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 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 <= 10 5 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 ...

Leetcode 347. Top K Frequent Elements. Python (Bubble Sort)

Top K Frequent Elements Given an integer array nums and an integer k , return the k most frequent elements . You may return the answer in any order . Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Example 2: Input: nums = [1], k = 1 Output: [1] Constraints: 1 <= nums.length <= 10 5 -10 4 <= nums[i] <= 10 4 k is in the range [1, the number of unique elements in the array] . It is guaranteed that the answer is unique . Follow up: Your algorithm's time complexity must be better than O(n log n) , where n is the array's size. Solution : class Solution: def topKFrequent(self, n: List[int], k: int) -> List[int]: # [1,1,1,2,2,3] & k = 2 f = [[] for i in range(len(n) + 1)] # f = [...

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