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 217. Contains Duplicate. Python (Easiest Approach ✅)

217 . Contains Duplicate Given an integer array nums , return true if any value appears at least twice in the array, and return false if every element is distinct. Example 1: Input: nums = [1,2,3,1] Output: true Example 2: Input: nums = [1,2,3,4] Output: false Example 3: Input: nums = [1,1,1,3,3,4,3,2,4,2] Output: true Constraints: 1 <= nums.length <= 10 5 -10 9 <= nums[i] <= 10 9 class Solution: def containsDuplicate(self, nums: List[int]) -> bool: hs = set() for n in nums: if n in hs: return True hs.add(n) return False Explaination :

Leetcode 322. Coin Change. Python (Greedy? vs DP?)

322 . Coin Change You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount . If that amount of money cannot be made up by any combination of the coins, return -1 . You may assume that you have an infinite number of each kind of coin. Example 1: Input: coins = [1,2,5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1 Example 2: Input: coins = [2], amount = 3 Output: -1 Example 3: Input: coins = [1], amount = 0 Output: 0 Constraints: 1 <= coins.length <= 12 1 <= coins[i] <= 2 31 - 1 0 <= amount <= 10 4 Solution : class Solution: def coinChange(self, coins: List[int], amount: int) -> int: dp = [amount + 1] * (amount + 1) #[0...7] dp[0] = 0 ...

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