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First Bad Version

Published
2 min read
First Bad Version
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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.

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which returns whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

LeetCode Problem - 278: Link | Click Here

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        // Initialize low, high, and result variables
        int low = 1;
        int high = n;
        int result = n;

        // Perform binary search
        while (low <= high) {
            // Calculate mid point
            int mid = low + (high - low) / 2;

            // Check if the mid version is a bad version
            if (isBadVersion(mid)) {
                // Update the result to the current mid, and search in the left half
                result = mid;
                high = mid - 1;
            } else {
                // If the mid version is not bad, search in the right half
                low = mid + 1;
            }
        }

        // Return the result, which represents the first occurrence of a bad version
        return result;
    }
}

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