Leetcode 191. Number of 1 Bits. Python

191. Number of 1 Bits

Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).

Note:

Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It 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 3, the input represents the signed integer. -3.

Example 1:

Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Constraints:

The input must be a binary string of length 32.

class Solution:
    def hammingWeight(self, n: int) -> int:
        
        #trick 1
        
        ans = 0
        while n:
            ans += n % 2
            n = n >> 1
        return ans
    
    
        # trick 2
        
        ans = 0
        while n:
            n &=(n-1)
            ans += 1
        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 :

Number of Connected Components in an Undirected Graph (Python)

66. Number of Connected Components in an Undirected Graph Question Link : check here Givennnodes labeled from0ton - 1and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph. Example 1: 0 3 | | 1 --- 2 4 Givenn = 5andedges = [[0, 1], [1, 2], [3, 4]], return2. Example 2: 0 4 | | 1 --- 2 --- 3 Givenn = 5andedges = [[0, 1], [1, 2], [2, 3], [3, 4]], return1. Note: You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges. Solution : class Solution: def counComponents(self, n: int, edges : List[List[int]]) -> i...

LEETCODE SOLUTIONS

Search This Blog