ivy Sdet Automation Test Engineer Interview Questions and Answers

Q1. Work Experience?
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Q2. Questions Regarding CV ?
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Q3. Question on Xpath (Selenium)
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Q4. Questions on Python basic coding
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Q5. Questions on API Testing
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Q1) Write a functions to perform all CRUD in api.
Q2) Given two sorted list
ar1 =[1,2,3,4,7,7,12,18,19]
ar2 = [3, 4, 7, 7,14, 18, 19]

You can add and find largest sum but you can switch when ar1 and ar2 have same numbers.

Solution: you can base your solution on merge algo and recursively call your function to return largest sum possible.

Coding test on hackerrank. Solve 30 day code challange on haackerrank and you will be able to clear it.

Frequency array, java collection, missing number

General Aptitude test was conducted

Q1. String related problem and code the solution. Some Mongodb questions
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Q1. OOPS concept, testing methods and other QA related questions
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Q1. Test Scenarios discussion, Selenium basics and database problems
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Q1. In depth technical discussion of current projects
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Q1. Cultural fitment round and basic personal questions
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Logic based questions

Tough and easy at the same time

Aptitude and coding round, one coding question and others are computer fundamentals

Q1. Return true if Subarray is present whose sum is 0
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Q2. Project, oops, dbms
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Q1. Subarray sun equal to k
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Q2. Project, oops, dbms
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Q1. Did not got slected i answered all still
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Q1. Intro of yourself
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Q1. LOCATE X PATH , GET WINDOW HANDLE
Ans. 
Locate element using Xpath and retrieve window handle in automation testing.
• Use findElement method with By.xpath to locate the element using Xpath.

• Use getWindowHandle method to retrieve the window handle.
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Q2. TYPES OF WAITS WITH THEIR SYNTAX
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Types of waits in automation testing with their syntax
• Implicit Wait: driver.manage().timeouts().implicitlyWait(10, TimeUnit.SECONDS);

• Explicit Wait: WebDriverWait wait = new WebDriverWait(driver, 10); wait.until(ExpectedConditions.visibilityOfElementLocated(By.id("elementID")));

• Fluent Wait: Wait wait = new FluentWait(driver).withTimeout(Duration.ofSeconds(10)).pollingEvery(Duration.ofSeconds(2)).ignoring(NoSuchElementEx
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