# Question

Formatted question description: https://leetcode.ca/all/338.html

 338	Counting Bits

Given a non negative integer number num.
For every numbers i in the range 0 ≤ i ≤ num
calculate the number of 1's in their binary representation
and return them as an array.

Example 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [0,1,1,2,1,2]

It is very easy to come up with a solution with run time O(n*sizeof(integer)).
But can you do it in linear time O(n) /possibly in a single pass?
Space complexity should be O(n).

Can you do it like a boss?
Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

@tag-dp


# Algorithm

DP: countAtNum[i] = countAtNum[i>>1] + (i&1);

# Code

Java

• 
public class Counting_Bits {
class Solution {
public int[] countBits(int num) {
int[] countAtNum = new int[num + 1];

if (num <= 0) {
return countAtNum;
}

countAtNum[0] = 0; // to be explicit
countAtNum[1] = 1;

for (int i = 2; i <= num; i++) {
// @note @memorize : & operator lower than +, so should be in ()
countAtNum[i] = countAtNum[i>>1] + (i&1);
}

return countAtNum;
}
}
}

• // OJ: https://leetcode.com/problems/counting-bits
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> countBits(int n) {
vector<int> ans(n + 1);
for (int i = 1; i <= n; i *= 2) {
ans[i] = 1;
for (int j = 1; j < i && i + j <= n; ++j) ans[i + j] = ans[i] + ans[j];
}
return ans;
}
};

• class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
if num == 0:
return [0]
ans = [0, 1]
j = 0
for i in range(2, num + 1):
ans.append(ans[i & (i - 1)] + 1)
return ans