leetcode - 338. Counting Bits

Description

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++)?

Solution

$o(n\log n)$

Use n & (n - 1) to delete the rightest 1 from n, and count the number of 1 for every number.

Time complexity: $o(n\log n)$
Space complexity: $o(1)$

$o(n)$

For even numbers, the rightest digit is always 0. So if you move the number to right by 1 bit, the number of 1 doesn’t change. But if it’s an odd number, the rightest digit is 1, so add this 1 to the number.

If we use dp[i] to denote the number of 1 of number i, then: dp[i] = dp[i >> 1] + (i % 2)

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

$o(n\log n)$

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

$o(n)$

class Solution:
    def countBits(self, n: int) -> List[int]:
        dp = [0] * (n + 1)
        for i in range(1, n + 1):
            dp[i] = dp[i >> 1] + (i % 2)
        return dp