Leetcode: Integer Replacement

Given a positive integer n and you can do operations as follow:

If n is even, replace n with n/2.
If n is odd, you can replace n with either n + 1 or n - 1.
What is the minimum number of replacements needed for n to become 1?

Example 1:

Input:
8

Output:
3

Explanation:
8 -> 4 -> 2 -> 1
Example 2:

Input:
7

Output:
4

Explanation:
7 -> 8 -> 4 -> 2 -> 1
or
7 -> 6 -> 3 -> 2 -> 1

Refer to: https://discuss.leetcode.com/topic/58334/a-couple-of-java-solutions-with-explanations/2
When to add 1 instead of minus 1, here is an example:

Look at this example:

111011 -> 111010 -> 11101 -> 11100 -> 1110 -> 111 -> 1000 -> 100 -> 10 -> 1

And yet, this is not the best way because

111011 -> 111100 -> 11110 -> 1111 -> 10000 -> 1000 -> 100 -> 10 -> 1

See? Both

111011 -> 111010

and

111011 -> 111100

remove the same number of 1's, but the second way is better.

Indeed, if a number ends with 01, then certainly decrementing is the way to go. Otherwise, if it ends with 11, then certainly incrementing is at least as good as decrementing (

*011 -> *010 / *100

) or even better (if there are three or more 1's). This leads to the following solution:

So the logic is:

If

n

is even, halve it.

If

n=3

or

n ends with "01", decrease n

Otherwise, increment

n

.(ends with "11", "111" or even more "1111")

1 public class Solution {
2     public int integerReplacement(int n) {
3         int count = 0;
4         while (n != 1) {
5             if ((n & 1) == 0)
6                 n = n >>> 1;
7             else if (n==3 || (n & 0b11) == 1)
8                 n--;
9             else n++;
10             count++;
11         }
12         return count;
13     }
14 }

Another logic is:

So the logic is:

If

n

is even, halve it.

If

n=3

or

n-1

has less 1's than

n+1

, decrement

n

.

Otherwise, increment

n

.

1 public int integerReplacement(int n) {
2     int c = 0;
3     while (n != 1) {
4         if ((n & 1) == 0) {
5             n >>>= 1;
6         } else if (n == 3 || Integer.bitCount(n + 1) > Integer.bitCount(n - 1)) {
7             --n;
8         } else {
9             ++n;
10         }
11         ++c;
12     }
13     return c;
14 }