LeetCode 338. Counting Bits

Description

https://leetcode.com/problems/counting-bits/

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++)?

Explanation

Convert number to binary strings and count ‘1’s.

Python Solution

class Solution:
    def countBits(self, n: int) -> List[int]:
        
        results = []
        
        for i in range(n + 1):
            bin_str = bin(i)[2:]
            
            one_count = bin_str.count('1')
            results.append(one_count)
        
        return results    
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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