Leetcode 338. Counting Bits. Python (Bubble Sort)

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 :

class Solution:

def countBits(self, n: int) -> List[int]:

dp = [0] * (n+1)

c = 1

for i in range(1, n+1):

if c * 2 == i :

c = i

dp[i] = 1 + dp[i-c]

return dp

Explained :

Comments

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Longest Substring Without Repeating Characters - Leetcode 3 - Python

Given a string s, find the length of the longest substring without repeating characters. class Solution: def lengthOfLongestSubstring(self, s: str) -> int: charSet = set() left = 0 ans = 0 for right in range(len(s)): while s[right] in charSet: charSet.remove(s[left]) left+=1 charSet.add(s[right]) ans = max(ans, right-left+1) return ans Explained :

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