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Leetcode 3226. Number of Bit Changes to Make Two Integers Equal

Problem Statement

You are given two integers, start and goal, and you need to determine the number of bit changes required to convert start to goal. A bit change means flipping a bit from 0 to 1 or from 1 to 0.

Clarifying Questions

  1. Range of Values:
    • What is the range of values for start and goal?
      • Typically, they are within the range of integers in Java, i.e., -2,147,483,648 to 2,147,483,647.
  2. Binary Representation:
    • Do I need to consider the two numbers in their 32-bit binary form, including signs?
      • For simplicity, considering their 32-bit representation will suffice.
  3. Leading Zeros:
    • Should leading zeros be considered in the bit representations?
      • Yes, in a typical 32-bit representation, leading zeros should indeed be considered.

Strategy

  1. Bitwise XOR Operation:
    • Utilize the XOR operation between start and goal, as XOR between two bits results in 1 if they are different and 0 if they are the same.
    • Counting the number of 1s in the result of the XOR will give the number of bit changes required.
  2. Count Set Bits:
    • Implement a method to count the number of 1s (set bits) in a binary representation of the XOR result. This can be done using Integer.bitCount() in Java.

Code

Here’s the implementation in Java:

public class BitChangeCounter {

    public static int bitChangesToEqual(int start, int goal) {
        // XOR the two numbers; different bits between two numbers will be 1
        int xorResult = start ^ goal;
        
        // Count the number of 1s in the binary representation of xorResult
        return Integer.bitCount(xorResult);
    }

    public static void main(String[] args) {
        int start = 29;    // Example start value
        int goal = 15;     // Example goal value
        
        System.out.println("Number of bit changes required: " + bitChangesToEqual(start, goal));
    }
}

Explanation of the Code

  1. XOR Operation:
    • int xorResult = start ^ goal;
    • This line calculates the XOR of start and goal. Each bit in the result is 1 if the corresponding bits of start and goal differ.
  2. Counting Set Bits:
    • return Integer.bitCount(xorResult);
    • This line counts the number of 1s in the xorResult. Integer.bitCount() is a utility method in Java that efficiently counts the number of 1-bits.
  3. Main Method:
    • Contains a simple example to demonstrate the function.

Time Complexity

This solution is both efficient and straightforward due to the bitwise operations involved.

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