# 338. Counting Bits (Medium)

https://leetcode.com/problems/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.

## Solutions

``````class Solution {

// Following solution is ingenious. Through careful observation, we will find that current elements can
// make use of previous element results, say num using num>>1.

public int[] countBits(int num) {

// from 0 (inclusive) to num (inclusive)
int[] ans = new int[num + 1];
Arrays.fill(ans, 0);

if (num == 0) {
return ans;
}

for (int i = 1; i <= num; i++) {

// if ans[i] is even, share same quantity of 1s with ans[i>>1], also can be expressed ans[i / 2]
if ((i & 1) == 0) {
ans[i] = ans[i >> 1];
}
//if ans[i] is odd, the carries one more 1 than previous number ans[i-1]
else {
ans[i] = ans[i - 1] + 1;
}
}

return ans;
}
}
``````