LeetCode - Counting Bits

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

Follow up:

• 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.

Solution:

public class CountingBits {

    /**
     * Function II Dynamic Programming
     * @param num
     * @return bits
     */
    public int[] countBits(int num) {
        int[] bits = new int[num + 1];
        for (int i = 0; i <= num; i++) {
            if (i % 2 == 0) {   // Even Number
                bits[i] = bits[i / 2];
            } else {
                bits[i] = bits[i / 2] + 1; // Odd Number
            }
        }
        return bits;
    }

    /**
     * Function I Bit Manipulation
     * @param n input number
     * @return bit
     */
    private int countBit(int n) {
        int bit = 0;
        while (n > 0) {
            if ((n & 1) == 1)
                bit++;
            n = n >> 1;
        }
        return bit;
    }

}

GitHub:LeetCode