338. Counting Bits

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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:

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

Example 2:

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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++)?

Hints/Notes

  • 2025/01/06
  • bit operation
  • No solution from 0x3F

Solution

Language: C++

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class Solution {
public:
    vector<int> countBits(int n) {
        vector<int> res(n + 1, 0);
        for (int i = 1; i <= n; i++) {
            res[i] = res[(i >> 1)] + i % 2;
        }
        return res;
    }
};