วีดีโอ

You're a legend sir. Thank you so much. Fuck this bloody error

Cant we do 1st prob without hashmap

Question 49 & 50 are the couldn't cover unfortunately are they coding questions or just mcqs ?

Today i have completed coding round Not all the test cases passed for 2 codes 70 percent completed do i get nextround invite?

Is this mandatory to pass for offer, because i got cleared technical, manager and HR rounds and now received for codility test. Shall we do on mobile also?

Nice question

Clean solution :)

Did you pass?

import java.io.*; import java.util.*; class Result { /* * Complete the 'getProductSuggestions' function below. * * The function is expected to return a 2D_STRING_ARRAY. * The function accepts following parameters: * 1. STRING_ARRAY products * 2. STRING search */ public static List<List<String>> getProductSuggestions(List<String> products, String search) { // Sort the products list lexicographically Collections.sort(products); List<List<String>> suggestions = new ArrayList<>(); String prefix = ""; for (char c : search.toCharArray()) { prefix += c; List<String> matches = new ArrayList<>(); // Use binary search to quickly find the starting point of the prefix int start = Collections.binarySearch(products, prefix); // If exact match isn't found, binarySearch returns a negative value // The negative value gives the insertion point where the prefix would be if (start < 0) { start = -start - 1; } // Collect up to three lexicographically smallest matches for (int i = start; i < products.size() && matches.size() < 3; i++) { if (products.get(i).startsWith(prefix)) { matches.add(products.get(i)); } else { break; // Since the list is sorted, no need to check further if the prefix doesn't match } } suggestions.add(matches); } return suggestions; } } public class Solution { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); int productsCount = Integer.parseInt(bufferedReader.readLine().trim()); List<String> products = new ArrayList<>(); for (int i = 0; i < productsCount; i++) { String productsItem = bufferedReader.readLine(); products.add(productsItem); } String search = bufferedReader.readLine(); List<List<String>> result = Result.getProductSuggestions(products, search); for (int i = 0; i < result.size(); i++) { for (int j = 0; j < result.get(i).size(); j++) { bufferedWriter.write(result.get(i).get(j)); if (j != result.get(i).size() - 1) { bufferedWriter.write(" "); } } if (i != result.size() - 1) { bufferedWriter.write(" "); } } bufferedWriter.newLine(); bufferedReader.close(); bufferedWriter.close(); } } Explanation: Sorting the Products: The list of products is sorted lexicographically to ensure that we can easily find the smallest matches. Iterating Through Search Prefixes: For each character in the search string, we build a prefix and find the matching products that start with this prefix. Binary Search for Optimization: We use Collections.binarySearch to quickly locate the starting point for matching products with the current prefix. Adding Matches: We collect up to three matches for each prefix and add them to the result list. Output Formatting: Finally, we format the output as a list of lists where each inner list contains the matching products for the corresponding prefix of the search string. This solution is efficient and leverages the sorted nature of the products list to quickly find the smallest lexicographical matches.

Thank you for uploading this sirr

Do we have to keep our camera switched on while giving the test?

Intersting, for what headquarters did you apply? or what country?

class Solution { public int solution(String S) { int patches = 0; int i = 0; while (i < S.length()) { if (S.charAt(i) == 'X') { // Place a patch covering segments i, i+1, and i+2 patches++; i += 3; } else { i++; } } return patches; } } Explanation: Initialize patches to 0, which will count the number of patches used. Initialize i to 0, which will be used to traverse the string. While i is less than the length of the string S: If the current character S.charAt(i) is 'X', it means we need to place a patch. Increment the patches counter by 1 and move i by 3 positions to the right because the patch covers the current segment and the next two segments. If the current character is not 'X', just move i by 1 position to the right. This algorithm ensures that you cover all potholes with the minimum number of patches. The time complexity is O(N), where N is the length of the string S, making it efficient for the given constraints.

I love you brother ❤❤❤

your code is not running, whatever you saved in github. please share proper code. Showing nullpointerexception

You did not show test cases passed or not?

Thanks it helped alot!

In the foo(1234) question, the correct answer is 4321001234

YES IT WORK ❤❤❤❤

Is this for freshers too

Hii bro 1) ques 10 test case passed and for 2 ) 4 passed is there any chance for next round?

what was the time allocated for the test?

They didnt given any option to turn on the camera or mic,we are directly taking the test right .then how can this test is proctored

Tq brother 😢❤

Thank you it worked✌✌

Hello

Hi can you plz provide questions asked in next technical round

How much time was allocated for screening?

How much time it will take for result after test?

I have assessment for the role of application developer at IBM what coding question i can expect. please reply

Do they send any assessment completion mail after the submission?... I have given this assessment but i haven't received any completion mail (like the assessment is completed or something) and also the link for this test is opening again.

How to solve this on mac os

============================================================================== Task 1: ============================================================================== you are working on system that helps you to manage tasks on a todo list. your job is to implement an endpoint that allows these tasks to updated. In order to achieve this, you should use the Hibernate and spring frameworks. Application >Spring boot application version 3.0 >spring-boot-starter-data-jpa and spring-boot-starter-wed modules included. >Database access id fully configured. >Hibernate version latest Database Tasks are stored in a relational database. the database schema lookes like this CREATE TABLE task( id bigint NOT NULL, description varchar(200) NOT NULL, priority bigint, PRIMARY KEY(ID) ) ; Endpoint Sending a request: PUT /tasks/{id} with request body: { "description": "task's description" "priority": 5 } should update the description and priority of the task with id=ID. Changes should be stored in the database. your tasks: configure the tasks class as a Hibernate entity implement an endpoint as described above, Furthermore, it should meet the following requirements: >endpoint should return error code 404 if a task with the given id does not exist. In response , the server should return the following JSON: { "message": "cannot find the task with given id" "status":404 } >Endpoint should return error code 400 if the task description is null(or key "description" is not present in the request JSON). In response, the server should return the following JSON: { "message": " task description is required" "status":400 } >Endpoint should return error code 200 when the task is successfully updated. the response should be the same as the request body, for example: { "description": "task's description" "priority": 5 } ======================================================= Solution: ========================================================= Create the Task Entity: Annotate the Task class with JPA annotations to map it to the database table. Create the TaskRepository Interface: Extend JpaRepository to provide CRUD operations for the Task entity. Create the TaskController Class: Implement the PUT endpoint to update the task. Add error handling for the conditions mentioned (task not found and invalid request). Application Configuration: Ensure the application is correctly set up with the necessary Spring Boot and Hibernate configurations. =-----------------------------------------------------------==--------------------------------------------- 1. Task Entity Note: In my case I did not clear task one and was having errors plz you pleople do not copy paste. copy pasting is not allowed, even answer of task 1 is partially correct , modify it based on the requirement, and Task 2 is 100% correct import javax.persistence.Entity; import javax.persistence.GeneratedValue; import javax.persistence.GenerationType; import javax.persistence.Id; @Entity public class Task { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private Long id; private String description; private Long priority; // Getters and Setters public Long getId() { return id; } public void setId(Long id) { this.id = id; } public String getDescription() { return description; } public void setDescription(String description) { this.description = description; } public Long getPriority() { return priority; } public void setPriority(Long priority) { this.priority = priority; } } 2. TaskRepository import com.example.demo.entity.Task; import org.springframework.data.jpa.repository.JpaRepository; public interface TaskRepository extends JpaRepository<Task, Long> { } 3. TaskController import com.example.demo.entity.Task; import com.example.demo.repository.TaskRepository; import org.springframework.beans.factory.annotation.Autowired; import org.springframework.http.ResponseEntity; import org.springframework.web.bind.annotation.*; import java.util.Optional; @RestController @RequestMapping("/tasks") public class TaskController { @Autowired private TaskRepository taskRepository; @PutMapping("/{id}") public ResponseEntity<?> updateTask(@PathVariable Long id, @RequestBody Task task) { if (task.getDescription() == null) { return ResponseEntity.badRequest().body(new ErrorResponse("task description is required", 400)); } Optional<Task> existingTaskOptional = taskRepository.findById(id); if (!existingTaskOptional.isPresent()) { return ResponseEntity.status(404).body(new ErrorResponse("cannot find the task with given id", 404)); } Task existingTask = existingTaskOptional.get(); existingTask.setDescription(task.getDescription()); existingTask.setPriority(task.getPriority()); taskRepository.save(existingTask); return ResponseEntity.ok(task); } static class ErrorResponse { private String message; private int status; public ErrorResponse(String message, int status) { this.message = message; this.status = status; } public String getMessage() { return message; } public void setMessage(String message) { this.message = message; } public int getStatus() { return status; } public void setStatus(int status) { this.status = status; } } } 4. Application Configuration Ensure that your application.properties is correctly configured for database access, e.g.,properties spring.datasource.url=jdbc:mysql://localhost:3306/tododb spring.datasource.username=root spring.datasource.password=yourpassword spring.jpa.hibernate.ddl-auto=update ========================================================================================================================= Task 2 ========================================================================================================================= There is an array A, consisting of N integers, and two integers, L and R. your task is to find the shortest fragment of consecutive elements of A that contains every integer from L to R inclusive. Write a function: class Solution{ public int solution(int[] A, int L, int R); } that, given A,L and R, returns the length of the shortes fragment of A that contains all numbers from L to R, If no such fragment exists, the function should return -1. Examples: 1. Given A=[2,1,4,3,2,1,1,4], L=2,R=4, the function should return 3. The shortest fragment containing the numbers 2,3 and 4 starts at the third element of A:[2,1,4,2,1,1,4] 2.Given A=[10^9, 1,1,1,1,1,10^9-1],L=10^9-1 and R=10^9, the function should return 7. the fragment is the whole array. 3.GivenA=[1,3,5,7], L=3 and R=5, the function should return -1. Array A does not contain the number 4. Write an efficient algorithem for the following assumptions: .N is an integer within the range[1.. 100,000]; .1<=L<=R<=1000000000; .each element of Array A is an integer within the range[1..1000000000]. ================================================= Solution/Answer ============================================ class Solution { public int solution(int[] A, int L, int R) { int n = A.length; int minLength = Integer.MAX_VALUE; Map<Integer, Integer> countMap = new HashMap<>(); int uniqueCount = 0; int left = 0; for (int right = 0; right < n; right++) { int currentElement = A[right]; // Only consider elements in the range [L, R] if (currentElement >= L && currentElement <= R) { countMap.put(currentElement, countMap.getOrDefault(currentElement, 0) + 1); if (countMap.get(currentElement) == 1) { uniqueCount++; } } // Check if the current window contains all integers from L to R while (uniqueCount == (R - L + 1)) { minLength = Math.min(minLength, right - left + 1); int leftElement = A[left]; if (leftElement >= L && leftElement <= R) { countMap.put(leftElement, countMap.get(leftElement) - 1); if (countMap.get(leftElement) == 0) { uniqueCount--; } } left++; } } return minLength == Integer.MAX_VALUE ? -1 : minLength; } public static void main(String[] args) { Solution sol = new Solution(); System.out.println(sol.solution(new int[]{2, 1, 4, 3, 2, 1, 1, 4}, 2, 4)); // Output: 3 System.out.println(sol.solution(new int[]{1000000000, 1, 1, 1, 1, 1, 999999999}, 999999999, 1000000000)); // Output: 7 System.out.println(sol.solution(new int[]{1, 3, 5, 7}, 3, 5)); // Output: -1 } } Thanks, Please comment if anything info needed regarding the same.

very nice thank you

Which exam is it ?

Can we navigate to google to check the logics

Thank you so much Sir😃

Hello Techbull. These are experienced interview questions otherwise Freshers one.

What decisions will be made based on codility exam?

We had our codility today and in 1 question only 1 test case passed and in 2nd only 1 test case remained other passed...so did I cleared codility??

Hey what is the criteria to pass it??

thanks

i have completed the trait based assessment 2 months back but now there is no update so what should i do. and another small doubt im qualified or not for the further rounds

Bro are you free tomorrow? i have ibm exam...

Thank You So Much Sir 🙏🙏

Hi bro i have completed the assessment before 1 day in backend developer role which consists of 1 coding and i database question and i have passed all the test cases and i solve the query also ,my doubt is when i get the next round details either i will select or not And next one is what is the next round bro

Thank you sir this is very useful video with the help of this video I am solved my problem thank you so much

bro where is pom.xml?

Has anyone cleared ibm backend interview??

Sir thanks a lot sir 🙇 i spent more time for solve this error, finally i saw u r video it solved,❤