338-counting-bits

Try it on leetcode

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

 

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++)?

Solution(Python)

class Solution:
    def countBits(self, n: int) -> List[int]:
        return self.dynamicprogramming(n)

    # Time Complexity :O(nlogn)
    # Space Complexity: O(n)
    def naive(self, n: int) -> List[int]:
        return [self.countones(num) for num in range(0, n + 1)]

    # Time Complexity :O(n)
    # Space Complexity: O(n)
    def dynamicprogramming(self, n: int) -> List[int]:
        dp = [0] * (n + 1)
        offset = 1
        for index in range(1, n + 1):
            if index == offset * 2:
                offset *= 2
            dp[index] = dp[index - offset] + 1
        return dp

    def countones(self, num):
        cnt = 0
        while num:
            cnt += 1
            num &= num - 1
        return cnt