338. Counting Bits

Problem¶

Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1’s in the binary representation of i.

Example 1:

Input: n = 2
Output: [0,1,1]
Explanation:
0 –> 0
1 –> 1
2 –> 10

Example 2:

Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 –> 0
1 –> 1
2 –> 10
3 –> 11
4 –> 100
5 –> 101

Constraints:

• 0 <= n <= 10**5

Follow up:

• It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
• Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

Solve¶

Built-in bit_count() python¶

O(n log n)

Python number have built-in bit_count() that give us total 1 in bit representation of any number. We assuming this take log n to count it.

class Solution:
    def countBits(self, n: int) -> List[int]:
        return [i.bit_count() for i in range(n+1)]

Dynamic programming¶

O(n)

We can split any binary number into <right-shift> + <remain 1/0> , for example 100110111001 can be split into 10011011100 and 1:

bit_count(100110111001) = bit_count(10011011100) + bit_count(1)
                        = bit_count(100110111001 >> 1) + bit_count(1)

Formula:

bit_count(x) = bit_count(x>>1) + (x & 1)

With this formula, we can create a simple Dynamic programing solution:

class Solution:
    def countBits(self, n: int) -> List[int]:
        cache = [0]
        count = 0
        for i in range(1,n+1):
            cache.append(cache[(i >> 1)] + (i & 1))
        return cache