Summary of "July Exam 😎! Virtusa Important Coding Questions | Virtusa Online Assessment Test 2025 | Tekno UF"

Summary of the Video:

July Exam 😎! Virtusa Important Coding Questions | Virtusa Online Assessment Test 2025 | Tekno UF

Main Ideas and Concepts:

1. Purpose of the Video: The video is aimed at helping candidates prepare for the upcoming Virtusa online assessment test and other campus/off-campus drives by providing important technical and coding questions that have appeared in recent Virtusa drives.
2. Technical MCQ Questions:
   • The video shares examples of typical multiple-choice questions (MCQs) asked in Virtusa assessments.
   • Emphasis on understanding concepts related to:
     • Access specifiers in classes (public, private, protected, none)
     • Range of signed numbers (e.g., 8-bit signed number range: -128 to 127)
     • Queue operations and data structures (FIFO vs LIFO)
     • Data Structures and Algorithms (DSA) related topics such as trees, sorting, time complexity, in-order and post-order traversals.
3. Recommendation for Preparation: Candidates should focus on DSA MCQs and coding problems. The video suggests visiting a resource website (topman.io/techners/techno_uf" target="_blank" rel="noopener noreferrer">topman.io/techners/techno_uf) for mock tests, technical MCQs, coding questions, and interview experiences.
4. Coding Question Explanation: Problem: Given an integer array, a string of characters 'P' and 'N', and the size of the array/string, calculate the total electrostatic force. Rules:
   • For each element in the array, if the corresponding character in the string is 'P', add the element to a total sum.
   • If the character is 'N', subtract the element from the total sum.
   • Return the absolute value of the total sum multiplied by 100.
   Example:
   • Input: Array = [4, 3, 5], String = "PNP", Size = 3
   • Calculation: 4 (P) + (-3) (N) + 5 (P) = 6 → Output = 6 * 100 = 600
5. Methodology to Solve the Coding Problem:
   • Iterate through the array using a loop (for or while).
   • Use a conditional check or switch-case on the corresponding character of the string:
     • If 'P', add the array element to total.
     • If 'N', subtract the array element from total.
   • After iteration, return the absolute value of total multiplied by 100.
6. Code Implementation:
   • Python:
     • Use a for loop to iterate over array indices.
     • Use if-else to check characters and update total accordingly.
     • Return abs(total) * 100.
   • Java:
     • Use a for loop with switch-case on characters.
     • Add or subtract based on 'P' or 'N'.
     • Return absolute value multiplied by 100.
   • Both implementations are demonstrated with sample inputs and outputs.
7. Final Advice: The questions asked in Virtusa assessments are relatively straightforward if you understand the pattern and types of questions. Consistent practice and awareness of frequently asked questions will help crack the Virtusa exam.

Detailed Bullet Points:

• Technical MCQs Examples:
  • Access specifiers: Which access specifier allows direct access outside the class? (Answer: public)
  • Range of 8-bit signed number: (Answer: -128 to 127)
  • Queue operation type: FIFO (First In First Out) for queue, LIFO for stack
  • Importance of DSA topics: trees, sorting, traversal, time complexity
• Coding Question Task:
  • Inputs: integer array, string of 'P' and 'N', integer size
  • Process:
    • For each index:
      • If string character is 'P', add array element to sum
      • If 'N', subtract array element from sum
    • Return absolute value of sum * 100
• Solution Approach:
  • Loop through array indices
  • Use if-else or switch-case for character check
  • Update sum accordingly
  • Return final result
• Python Code Outline:
  • Define function with three parameters
  • Initialize total = 0
  • For i in range(size):
    • Check character at position i
    • Add or subtract array[i] from total
  • Return abs(total) * 100
• Java Code Outline:
  • Static method with three parameters
  • Initialize sum = 0
  • For loop through indices
  • Switch on character at index:
    • case 'P': sum += array[i]
    • case 'N': sum -= array[i]
  • Return Math.abs(sum) * 100
• Additional Tips:
  • Understand question patterns from previous exams

Category

Educational

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