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LTI Technical Interview Questions
LTI Technical Interview Questions and Answers 2023
Check out the most recent LTI Technical interview questions and the candidates’ responses.
Page Highlights:
- LTI Test Pattern
- LTI Eligibility Criteria
- LTI Technical Interview Questions
LTI Test Pattern
LTI Cognitive Assessment Pattern
Section | No. of Questions | Time Allotted |
---|---|---|
Verbal English | 12 Questions | 15 mins |
Quantitative Aptitude | 12 Questions | 15 mins |
Logical Reasoning | 12 Questions | 15 mins |
CMCQs | 20 Questions | 35 mins |
Pseudo Code | 20 Questions | 20 mins |
Automata Fix | 2 Questions | 40 mins |
Communication Ability | 30-40 Questions | 20mins |
LTI Technical Interview Questions
Question 1: What is top-n analysis?
Answer:
TOP-N analysis is used to reduce the number of rows returned from a set of sorted data items. The N lowest or highest rows may be returned with this, which is quite useful.
Question 2: Explain different file systems and database management systems.
FILE SYSTEM | DBMS |
---|---|
is used to control and arrange files stored on the computer’s hard drive | a program for storing and retrieving user data |
There is data that is redundant | there is no redundant data present. |
Processing queries are not very effective. | Processing of queries is fast. |
Low data consistency | The normalization procedure ensures high data consistency. |
lowered security | additional security methods are supported |
cheaper than DBMS in terms of price | more expensive than the File system |
not compatible with crash recovery | Highly supported crash recovery mechanism |
Question 3:What is Object Oriented Programming features?
Answer:
Object-oriented programming (OOP) is a programming technique that organizes software design around data rather than functions and logic. A data field with unique characteristics and behavior is called an object.
Questions 4: What is the difference between Java and C++?
Answer:
Java | C++ |
---|---|
Pointers, unions, operator overloading, and structure are not supported by Java. | Pointers, unions, operator overloading, and structure are supported in C++. |
Java has garbage collection support. | There is no trash collection support in C++. |
Java works across all platforms. | Platforms differ while using C++. |
Except for multiple inheritances, Java permits inheritance. | Inheritance, including multiple inheritances, is supported in C++. |
Java can be interpreted. | C++ can be compiled. |
Destructors are not supported by Java. | Destructors are supported by C++. |
Question 5: What is the minimum diameter of the longitudinal bar of a column?
Answer:
The minimum diameter of the longitudinal bar of a column is 12mm.
Question 6: What are the various levels of testing?
Answer:
There are typically four levels of testing that are accepted:
- Unit/component testing
- Integration testing
- System testing
- Acceptance testing.
Questions 7: What do you mean by hashing?
Answer:
Hashing is the process of transforming any given key or a string of characters into another value. This is usually represented by a shorter, fixed-length value or key that represents and makes it easier to find or employ the original string. The most popular use for hashing is the implementation of hash tables.
Question 8: What is data structure?
Answer:
A data structure is a specific type of format used to arrange, analyze, retrieve, and store data. Data structures come in a variety of simple and complex forms and are all made to organize data in a way that is appropriate for a certain function. Users find it simple to get the data they need and use it appropriately thanks to data structures.
Question 9: Write a program to check whether or not a number is a perfect number.
n = 28 sum = 0 for i in range(1, n): if n % i == 0: sum = sum + i if sum == n: print("The number is a Perfect number") else: print("The number is not a Perfect number")
Question 10: Write a program for printing the reverse of a Linked List without actually reversing.
#include<stdio.h> #include<stdlib.h> struct node //code for making a node { int data; struct node *next; }; void display(struct node *head) //method for reverse display nodes { if (head == NULL) return; // print the list after head node display(head->next); // After everything else is printed, print head printf("%d ", head->data); } int main() { struct node *prev,*head, *p; int n,i; printf ("Enter size of Linked List: "); scanf("%d",&n); head=NULL; for(i=0;i<n;i++) { p=malloc(sizeof(struct node)); printf ("Enter the data: "); scanf("%d",&p->data); p->next=NULL; if(head==NULL) head=p; else prev->next=p; prev=p; } display(head); return 0; }
Question 11: Write a program for decimal to binary conversion.
def convertBinary(num): binaryArray = [] while num>0: binaryArray.append(num%2) num = num//2 for j in binaryArray: print(j, end="") decimal_num = 21 convertBinary(decimal_num)
Question 12: What is operator loading?
Answer:
Operator overloading is a programming language method where operators are implemented in user-defined types with specific logic dependent on the types of given arguments.
Question 13: What is a copy constructor?
Answer:
The copy constructor produces an object by initializing it using a different object of the same class that has already been constructed. One object of the same type is initialized from another using the copy constructor.
Question 14: What are the advantages of using OOPS?
Answer:
The advantages of using OOPS are:
- The OOP language makes troubleshooting simpler.
- Reusability of Code
- Productivity solving issues
- Security
- Redundancy in Data
- Flexible Codes
Question 15: What is recursion?
Answer:
Recursion is a programming approach that uses a function or algorithm that repeatedly calls itself up until a certain condition is satisfied, at which point the remainder of each iteration is processed starting from the last call.
Question 16: What is RDBMS?
Answer:
RDBMSs are database management programs that keep track of the records and indexes of data in tables. Relationships may be established and maintained between the data and tables as well as between them. Tables in a relational database are used to represent the relationships between data elements. Instead of using pointers, these tables’ interdependencies are expressed by data values.
Question 17: What is an Authorization and Integrity manager?
Answer:
It is the software module that evaluates if integrity requirements are satisfied and determine whether the user has the right to access the data.
Question 18: Write a program to print the spiral traversal of the matrix.
#include <stdio.h> #define r 4 #define c 4 int main() { int a[4][4] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; int i, left = 0, right = c-1, top = 0, bottom = r-1; while (left <= right && top <= bottom) { /* Print the first row from the remaining rows */ for (i = left; i <= right; ++i) { printf("%d ", a[top][i]); } top++; /* Print the last column from the remaining columns */ for (i = top; i <= bottom; ++i) { printf("%d ", a[i][right]); } right--; /* Print the last row from the remaining rows */ if (top <= bottom) { for (i = right; i >= left; --i) { printf("%d ", a[bottom][i]); } bottom--; } /* Print the first column from the remaining columns */ if (left <= right) { for (i = bottom; i >= top; --i) { printf("%d ", a[i][left]); } left++; } } return 0; }
Question 19: Write a Program to Print Prime numbers in a given range.
#include <stdio.h> int checkPrime(int num) { // 0, 1 and negative numbers are not prime if(num < 2){ return 0; } else{ // no need to run loop till num-1 as for any number x the numbers in // the range(num/2 + 1, num) won't be divisible anyways. // Example 36 wont be divisible by anything b/w 19-35 int x = num/2; for(int i = 2; i < x; i++) { if(num % i == 0) { return 0; } } } // the number would be prime if we reach here return 1; } int main() { int a=10, b=20; for(int i=a; i <= b; i++){ if(checkPrime(i)) printf("%d ",i); } return 0; } //Time Complexity: O(N^2) //Space Complexity O(1)
Question 20: Write a program to find the minimum scalar product.
#include<bits/stdc++.h> using namespace std; int main(){ int arr1[] = {1, 2, 6, 3, 7}; int arr2[] = {10, 7, 45, 3, 7}; int n = sizeof(arr1)/sizeof(arr1[0]); //Sort arr1 in ascending order for(int i=0; i<n; i++){ for(int j=i+1; j<n; j++){ if(arr1[i]>arr1[j]){ int temp = arr1[i]; arr1[i] = arr1[j]; arr1[j] = temp; } } } //Sort arr2 in descending order for(int i=0; i<n; i++){ for(int j=i+1; j<n; j++){ if(arr2[i]<arr2[j]){ int temp = arr2[i]; arr2[i] = arr2[j]; arr2[j] = temp; } } } int product = 0; for(int i=0; i<n; i++) product += arr1[i]*arr2[i]; cout<< product; }
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