Below is a collection of real-world, scenario-based JavaScript array programs designed for QA/SDET training, coding practice, and interviews. They progress from beginner to advanced.
1. Find the Highest Product Price
Scenario
An e-commerce application stores product prices. Find the most expensive product.
const prices = [1200, 4500, 2300, 8900, 1500];
let maxPrice = prices[0];
for (let price of prices) {
if (price > maxPrice) {
maxPrice = price;
}
}
console.log("Highest Price:", maxPrice);Expected Output:
Highest Price: 89002. Find the Lowest Product Price
const prices = [1200, 4500, 2300, 8900, 1500];
let minPrice = prices[0];
for (let price of prices) {
if (price < minPrice) {
minPrice = price;
}
}
console.log("Lowest Price:", minPrice);3. Calculate Total Shopping Cart Amount
Scenario
A shopping cart contains multiple product prices. Calculate the total.
const cart = [499, 1299, 799, 2499];
let total = 0;
for (let price of cart) {
total += price;
}
console.log("Cart Total:", total);Output:
Cart Total: 50964. Find Products Above ₹1000
const prices = [500, 1200, 2500, 700, 1800, 450];
const result = [];
for (let price of prices) {
if (price > 1000) {
result.push(price);
}
}
console.log(result);Output:
[1200, 2500, 1800]5. Remove Duplicate Product IDs
Scenario
During automation testing, duplicate product IDs are received from an API.
const productIds = [101, 102, 103, 101, 104, 102, 105];
const uniqueIds = [];
for (let id of productIds) {
if (!uniqueIds.includes(id)) {
uniqueIds.push(id);
}
}
console.log(uniqueIds);Output:
[101, 102, 103, 104, 105]6. Find Duplicate Values
const ids = [101, 102, 103, 101, 104, 102, 105];
const duplicates = [];
for (let i = 0; i < ids.length; i++) {
for (let j = i + 1; j < ids.length; j++) {
if (ids[i] === ids[j] && !duplicates.includes(ids[i])) {
duplicates.push(ids[i]);
}
}
}
console.log("Duplicates:", duplicates);Output:
Duplicates: [101, 102]7. Find Second Highest Salary
Scenario
An HR application stores employee salaries. Find the second-highest salary without sorting.
const salaries = [45000, 75000, 55000, 90000, 65000];
let highest = -Infinity;
let secondHighest = -Infinity;
for (let salary of salaries) {
if (salary > highest) {
secondHighest = highest;
highest = salary;
}
else if (salary > secondHighest && salary !== highest) {
secondHighest = salary;
}
}
console.log("Highest:", highest);
console.log("Second Highest:", secondHighest);Output:
Highest: 90000
Second Highest: 750008. Count Passed and Failed Students
Scenario
A training institute stores student marks. Determine how many students passed.
const marks = [85, 45, 72, 30, 90, 55, 28];
let passed = 0;
let failed = 0;
for (let mark of marks) {
if (mark >= 40) {
passed++;
} else {
failed++;
}
}
console.log("Passed:", passed);
console.log("Failed:", failed);9. Calculate Average Marks
const marks = [80, 75, 90, 65, 85];
let total = 0;
for (let mark of marks) {
total += mark;
}
const average = total / marks.length;
console.log("Average:", average);10. Find Students Who Scored Above Average
const marks = [80, 45, 90, 65, 85];
let total = 0;
for (let mark of marks) {
total += mark;
}
const average = total / marks.length;
const aboveAverage = [];
for (let mark of marks) {
if (mark > average) {
aboveAverage.push(mark);
}
}
console.log("Average:", average);
console.log("Above Average:", aboveAverage);11. Find Missing Test Case IDs
Scenario
A QA automation suite should execute test cases from 1 to 10, but some test cases are missing.
const executedTests = [1, 2, 3, 5, 6, 8, 10];
const missingTests = [];
for (let i = 1; i <= 10; i++) {
if (!executedTests.includes(i)) {
missingTests.push(i);
}
}
console.log("Missing Tests:", missingTests);Output:
Missing Tests: [4, 7, 9]12. Count Even and Odd Numbers
Scenario
An application receives transaction IDs and needs to classify them.
const transactionIds = [101, 202, 303, 404, 505, 606];
let even = 0;
let odd = 0;
for (let id of transactionIds) {
if (id % 2 === 0) {
even++;
} else {
odd++;
}
}
console.log("Even:", even);
console.log("Odd:", odd);13. Reverse an Array Without reverse()
const users = ["John", "David", "Mike", "Alex"];
const reversed = [];
for (let i = users.length - 1; i >= 0; i--) {
reversed.push(users[i]);
}
console.log(reversed);Output:
["Alex", "Mike", "David", "John"]14. Find a Specific User
const users = ["John", "David", "Mike", "Alex"];
const searchUser = "Mike";
if (users.includes(searchUser)) {
console.log("User Found");
} else {
console.log("User Not Found");
}15. Count Occurrence of a Value
Scenario
Find how many times a particular product was purchased.
const products = [
"Laptop",
"Mobile",
"Laptop",
"Tablet",
"Laptop",
"Mobile"
];
const searchProduct = "Laptop";
let count = 0;
for (let product of products) {
if (product === searchProduct) {
count++;
}
}
console.log("Laptop purchased:", count, "times");16. Find Common Elements Between Two Arrays
Scenario
Find users who are present in both applications.
const app1Users = ["John", "David", "Mike", "Alex"];
const app2Users = ["Mike", "Alex", "Robert", "Sam"];
const commonUsers = [];
for (let user of app1Users) {
if (app2Users.includes(user)) {
commonUsers.push(user);
}
}
console.log(commonUsers);Output:
["Mike", "Alex"]17. Find Unique Elements From Two Arrays
const teamA = ["John", "David", "Mike"];
const teamB = ["Mike", "Alex", "David"];
const result = [];
for (let user of [...teamA, ...teamB]) {
if (!result.includes(user)) {
result.push(user);
}
}
console.log(result);18. Find Failed Test Cases
Scenario
A Playwright test execution produces test statuses.
const statuses = [
"passed",
"failed",
"passed",
"skipped",
"failed",
"passed"
];
const failedTests = [];
for (let status of statuses) {
if (status === "failed") {
failedTests.push(status);
}
}
console.log("Failed Tests:", failedTests.length);19. Separate Positive and Negative Numbers
const numbers = [10, -5, 20, -8, 15, -2];
const positive = [];
const negative = [];
for (let number of numbers) {
if (number >= 0) {
positive.push(number);
} else {
negative.push(number);
}
}
console.log("Positive:", positive);
console.log("Negative:", negative);20. Move All Zeros to the End
Scenario
An API returns an array containing zero values. Move all zero values to the end.
const numbers = [0, 5, 0, 3, 8, 0, 2];
const result = [];
let zeroCount = 0;
for (let number of numbers) {
if (number === 0) {
zeroCount++;
} else {
result.push(number);
}
}
for (let i = 0; i < zeroCount; i++) {
result.push(0);
}
console.log(result);Output:
[5, 3, 8, 2, 0, 0, 0]21. Find First Non-Repeated Element
Scenario
Find the first unique transaction ID.
const ids = [101, 102, 101, 103, 102, 104];
for (let id of ids) {
let count = 0;
for (let value of ids) {
if (id === value) {
count++;
}
}
if (count === 1) {
console.log("First non-repeated ID:", id);
break;
}
}Output:
First non-repeated ID: 10322. Find Maximum Consecutive Number
const numbers = [10, 20, 30, 25, 50, 60];
let maxDifference = 0;
let firstNumber;
let secondNumber;
for (let i = 0; i < numbers.length - 1; i++) {
const difference = numbers[i + 1] - numbers[i];
if (difference > maxDifference) {
maxDifference = difference;
firstNumber = numbers[i];
secondNumber = numbers[i + 1];
}
}
console.log(firstNumber, secondNumber);23. Pagination Scenario
Scenario
An API returns 50 records. Display records for page 3 where each page contains 10 records.
const users = Array.from({ length: 50 }, (_, i) => `User-${i + 1}`);
const page = 3;
const pageSize = 10;
const startIndex = (page - 1) * pageSize;
const pageData = users.slice(
startIndex,
startIndex + pageSize
);
console.log(pageData);Output:
[
"User-21",
"User-22",
...
"User-30"
]24. Search Products by Keyword
const products = [
"iPhone 15",
"Samsung Galaxy",
"MacBook Pro",
"iPad Air",
"Samsung TV"
];
const keyword = "Samsung";
const result = products.filter(product =>
product.toLowerCase().includes(keyword.toLowerCase())
);
console.log(result);25. Apply Discount to Product Prices
Scenario
Apply a 10% discount to all products.
const prices = [1000, 2500, 5000, 7500];
const discountedPrices = prices.map(price => {
return price - (price * 10 / 100);
});
console.log(discountedPrices);Output:
[900, 2250, 4500, 6750]26. Find Products Within a Price Range
const prices = [500, 1200, 2500, 3500, 4500, 6000];
const min = 1000;
const max = 4000;
const result = prices.filter(price =>
price >= min && price <= max
);
console.log(result);Output:
[1200, 2500, 3500]27. Find the Most Frequent Element
Scenario
Find the product that appears most frequently in orders.
const products = [
"Laptop",
"Mobile",
"Laptop",
"Tablet",
"Mobile",
"Laptop"
];
let maxCount = 0;
let mostFrequent;
for (let product of products) {
let count = 0;
for (let value of products) {
if (product === value) {
count++;
}
}
if (count > maxCount) {
maxCount = count;
mostFrequent = product;
}
}
console.log("Most Frequent:", mostFrequent);
console.log("Count:", maxCount);28. Compare Expected and Actual Results
Scenario
This is especially useful for API/UI automation validation.
const expected = ["Login", "Dashboard", "Logout"];
const actual = ["Login", "Dashboard", "Logout"];
let isMatching = true;
if (expected.length !== actual.length) {
isMatching = false;
} else {
for (let i = 0; i < expected.length; i++) {
if (expected[i] !== actual[i]) {
isMatching = false;
break;
}
}
}
console.log("Result:", isMatching);29. Find Missing Values Between Two Arrays
const expected = [101, 102, 103, 104, 105];
const actual = [101, 103, 105];
const missing = [];
for (let id of expected) {
if (!actual.includes(id)) {
missing.push(id);
}
}
console.log("Missing:", missing);Output:
Missing: [102, 104]30. Flatten Nested Array
Scenario
An API returns nested categories.
const categories = [
["Electronics", "Mobile"],
["Furniture", "Chair"],
["Books", "Novel"]
];
const result = categories.flat();
console.log(result);Output:
[
"Electronics",
"Mobile",
"Furniture",
"Chair",
"Books",
"Novel"
]31. QA/SDET Scenario: Validate API Response IDs
const apiResponse = [
{ id: 101, name: "John" },
{ id: 102, name: "David" },
{ id: 103, name: "Mike" }
];
const ids = apiResponse.map(user => user.id);
const expectedIds = [101, 102, 103];
console.log(
JSON.stringify(ids) === JSON.stringify(expectedIds)
? "Test Passed"
: "Test Failed"
);32. QA/SDET Scenario: Find Failed Test Names
const testResults = [
{ name: "Login Test", status: "passed" },
{ name: "Search Test", status: "failed" },
{ name: "Checkout Test", status: "passed" },
{ name: "Payment Test", status: "failed" }
];
const failedTests = testResults
.filter(test => test.status === "failed")
.map(test => test.name);
console.log(failedTests);Output:
["Search Test", "Payment Test"]33. QA/SDET Scenario: Calculate Test Execution Summary
const results = [
"passed",
"passed",
"failed",
"skipped",
"passed",
"failed",
"passed"
];
const summary = {
passed: 0,
failed: 0,
skipped: 0
};
for (let result of results) {
summary[result]++;
}
console.log(summary);Output:
{
passed: 4,
failed: 2,
skipped: 1
}34. Find Duplicate Test Case IDs
const testCases = [
"TC001",
"TC002",
"TC003",
"TC001",
"TC004",
"TC002"
];
const duplicates = [];
for (let i = 0; i < testCases.length; i++) {
for (let j = i + 1; j < testCases.length; j++) {
if (
testCases[i] === testCases[j] &&
!duplicates.includes(testCases[i])
) {
duplicates.push(testCases[i]);
}
}
}
console.log("Duplicate Test Cases:", duplicates);Output:
Duplicate Test Cases: ["TC001", "TC002"]35. Advanced Interview Scenario: Find Two Numbers Whose Sum Equals Target
Scenario
Find two product prices whose total is ₹3000.
const prices = [500, 1200, 1800, 2500, 1500];
const target = 3000;
for (let i = 0; i < prices.length; i++) {
for (let j = i + 1; j < prices.length; j++) {
if (prices[i] + prices[j] === target) {
console.log(
prices[i],
"+",
prices[j],
"=",
target
);
}
}
}Output:
1200 + 1800 = 3000
1500 + 1500 = 3000Interview Practice Set
For your JavaScript/Playwright/SDET training sessions, these are particularly good interview exercises:
Beginner
- Find maximum number.
- Find minimum number.
- Calculate array sum.
- Calculate average.
- Count even and odd numbers.
- Reverse an array.
- Search an element.
- Find positive and negative numbers.
- Remove duplicates.
- Count occurrences.
Intermediate
- Find second maximum without sorting.
- Find duplicate elements.
- Find missing numbers.
- Find common elements between arrays.
- Find unique elements.
- Move zeros to the end.
- Find first non-repeated element.
- Find most frequent element.
- Find elements above average.
- Find values within a range.
Advanced
- Two-sum problem.
- Compare two arrays.
- Find array intersection.
- Find array difference.
- Flatten nested arrays.
- Group test results.
- Validate API response arrays.
- Find duplicate test case IDs.
- Implement pagination using arrays.
- Process and summarize test execution results.