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Minimum Bit Flips to Convert Number

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Minimum Bit Flips to Convert Number
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As a Systems Engineer at Tata Consultancy Services, I deliver exceptional software products for mobile and web platforms, using agile methodologies and robust quality maintenance. I am experienced in performance testing, automation testing, API testing, and manual testing, with various tools and technologies such as Jmeter, Azure LoadTest, Selenium, Java, OOPS, Maven, TestNG, and Postman.

I have successfully developed and executed detailed test plans, test cases, and scripts for Android and web applications, ensuring high-quality standards and user satisfaction. I have also demonstrated my proficiency in manual REST API testing with Postman, as well as in end-to-end performance and automation testing using Jmeter and selenium with Java, TestNG and Maven. Additionally, I have utilized Azure DevOps for bug tracking and issue management.

A bit flip of a number x is choosing a bit in the binary representation of x and flipping it from either 0 to 1 or 1 to 0.

  • For example, for x = 7, the binary representation is 111 and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get 110, flip the second bit from the right to get 101, flip the fifth bit from the right (a leading zero) to get 10111, etc.

Given two integers start and goal, return the minimum number of bit flips to convert start to goal.

LeetCode Problem - 2220

class Solution {
    // Method to determine the minimum number of bit flips required to convert 'start' to 'goal'
    public int minBitFlips(int start, int goal) {
        // Calculate the XOR of 'start' and 'goal'
        // XOR operation will result in a number where each bit is set to 1 if the corresponding bits of 'start' and 'goal' are different
        int xor = start ^ goal;

        // Count the number of 1s in the XOR result
        // Each 1 represents a bit that needs to be flipped to transform 'start' into 'goal'
        int bitFlips = Integer.bitCount(xor);

        // Return the number of bit flips required
        return bitFlips;
    }
}

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