### java

#### Big O notation and not understading from class lecture

```I have this problem where in class my professor said that the below statement is O(log(n)) where I thought it was O(n). Could someone please clarify how it is O(log(n))?
Printing a number of magnitude n in binary. Assume that printing each bit requires constant time.
```
```You should work out some examples. Write some numbers in binary. For example, how many bits are there in 63, 255, and 511? Notice that the number of bits does not grow nearly as quickly as the number itself.
```
```It's O(log(n)) because you have to divide by 2 every time you are going to print a 0 or 1.
For example, to print 256 in binary, you'd have to divide by 2 starting from 256 and print the result of % 2 every time.
256 % 2 -> 0
64% 2 -> 0
32 % 2 -> 0
16 % 2 -> 0
8 % 2 -> 0
4 % 2 -> 0
2 % 2 -> 0
1 % 2 -> 1
So, for a number of magnitude 256 you would have to iterate 8 times which is equals to log 256.
```
```O(log(n)) is all about cutting data by half.
When each step of an algorithm rules out a fraction of the remaining input -- e.g. you always cut the space in half, or by a third, or even to 99/100 of the previous step -- that algorithm runs in O(log(n)) time.
hope this helps```

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