• Question
• Solution

Question

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.

Hint:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

Solution

Result: Accepted Time: 40 ms

Here should be some explanations.

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* countBits(int num, int* returnSize) {
    int * ary = malloc((num+2)*sizeof(int));
    *returnSize = num + 1;
    int ptr = 1;
    int cnt = 2;
    ary[0] = 0;
    ary[1] = 1;
    int limit = num / 2;
    while(ptr <= limit)
    {
        ary[(ptr<<1)] = ary[ptr];
        ary[(ptr<<1)|1] = ary[ptr]+1;
        ptr++;
    }
    return ary;
}

Complexity Analytics

• Time Complexity: $O(n)$
• Space Complexity: $O(n)$