Counting Bits(Medium)

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:

For num = 5 you should return [0,1,1,2,1,2].

Follow UP:

Space complexity should be O(n).

public int[] countBits(int num) {
        int dp[] = new int[num + 1];

        // dp[0] is initialized as 0
        dp[0] = 0;

        // If i is even then number of bits is same as i/2
        // Else if i is odd then number of bits is +1 as that of i/2
        for (int i = 1 ; i <= num ; i++) {
            if (i % 2 == 0) {
                dp[i] = dp[i/2];
            } else {
                dp[i] = dp[i/2] + 1;
            }
        }

        return dp;
    }