auxiliary Space: O(1)
More efficient solution
- Flip the LSB of b.
- 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<bits/stdc++.h>
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;
}
// 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
# 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<<i))
return y
# Driver code
x, y = 10, 15
print(andOperator(x, y))
# This code is contributed by Pushpesh Raj
// 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
// 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<Math.log2(y)+1;i++)
{
//repeat till x >= 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
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.
- Run a while loop as long as a != b and a > 0
- Right shift a by 1 and right shift b by 1
- increment shiftcount
- after while loop return left * 2^(shiftcount)
C++
// An efficient C++ program to find bit-wise & of all
// numbers from x to y.
#include<bits/stdc++.h>
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;
}
// 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
# 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<<shiftcount
# Driver code
a, b =10, 15
print(andOperator(a, b))
# This code is contributed by Pushpesh Raj
// 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
// 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
Time Complexity: O(log(max(x, y)))
Auxiliary Space: O(1)
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