338. Counting Bits

Problem

Tags: Dynamic Programming, Bit Manipulation

Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i.

Example 1:

Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10

Example 2:

Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

Constraints:

• 0 <= n <= 10^5

Follow up:

• It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
• Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

Code

C

// 338. Counting Bits (6/11/54133)
// Runtime: 44 ms (86.91%) Memory: 10.69 MB (0.00%) 

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
typedef int32_t i32;

i32* countBits (i32 n, i32* size) {
    *size = n + 1;
    
    i32* results = malloc(*size * sizeof(i32));
    for (i32 i = 0; i <= n; i++) {
        results[i] = __builtin_popcount(i);
    }
    
    return results;
}