0338.Counting_Bits.py

"""
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 <= 105
 

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++)?
"""

"""
1.define state: dp[i] represent the number of 1's in the binary representation of i
2.get dp[n]
3.initialize dp[0] = 0 dp[1] = 1
4.dp[i] = dp[i//2] + i%2
"""
class Solution:
    def countBits(self, n: int) -> List[int]:
        dp = [0] * (n+1)
        for i in range(1, n+1):
            dp[i] = dp[i//2] + i%2
        return dp