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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0191_number_of_1_bits;

// #Easy #Top_Interview_Questions #Bit_Manipulation #Algorithm_I_Day_13_Bit_Manipulation
// #Programming_Skills_I_Day_2_Operator #Udemy_Bit_Manipulation
// #2022_06_28_Time_1_ms_(84.87%)_Space_41.8_MB_(10.40%)

/**
 * 191 - Number of 1 Bits\.
 *
 * Easy
 *
 * Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the [Hamming weight](http://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Two%27s_complement). 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`.
 *
 * **Follow up:** If this function is called many times, how would you optimize it?
**/
public class Solution {
    public int hammingWeight(int n) {
        int sum = 0;
        boolean flag = false;
        if (n < 0) {
            flag = true;
            n = n - Integer.MIN_VALUE;
        }
        while (n > 0) {
            int k = n % 2;
            sum += k;
            n /= 2;
        }
        return flag ? sum + 1 : sum;
    }
}