Description
https://leetcode.com/problems/counting-bits/
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 timeO(n)
and possibly in a single pass? - Can you do it without using any built-in function (i.e., like
__builtin_popcount
in C++)?
Explanation
Convert number to binary strings and count ‘1’s.
Python Solution
class Solution:
def countBits(self, n: int) -> List[int]:
results = []
for i in range(n + 1):
bin_str = bin(i)[2:]
one_count = bin_str.count('1')
results.append(one_count)
return results
- Time Complexity: O(N).
- Space Complexity: O(N).