counting-bits.cpp

//question//

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 <= 105
 

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++)?


/solutuion//

class Solution {
public:
    int counting(int n)
    {
        int count=0;
        while(n>0)
        {
            int rem = n%2;
            if(rem==1)
            {
                count++;
            }
            n = n/2;
        }
        return count;
    }
    vector<int> countBits(int n) {
        vector<int> v;
        for(int i=0;i<=n;i++)
        {
            v.push_back(counting(i));
        }
        return v;
    }
};