# 338. Counting Bits

## Description

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++)?

## Solutions

• class Solution {
public int[] countBits(int n) {
int[] ans = new int[n + 1];
for (int i = 1; i <= n; ++i) {
ans[i] = ans[i & (i - 1)] + 1;
}
return ans;
}
}

• class Solution {
public:
vector<int> countBits(int n) {
vector<int> ans(n + 1);
for (int i = 0; i <= n; ++i) {
ans[i] = __builtin_popcount(i);
}
return ans;
}
};

• class Solution:
def countBits(self, n: int) -> List[int]:
ans = [0] * (n + 1)
for i in range(1, n + 1):
ans[i] = ans[i & (i - 1)] + 1
return ans


• func countBits(n int) []int {
ans := make([]int, n+1)
for i := 1; i <= n; i++ {
ans[i] = ans[i&(i-1)] + 1
}
return ans
}

• function countBits(n: number): number[] {
const ans: number[] = Array(n + 1).fill(0);
for (let i = 1; i <= n; ++i) {
ans[i] = ans[i & (i - 1)] + 1;
}
return ans;
}