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# Counting Bits LeetCode Solution

## Problem – Counting Bits LeetCode Solution

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`

• 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++)?

## Counting Bits LeetCode Solution in C++

``````class Solution {
public:
vector<int> countBits(int n) {
vector<int> ans;

// iterating fromt 0 to n
for(int i = 0; i<=n; i++)
{
// sum is initialised as 0
int sum = 0;
int num = i;
// while num not equals zero
while(num != 0)
{
// we have to count 1's in binary representation of i, therefore % 2
sum += num%2;
num = num/2;
}
// add sum to ans vector
ans.push_back(sum);
}
// return
return ans;
}
};
``````

## Counting Bits LeetCode Solution in Java

``````public int[] countBits(int num) {
int[] f = new int[num + 1];
for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
return f;
}``````

## Counting Bits LeetCode Solution in Python

``````def countBits(self, num: int) -> List[int]:
counter = 
for i in range(1, num+1):
counter.append(counter[i >> 1] + i % 2)
return counter
``````
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