Binary Watch Problem LeetCode

Problem: A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59). Each LED represents a zero or one, with the least significant bit on the right.

• For example, the below binary watch reads "4:51".

Given an integer turnedOn which represents the number of LEDs that are currently on, return all possible times the watch could represent. You may return the answer in any order.

The hour must not contain a leading zero.

• For example, "01:00" is not valid. It should be "1:00".

The minute must be consist of two digits and may contain a leading zero.

• For example, "10:2" is not valid. It should be "10:02".

 Example 1:

Input: turnedOn = 1
Output: ["0:01","0:02","0:04","0:08","0:16","0:32","1:00","2:00","4:00","8:00"]

Example 2:

Input: turnedOn = 9
Output: []

Constraints:

• 0 <= turnedOn <= 10

Solution:

LeetCode: Hamming Distance

// The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

// Given two integers x and y, calculate the Hamming distance.

// Note:
// 0 ≤ x, y < 2^31.

// Example:

// Input: x = 1, y = 4

// Output: 2

// Explanation:
// 1   (0 0 0 1)
// 4   (0 1 0 0)
//        ↑   â†‘

// The above arrows point to positions where the corresponding bits are different.

public class HammingDistance {
    public int hammingDistance(int x, int y) {
        return Integer.bitCount(x ^ y);
    }
}

LeetCode: Number of One Bits

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

// For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.

public class NumberOfOneBits {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        if(n == 0) {
            return 0;
        }
     
        int count = 0;
     
        while(n != 0) {
            count += (n) & 1;
            n >>>= 1;
        }
     
        return count;
    }
}

LeetCode: Sum of Two Integers

// Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

// Example:
// Given a = 1 and b = 2, return 3.

public class SumOfTwoIntegers {
    public int getSum(int a, int b) {
        if(a == 0) {
            return b;
        }

        if(b == 0) {
            return a;
        }
     
        while(b != 0) {
            int carry = a & b;
            a = a ^ b;
            b = carry << 1;
        }
     
        return a;
    }
}