LeetCode in Elixir
338. Counting Bits
Easy
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_popcountin C++)?
Solution
defmodule Solution do
@spec count_bits(n :: integer) :: [integer]
def count_bits(n), do: do_count_bits(1, n, Map.new(0..div(n,2), &({&1,0})), [0])
defp do_count_bits(i, n, map, acc) when i > n, do: Enum.reverse(acc)
defp do_count_bits(i, n, map, acc) do
i_bits = map[div(i,2)] + rem(i,2)
new_map = if i > div(n,2), do: map, else: Map.put(map, i, i_bits)
do_count_bits(i + 1, n, new_map, [i_bits | acc])
end
end