## Sunday, November 13, 2016

### Bitwise and (or &) of a range - GeeksforGeeks

Bitwise and (or &) of a range - GeeksforGeeks
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

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.
`// 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`
`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;`
`}`