Leetcode 190. Reverse Bits. Python. Bit Manipulation

190. Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

Note:

Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

Example 1:

Input: n = 00000010100101000001111010011100
Output:    964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input: n = 11111111111111111111111111111101
Output:   3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

Constraints:

The input must be a binary string of length 32

Follow up: If this function is called many times, how would you optimize it?

class Solution:
    def reverseBits(self, n: int) -> int:
        ans = 0
        
        for i in range(32):
            b = (n>>i) & 1
            ans = ans | (b << (31-i))
        
        return ans

Explained :

Comments

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 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 ...

Container with Most Water - Leetcode 11 - Python

You are given an integer array height of length n . There are n vertical lines drawn such that the two endpoints of the i th line are (i, 0) and (i, height[i]) . Find two lines that together with the x-axis form a container, such that the container contains the most water. Return the maximum amount of water a container can store . Notice that you may not slant the container. Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49. class Solution: def maxArea(self, height: List[int]) -> int: # ans = 0 # for l in range(len(height)): # for r in range(l+1, len(height)): # ...

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