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

Leetcode 424. Longest Repeating Character Replacement. Python (Sliding Window)

424 . Longest Repeating Character Replacement You are given a string s and an integer k . You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times. Return the length of the longest substring containing the same letter you can get after performing the above operations . Example 1: Input: s = "ABAB", k = 2 Output: 4 Explanation: Replace the two 'A's with two 'B's or vice versa. Example 2: Input: s = "AABABBA", k = 1 Output: 4 Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA". The substring "BBBB" has the longest repeating letters, which is 4. Constraints: 1 <= s.length <= 10 5 s consists of only uppercase English letters. 0 <= k <= s.length Solution : class Solution: def characterReplacement(self, s: str, k: int) -> int: hm = {} ans = 0

Leetcode 371. Sum of Two Integers. C++ / Java

371 . Sum of Two Integers Given two integers a and b , return the sum of the two integers without using the operators + and - . Example 1: Input: a = 1, b = 2 Output: 3 Example 2: Input: a = 2, b = 3 Output: 5 Constraints: -1000 <= a, b <= 1000 Solution : C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java : class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :

Leetcode 242. Valid Anagram. Python (3 Ways)

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