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# bitwise-and-number-range

## Problem Description

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
Example 1:
Input: [5,7]
Output: 4
Example 2:
Input: [0,1]
Output: 0

## Solution

This is a Bit operation problem, and I feel helpless every time when I see bit operation... anyway, look at this problem, we can find out that it needed to bit AND to all the number for given range, for example [4,5], it asking we do `4 (100) AND 5 (101) = 100 4(100)`, bit AND, only 1 & 1 = 1, otherwise is 0. for we can see we need to find out the highest common 1s(left most common 1s) in the range.
i.e. [4,6] -- the left most common 1s is 100 = 4.
4 5 6
100 101 110
How to find the highest common 1s.
• if `m != n`, left shift m and n, `m>>=1, n>>=1`
• count shifts for each shift
• until `m==n`, found left most common 1s,
• the result is left most common 1s right shift shiftCounts.

### Code

class Solution {
public int rangeBitwiseAnd(int m, int n) {
if (m == 0) return 0;
int shift = 0;
while (m != n) {
m >>= 1;
n >>= 1;
shift++;
}
return m << shift;
}
}