Number of 1 Bits LeetCode

 Given a positive integer n, write a function that returns the number of set bits in its binary representation (also known as the Hamming weight).


 


Example 1:


Input: n = 11


Output: 3


Explanation:


The input binary string 1011 has a total of three set bits.


Example 2:


Input: n = 128


Output: 1

Explanation:

The input binary string 10000000 has a total of one set bit.

Example 3:

Input: n = 2147483645


Output: 30


Explanation:

The input binary string 1111111111111111111111111111101 has a total of thirty set bits.

Constraints:

1 <= n <= 231 - 1

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



Approach 1: Using Built in functiion 

class Solution {
    public int hammingWeight(int n) {
       
         int count = Integer.bitCount(n);
         return count;
    }
}


Approach 2 :


📘 Explanation:

TermMeaning
1 << iShifts 1 to the left by i bits to create a mask like 00001000
n & maskChecks if the i-th bit in n is 1
!= 0If the result is not zero, it means that bit was a 1
class Solution {
    public int hammingWeight(int n) {
        int count = 0;  // To keep track of number of 1s

        // Loop through all 32 bits of the integer
        for (int i = 0; i < 32; i++) {

            // Create a mask by left-shifting 1 by i positions
            // This helps check each bit from 0th to 31st
            int mask = 1 << i;

            // Use bitwise AND to check if the ith bit of n is 1
            if ((n & mask) != 0) {
                count++;  // If the result is not 0, that bit is set (1)
            }
        }

        // Return the total number of set bits
        return count;
    }
}


public class codefile{
    static int solve(String s){
       
int n = Integer.parseUnsignedInt(s, 2);
//return Integer.bitCount(n); // using built in function
int count=0;
for(int i=0;i<32;i++)
{
    int mask=1<<i;

    if((mask & n) !=0)
    {
        count++;
    }
}
return count;
    }
}

Comments

Post a Comment

Popular posts from this blog

Two Sum II - Input Array Is Sorted

 Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order , find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length. Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2. The tests are generated such that there is exactly one solution. You may not use the same element twice. Your solution must use only constant extra space .   Example 1: Input: numbers = [2,7,11,15], target = 9 Output: [1,2] Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2]. Example 2: Input: numbers = [2,3,4], target = 6 Output: [1,3] Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3]. Example 3: Input: numbers = [-1,0], target = -1 Output: [1,2] Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1,...
Read more

Comparable Vs. Comparator in Java

  In Java , Comparable and Comparator both are interfaces used to sort objects. Comparable It defines  natural ordering  within a class. It is implemented  inside  the class being sorted. It uses the compareTo() method. It allows sorting  by one one . Comparable provides a  single sorting sequence . In other words, we can sort the collection on the basis of a single element such as id, name, and price. Comparable  affects the original class , i.e., the actual class is modified. Comparable is present in  java.lang  package. We can sort the list elements of Comparable type by  Collections.sort(List)  method. Only one compareTo() method can be implemented, restricting flexibility. Comparator It defines  custom sorting logic It is implemented  in a separate class  or using lambda expressions . It uses the compare() method. It allows sorting  by multiple criteria . The Comparator provides  multiple sorting...
Read more

Increasing Triplet Subsequence

 Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.   Example 1: Input: nums = [1,2,3,4,5] Output: true Explanation: Any triplet where i < j < k is valid. Example 2: Input: nums = [5,4,3,2,1] Output: false Explanation: No triplet exists. Example 3: Input: nums = [2,1,5,0,4,6] Output: true Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6. Brute Force : Failed for some testcase class Solution {     public boolean increasingTriplet ( int [] nums ) {         int n = nums . length ;         int i = 0 ;         int j =i+ 1 ;         int k =j+ 1 ;         while (i <n && j<n && k<n)         {         if (nums...
Read more