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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 timeO(n)
and possibly in a single pass? - Can you do it without using any built-in function (ie, like
__builtin_popcount
in C++)?
Python
from typing import List class Solution: def countBits(self, n: int) -> List[int]: result = [] for i in range(n+1): result.append(str(bin(i)).split ("0b")[1].count('1')) return result print(Solution().countBits(5))