Given a range $[m, n]$ where $0 \leq m \leq n \leq 2^{31} - 1$, return the bitwise AND of all numbers in this range, inclusive.

## Discussion

The problem is trivial to solve with a loop but there’s an interesting idea lurking around the corner, so we will avoid doing this via a straightforward loop.

Take a number, $m = 14$. Lets examine what happens in binary as we move forward from 14

(14) .... 0000 1110
(15) .... 0000 1111
(16) .... 0001 0000 <-
(17) .... 0001 0001
(18) .... 0001 0010
(19) .... 0001 0011
(20) .... 0001 0100
(21) .... 0001 0101
(22) .... 0001 0110
(23) .... 0001 0111
(24) .... 0001 1000
(25) .... 0001 1001
(26) .... 0001 1010
(27) .... 0001 1011
(28) .... 0001 1100
(29) .... 0001 1101


Looking from the right hand side, here’s some properties to keep in mind:

1. $n$ is likely to have more significant bits set than $m$. So the AND of all contiguous elements in this range can only be as large as $m$
2. If $m < 2^k < n$, then the result is always a $0$, since the power of $2$’s leftmost set bit will be further right than $m$’s leftmost set bit
3. If $n > m$, the rightmost bit for this range is $0$

Property $3$ can be used with a little more insight:

Notice what happens if we replace the rightmost bit from the sequence with X because we don’t care what this value is.

(14) .... 0000 111X
(15) .... 0000 111X
(16) .... 0001 000X
(17) .... 0001 000X
(18) .... 0001 001X
(19) .... 0001 001X
(20) .... 0001 010X
(21) .... 0001 010X
(22) .... 0001 011X
(23) .... 0001 011X
(24) .... 0001 100X
(25) .... 0001 100X
(26) .... 0001 101X
(27) .... 0001 101X
(28) .... 0001 110X
(29) .... 0001 110X


Ignoring the $X$, alternating numbers now form an increasing sequence, how long can we do this? We ignored the right most bit because this changed constantly, so we’ll stop ignoring it when it stops changing. When it does stop changing, no bit to it’s left will change because contiguous binary sequences bits change left to right, one at a time.

So AND-ing the entire sequence is the same as finding the common left bits of all numbers in that range. That value doesn’t depend on all the numbers, but just the endpoints.

This is what the method looks like:

(17) .... 0001 0001
(26) .... 0001 1010
---
(08) .... 0001 000X
(13) .... 0001 101X
---
(04) .... 0001 00XX
(06) .... 0001 10XX
---
(02) .... 0001 0XXX
(03) .... 0001 1XXX
---
(01) .... 0001 XXXX
(01) .... 0001 XXXX


It doesn’t necessarily have to be $1$, but it has to end when right shifting the endpoints causes them to become equal.

## Code

This here is a C++ implementation of the idea above

class Solution
{
public:
int rangeBitwiseAnd(int m, int n)
{
int rightMask = 1;
while(n != m)
{
m /= 2;
n /= 2;