Table of Contents

Infosys Coding And Pseudo-code Questions 2024

Practice the top 10 Infosys Coding and Pseudo-code Questions uploaded to crack the Infosys exam on the first attempt. These coding questions will evaluate a candidate’s problem-solving abilities, logical thinking, and proficiency in programming languages. The questions range from simple data structure and algorithm problems to more complex ones involving real-world scenarios. This blog focuses on the Top 13 Infosys Coding Questions and Answers for the examination and 10 Coding Questions for specialist programmers in 2024. We have also uploaded Infosys Pseudocode questions with answers here.

Top IT firms like Infosys require coding skills and problem-solving skills for their hiring process for 2024. These skills are mandatory for the candidates applying for the job in Infosys and are used to evaluate during the Infosys interview and online test. The Pseudocode round is one of the important rounds in the Infosys coding questions in 2024. Through this Pseudocode round, they will analyze the candidate’s ability to write the algorithmic solutions with the simplified code.

Infosys coding questions often test a candidate’s understanding of concepts like arrays, strings, linked lists, trees, recursion, and dynamic programming. Solving these questions efficiently and effectively is crucial for aspiring candidates to showcase their coding skills and stand out in the competitive hiring process.

Infosys Pseudocode Questions 2024

Pseudo-code in data science, is a step-by-step description of the computer algorithm. In short, It is the high-level, language-agnostic representation of the algorithm. It indicates how to write down the algorithm’s steps in a language-agnostic manner.

Furthermore, it is useful in the early stages of software development and aids the programmers to organize and structure the programs before writing the actual code. This will also help developers focus on the program’s logic and flow. Pseudo-code will use programming language such as loops, conditionals and variables, and English words and phrases.

The Infosys Pseudo-code round has 5 questions, to be completed in 10 minutes. Pseudo-code is considered the representation of the algorithm code in Java, C, and C++. The difficulty level of the Infosys Pseudo-code Questions is moderate-high. Pseudo-code refers to the algorithm that is displayed in the coding format.

Check out the Top 10 Infosys Pseudo-code questions with answers in 2024. Solve these pseudo-code questions and crack the Infosys interview and exam.

10 Infosys Pseudo-code Questions With Answers 2024

The top 10 Infosys Pseudo-code questions are added here for your reference. Candidates preparing for the Infosys interview can understand and practice the Infosys Pseudo-code questions provided below.

1. What will be the output of the pseudocode given here?

public class Main {
    public static void main(String[] args) {
        String names[] = new String[5];
        for(int x = 0; x < args.length && x < names.length; x++) {
            names[x] = args[x];
        }
        if (args.length >= 3) {
            System.out.println(names[2]);
        } else {
            System.out.println("Not enough arguments provided");
        }
    }
}

Ans.

Scenario 1: Exactly 5 Arguments

Command:

java Main one two three four five

Output:

three

Scenario 2: Fewer Than 3 Arguments

Command:

java Main one two

Output:

Not enough arguments provided

Scenario 3: More Than 5 Arguments

Command:

java Main one two three four five six

Output:

three

Scenario 4: Exactly 3 Arguments

Command:

java Main one two three

Output:

three

Scenario 5: No Arguments

Command:

java Main

Output:

Not enough arguments provided

These outputs correspond to the improved version of the program which includes a check to ensure there are at least 3 arguments before accessing names[2].

2. What will be the output of the Pseudo Code provided below?

#include <stdio.h>
int main()
{
    float i;
    i = 1;
    printf("%f", i);
    return 0;
}

Ans. Output

1.000000

3. What will be the output of the pseudocode given here?

#include <stdio.h>
int main()
{
    int num = 987;
    int rem;
    while(num!=0)
    {
        rem = num % 4;
        num = num / 10;
    }
    printf("%d",rem);
}

Ans. Output

1

4. What will be the output of the pseudo-code provided here?

#include<stdio.h>
int main(){
    float x = 0.0;
    long int y = 10;
    printf("%d", sizeof(x) == sizeof(x+y));
    return 0;
}

Ans. Output

1

5. What will be the output of the pseudocode below?

#include<stdio.h>
int main()
{
int go = 5.0, num = 1*10;
do 
{
num /= go;
} while(go--);
printf ("%d\n", num);
return 0;
}

Ans. Output

0

6. What is the output of the program given below?

#include<stdio.h>
int main()
{
    int any = ' ' * 10;
    printf("%d", any);
    return 0;
}

Ans. Output

320

7. Write the output for the given program.

#include<stdio.h>
int main(){
    float x = 0.0;
    long int y = 10;
    printf("%d", sizeof(y) == sizeof(x+y));
    return 0;
}

Ans.

0

8. What is the output of the following program?

#include<stdio.h>
int main() 
{
    typedef int num;
    num bunk = 0.00;    
    printf("%d", bunk);
    return 0;
    
}

Ans.

0

9. What is the output?

#include<stdio.h>
int main()
{
    int a = 100;
    printf("%0 %x", a);
}

Ans. Output

100 64 

%d: 100 is printed in decimal format.%x: 100 is printed in hexadecimal format. In hexadecimal, 100 is represented as 64.

10. What is the output?

#include<stdio.h>
int main()
{
    for (int x = 10; x >= 0; x--) {
        int z = x & (x >> 1);
        if (z)
            printf("%d ", x);
     }
    
}

Ans.

Output

10 9 8 7 6 5 4 3 2 1

13 Infosys Coding Questions and Answers 2024

In Infosys Exam, they have 3 Infosys coding questions, to be completed in 3 hours. So, candidates should practice the Infosys coding questions and answers for the Infosys Exam. Check out the table below to learn more about the difficulty level of the Infosys Coding Question.

Applicants of the Infosys examination can refer to the coding questions and answers added here. Check the Infosys coding questions and answers, and improve your coding skills.

1. Write a program to arrange the given number to form the biggest number.

Ans.

Python

from functools import cmp_to_key

def compare(x, y):
    # Compare combined numbers to determine order
    if x + y > y + x:
        return -1
    elif x + y < y + x:
        return 1
    else:
        return 0

def largest_number(nums):
    # Convert numbers to strings for easy concatenation
    str_nums = [str(num) for num in nums]
    # Sort numbers using custom comparator
    str_nums.sort(key=cmp_to_key(compare))
    # Join sorted numbers to form the largest number
    largest_num = ''.join(str_nums)
    # Edge case for leading zeros
    return largest_num if largest_num[0] != '0' else '0'

# Example usage
numbers = [3, 30, 34, 5, 9]
print("The largest number is:", largest_number(numbers))

Output

The largest number is: 9534330

Java

import java.util.Arrays;
import java.util.Comparator;

public class LargestNumber {

    // Custom comparator to decide the order based on concatenation
    private static class NumberComparator implements Comparator<String> {
        @Override
        public int compare(String x, String y) {
            String order1 = x + y;
            String order2 = y + x;
            return order2.compareTo(order1); // reverse order for descending
        }
    }

    public static String largestNumber(int[] nums) {
        // Convert int array to String array
        String[] strNums = Arrays.stream(nums)
                .mapToObj(String::valueOf)
                .toArray(String[]::new);

        // Sort array using custom comparator
        Arrays.sort(strNums, new NumberComparator());

        // Edge case: Check if the largest number is 0
        if (strNums[0].equals("0")) {
            return "0";
        }

        // Join sorted array to form the largest number
        StringBuilder largestNum = new StringBuilder();
        for (String num : strNums) {
            largestNum.append(num);
        }

        return largestNum.toString();
    }

    public static void main(String[] args) {
        int[] numbers = {3, 30, 34, 5, 9};
        System.out.println("The largest number is: " + largestNumber(numbers));
    }
}

Output

The largest number is: 9534330

.NET

using System;
using System.Linq;

public class LargestNumber
{
    // Custom comparator to decide the order based on concatenation
    public class NumberComparer : IComparer<string>
    {
        public int Compare(string x, string y)
        {
            string order1 = x + y;
            string order2 = y + x;
            return order2.CompareTo(order1); // reverse order for descending
        }
    }

    public static string LargestNumberFromArray(int[] nums)
    {
        // Convert int array to string array
        var strNums = nums.Select(num => num.ToString()).ToArray();

        // Sort array using custom comparator
        Array.Sort(strNums, new NumberComparer());

        // Edge case: Check if the largest number is 0
        if (strNums[0] == "0")
        {
            return "0";
        }

        // Join sorted array to form the largest number
        return string.Join("", strNums);
    }

    public static void Main(string[] args)
    {
        int[] numbers = { 3, 30, 34, 5, 9 };
        Console.WriteLine("The largest number is: " + LargestNumberFromArray(numbers));
    }
}

Output

The largest number is: 9534330

2. Find the next permutation of the given string in the Program

Ans.

Python

def next_permutation(s):
    # Convert string to list for mutability
    arr = list(s)
    n = len(arr)

    # Step 1: Find the largest index i such that arr[i-1] < arr[i]
    i = n - 1
    while i > 0 and arr[i - 1] >= arr[i]:
        i -= 1

    # If no such index exists, the permutation is the last permutation
    if i == 0:
        return ''.join(sorted(arr))

    # Step 2: Find the largest index j such that arr[j] > arr[i-1]
    j = n - 1
    while arr[j] <= arr[i - 1]:
        j -= 1

    # Step 3: Swap arr[i-1] with arr[j]
    arr[i - 1], arr[j] = arr[j], arr[i - 1]

    # Step 4: Reverse the suffix starting from index i
    arr[i:] = arr[n - 1:i - 1:-1]

    return ''.join(arr)

# Example usage
s = "123"
print("Next permutation of", s, ":", next_permutation(s))

Output

Next permutation is: 536479

Java

import java.util.Arrays;

public class NextPermutation {

    public static String nextPermutation(String s) {
        char[] arr = s.toCharArray();
        int n = arr.length;

        // Step 1: Find the largest index i such that arr[i-1] < arr[i]
        int i = n - 1;
        while (i > 0 && arr[i - 1] >= arr[i]) {
            i--;
        }

        // If no such index exists, the permutation is the last permutation
        if (i == 0) {
            Arrays.sort(arr);
            return new String(arr);
        }

        // Step 2: Find the largest index j such that arr[j] > arr[i-1]
        int j = n - 1;
        while (arr[j] <= arr[i - 1]) {
            j--;
        }

        // Step 3: Swap arr[i-1] with arr[j]
        char temp = arr[i - 1];
        arr[i - 1] = arr[j];
        arr[j] = temp;

        // Step 4: Reverse the sequence from arr[i] to arr[n-1]
        reverse(arr, i, n - 1);

        return new String(arr);
    }

    private static void reverse(char[] arr, int start, int end) {
        while (start < end) {
            char temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }

    public static void main(String[] args) {
        String s = "534976";
        System.out.println("Next permutation is: " + nextPermutation(s));
    }
}

Output

Next permutation is: 536479

.NET

using System;

public class NextPermutation
{
    public static string GetNextPermutation(string s)
    {
        char[] arr = s.ToCharArray();
        int n = arr.Length;

        // Step 1: Find the largest index i such that arr[i-1] < arr[i]
        int i = n - 1;
        while (i > 0 && arr[i - 1] >= arr[i])
        {
            i--;
        }

        // If no such index exists, the permutation is the last permutation
        if (i == 0)
        {
            Array.Sort(arr);
            return new string(arr);
        }

        // Step 2: Find the largest index j such that arr[j] > arr[i-1]
        int j = n - 1;
        while (arr[j] <= arr[i - 1])
        {
            j--;
        }

        // Step 3: Swap arr[i-1] with arr[j]
        char temp = arr[i - 1];
        arr[i - 1] = arr[j];
        arr[j] = temp;

        // Step 4: Reverse the sequence from arr[i] to arr[n-1]
        Array.Reverse(arr, i, n - i);

        return new string(arr);
    }

    public static void Main(string[] args)
    {
        string s = "534976";
        Console.WriteLine("Next permutation is: " + GetNextPermutation(s));
    }
}

Output

Next permutation is: 536479

3. How do you rotate a matrix by 90 degrees? Write a program for this.

Ans.

Python

def rotate_matrix(matrix):
    n = len(matrix)
    # Transpose the matrix
    for i in range(n):
        for j in range(i, n):
            matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
    # Reverse each row
    for i in range(n):
        matrix[i].reverse()

# Example usage
matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]
rotate_matrix(matrix)
print("Rotated matrix:")
for row in matrix:
    print(row)

Output

Rotated matrix:
[7, 4, 1]
[8, 5, 2]
[9, 6, 3]

Java

public class RotateMatrix {
    public static void rotate(int[][] matrix) {
        int n = matrix.length;
        // Transpose the matrix
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
        // Reverse each row
        for (int i = 0; i < n; i++) {
            int left = 0;
            int right = n - 1;
            while (left < right) {
                int temp = matrix[i][left];
                matrix[i][left] = matrix[i][right];
                matrix[i][right] = temp;
                left++;
                right--;
            }
        }
    }

    public static void main(String[] args) {
        int[][] matrix = {
            {1, 2, 3},
            {4, 5, 6},
            {7, 8, 9}
        };
        rotate(matrix);
        System.out.println("Rotated matrix:");
        for (int[] row : matrix) {
            for (int num : row) {
                System.out.print(num + " ");
            }
            System.out.println();
        }
    }
}

Output

Rotated matrix:
[7, 4, 1]
[8, 5, 2]
[9, 6, 3]

.NET

using System;

public class RotateMatrix
{
    public static void Rotate(int[][] matrix)
    {
        int n = matrix.Length;
        // Transpose the matrix
        for (int i = 0; i < n; i++)
        {
            for (int j = i; j < n; j++)
            {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
        // Reverse each row
        for (int i = 0; i < n; i++)
        {
            Array.Reverse(matrix[i]);
        }
    }

    public static void Main(string[] args)
    {
        int[][] matrix = new int[][]
        {
            new int[] {1, 2, 3},
            new int[] {4, 5, 6},
            new int[] {7, 8, 9}
        };
        Rotate(matrix);
        Console.WriteLine("Rotated matrix:");
        foreach (var row in matrix)
        {
            Console.WriteLine(string.Join(", ", row));
        }
    }
}

Output

Rotated matrix:
7, 4, 1
8, 5, 2
9, 6, 3

4. How do you find the missing characters to make a string pangram? Write a program.

Ans.

Python

def find_missing_characters(s):
    alphabet = set('abcdefghijklmnopqrstuvwxyz')
    s = s.lower()
    present_chars = set(s)
    missing_chars = alphabet - present_chars
    return ''.join(sorted(missing_chars))

# Example usage
input_string = "The quick brown fox jumps over a lazy dog"
missing_chars = find_missing_characters(input_string)
print(f"Missing characters to make the string a pangram: '{missing_chars}'")

Output

Missing characters to make the string a pangram: ''

Java

import java.util.HashSet;
import java.util.Set;

public class PangramChecker {

    public static String findMissingCharacters(String s) {
        Set<Character> alphabet = new HashSet<>();
        for (char c = 'a'; c <= 'z'; c++) {
            alphabet.add(c);
        }

        s = s.toLowerCase();
        Set<Character> presentChars = new HashSet<>();
        for (char c : s.toCharArray()) {
            if (Character.isLetter(c)) {
                presentChars.add(c);
            }
        }

        alphabet.removeAll(presentChars);

        StringBuilder missingChars = new StringBuilder();
        for (char c : alphabet) {
            missingChars.append(c);
        }

        return missingChars.toString();
    }

    public static void main(String[] args) {
        String inputString = "The quick brown fox jumps over a lazy dog";
        String missingChars = findMissingCharacters(inputString);
        System.out.println("Missing characters to make the string a pangram: '" + missingChars + "'");

        // For a non-pangram example
        inputString = "Hello World";
        missingChars = findMissingCharacters(inputString);
        System.out.println("Missing characters to make the string a pangram: '" + missingChars + "'");
    }
}

Output

Missing characters to make the string a pangram: ''
Missing characters to make the string a pangram: 'abcfijkmnpqstuvxyz'

.NET

using System;
using System.Collections.Generic;

public class PangramChecker
{
    public static string FindMissingCharacters(string s)
    {
        HashSet<char> alphabet = new HashSet<char>("abcdefghijklmnopqrstuvwxyz");
        s = s.ToLower();
        HashSet<char> presentChars = new HashSet<char>();

        foreach (char c in s)
        {
            if (char.IsLetter(c))
            {
                presentChars.Add(c);
            }
        }

        alphabet.ExceptWith(presentChars);
        List<char> missingChars = new List<char>(alphabet);
        missingChars.Sort();

        return new string(missingChars.ToArray());
    }

    public static void Main(string[] args)
    {
        string inputString = "The quick brown fox jumps over a lazy dog";
        string missingChars = FindMissingCharacters(inputString);
        Console.WriteLine("Missing characters to make the string a pangram: '" + missingChars + "'");

        // For a non-pangram example
        inputString = "Hello World";
        missingChars = FindMissingCharacters(inputString);
        Console.WriteLine("Missing characters to make the string a pangram: '" + missingChars + "'");
    }
}

Output

Missing characters to make the string a pangram: ''
Missing characters to make the string a pangram: 'abcfijkmnpqstuvxyz'

5. How do you find the number of unique numbers in a given string? Write a program.

Ans.

Python

import re

def count_unique_numbers(s):
    # Find all numbers in the string using regex
    numbers = re.findall(r'\d+', s)
    # Convert to integers to handle leading zeros and find unique numbers
    unique_numbers = set(int(num) for num in numbers)
    return len(unique_numbers)

# Example usage
input_string = "abc123def456abc123def7890123"
unique_count = count_unique_numbers(input_string)
print(f"Number of unique numbers in the string: {unique_count}")

Output

Number of unique numbers in the string: 3

Java

import java.util.HashSet;
import java.util.Set;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class UniqueNumbers {

    public static int countUniqueNumbers(String s) {
        // Create a pattern to find all numbers
        Pattern pattern = Pattern.compile("\\d+");
        Matcher matcher = pattern.matcher(s);

        // Use a set to store unique numbers
        Set<Integer> uniqueNumbers = new HashSet<>();

        // Find all matches and add to the set as integers
        while (matcher.find()) {
            uniqueNumbers.add(Integer.parseInt(matcher.group()));
        }

        return uniqueNumbers.size();
    }

    public static void main(String[] args) {
        String inputString = "abc123def456abc123def7890123";
        int uniqueCount = countUniqueNumbers(inputString);
        System.out.println("Number of unique numbers in the string: " + uniqueCount);
    }
}

Output

Number of unique numbers in the string: 3

.NET

using System;
using System.Collections.Generic;
using System.Text.RegularExpressions;

public class UniqueNumbers
{
    public static int CountUniqueNumbers(string s)
    {
        // Create a pattern to find all numbers
        Regex regex = new Regex(@"\d+");
        MatchCollection matches = regex.Matches(s);

        // Use a set to store unique numbers
        HashSet<int> uniqueNumbers = new HashSet<int>();

        // Find all matches and add to the set as integers
        foreach (Match match in matches)
        {
            uniqueNumbers.Add(int.Parse(match.Value));
        }

        return uniqueNumbers.Count;
    }

    public static void Main(string[] args)
    {
        string inputString = "abc123def456abc123def7890123";
        int uniqueCount = CountUniqueNumbers(inputString);
        Console.WriteLine("Number of unique numbers in the string: " + uniqueCount);
    }
}

6. Write a program for the Subtraction of two Matrices.

Ans.

Python

def subtract_matrices(matrix1, matrix2):
    if len(matrix1) != len(matrix2) or len(matrix1[0]) != len(matrix2[0]):
        raise ValueError("Matrices must have the same dimensions for subtraction")

    result = []
    for i in range(len(matrix1)):
        row = []
        for j in range(len(matrix1[0])):
            row.append(matrix1[i][j] - matrix2[i][j])
        result.append(row)
    
    return result

# Example usage
matrix1 = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]
matrix2 = [
    [9, 8, 7],
    [6, 5, 4],
    [3, 2, 1]
]
result = subtract_matrices(matrix1, matrix2)
print("Result of subtraction:")
for row in result:
    print(row)
    

Output

Result of subtraction:
[-8, -6, -4]
[-2, 0, 2]
[4, 6, 8]

Java

public class MatrixSubtraction {

    public static int[][] subtractMatrices(int[][] matrix1, int[][] matrix2) {
        if (matrix1.length != matrix2.length || matrix1[0].length != matrix2[0].length) {
            throw new IllegalArgumentException("Matrices must have the same dimensions for subtraction");
        }

        int[][] result = new int[matrix1.length][matrix1[0].length];
        for (int i = 0; i < matrix1.length; i++) {
            for (int j = 0; j < matrix1[0].length; j++) {
                result[i][j] = matrix1[i][j] - matrix2[i][j];
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[][] matrix1 = {
            {1, 2, 3},
            {4, 5, 6},
            {7, 8, 9}
        };
        int[][] matrix2 = {
            {9, 8, 7},
            {6, 5, 4},
            {3, 2, 1}
        };
        int[][] result = subtractMatrices(matrix1, matrix2);
        System.out.println("Result of subtraction:");
        for (int[] row : result) {
            for (int num : row) {
                System.out.print(num + " ");
            }
            System.out.println();
        }
    }
}

Output

Result of subtraction:
-8 -6 -4 
-2 0 2 
4 6 8 

.NET

using System;

public class MatrixSubtraction
{
    public static int[,] SubtractMatrices(int[,] matrix1, int[,] matrix2)
    {
        if (matrix1.GetLength(0) != matrix2.GetLength(0) || matrix1.GetLength(1) != matrix2.GetLength(1))
        {
            throw new ArgumentException("Matrices must have the same dimensions for subtraction");
        }

        int rows = matrix1.GetLength(0);
        int cols = matrix1.GetLength(1);
        int[,] result = new int[rows, cols];

        for (int i = 0; i < rows; i++)
        {
            for (int j = 0; j < cols; j++)
            {
                result[i, j] = matrix1[i, j] - matrix2[i, j];
            }
        }

        return result;
    }

    public static void Main(string[] args)
    {
        int[,] matrix1 = {
            {1, 2, 3},
            {4, 5, 6},
            {7, 8, 9}
        };
        int[,] matrix2 = {
            {9, 8, 7},
            {6, 5, 4},
            {3, 2, 1}
        };
        int[,] result = SubtractMatrices(matrix1, matrix2);
        Console.WriteLine("Result of subtraction:");
        for (int i = 0; i < result.GetLength(0); i++)
        {
            for (int j = 0; j < result.GetLength(1); j++)
            {
                Console.Write(result[i, j] + " ");
            }
            Console.WriteLine();
        }
    }
}

Output

Result of subtraction:
-8 -6 -4
-2 0 2
4 6 8

7. How do you multiply two matrices and show results through another matrix? Write a program.

Ans.

Python

def multiply_matrices(matrix1, matrix2):
    if len(matrix1[0]) != len(matrix2):
        raise ValueError("Number of columns in first matrix must be equal to number of rows in second matrix")

    result = [[0 for _ in range(len(matrix2[0]))] for _ in range(len(matrix1))]

    for i in range(len(matrix1)):
        for j in range(len(matrix2[0])):
            for k in range(len(matrix2)):
                result[i][j] += matrix1[i][k] * matrix2[k][j]

    return result

# Example usage
matrix1 = [
    [1, 2, 3],
    [4, 5, 6],
]
matrix2 = [
    [7, 8],
    [9, 10],
    [11, 12]
]
result = multiply_matrices(matrix1, matrix2)
print("Result of matrix multiplication:")
for row in result:
    print(row)

Output

Result of matrix multiplication:
[58, 64]
[139, 154]

Java

public class MatrixMultiplication {

    public static int[][] multiplyMatrices(int[][] matrix1, int[][] matrix2) {
        if (matrix1[0].length != matrix2.length) {
            throw new IllegalArgumentException("Number of columns in first matrix must be equal to number of rows in second matrix");
        }

        int[][] result = new int[matrix1.length][matrix2[0].length];

        for (int i = 0; i < matrix1.length; i++) {
            for (int j = 0; j < matrix2[0].length; j++) {
                for (int k = 0; k < matrix2.length; k++) {
                    result[i][j] += matrix1[i][k] * matrix2[k][j];
                }
            }
        }

        return result;
    }

    public static void main(String[] args) {
        int[][] matrix1 = {
            {1, 2, 3},
            {4, 5, 6}
        };
        int[][] matrix2 = {
            {7, 8},
            {9, 10},
            {11, 12}
        };
        int[][] result = multiplyMatrices(matrix1, matrix2);
        System.out.println("Result of matrix multiplication:");
        for (int[] row : result) {
            for (int num : row) {
                System.out.print(num + " ");
            }
            System.out.println();
        }
    }
}

Output

Result of matrix multiplication:
58 64 
139 154 

.NET

using System;

public class MatrixMultiplication
{
    public static int[,] MultiplyMatrices(int[,] matrix1, int[,] matrix2)
    {
        if (matrix1.GetLength(1) != matrix2.GetLength(0))
        {
            throw new ArgumentException("Number of columns in first matrix must be equal to number of rows in second matrix");
        }

        int[,] result = new int[matrix1.GetLength(0), matrix2.GetLength(1)];

        for (int i = 0; i < matrix1.GetLength(0); i++)
        {
            for (int j = 0; j < matrix2.GetLength(1); j++)
            {
                for (int k = 0; k < matrix2.GetLength(0); k++)
                {
                    result[i, j] += matrix1[i, k] * matrix2[k, j];
                }
            }
        }

        return result;
    }

    public static void Main(string[] args)
    {
        int[,] matrix1 = {
            {1, 2, 3},
            {4, 5, 6}
        };
        int[,] matrix2 = {
            {7, 8},
            {9, 10},
            {11, 12}
        };
        int[,] result = MultiplyMatrices(matrix1, matrix2);
        Console.WriteLine("Result of matrix multiplication:");
        for (int i = 0; i < result.GetLength(0); i++)
        {
            for (int j = 0; j < result.GetLength(1); j++)
            {
                Console.Write(result[i, j] + " ");
            }
            Console.WriteLine();
        }
    }
}

Output

Result of matrix multiplication:
58 64 
139 154 

8. How do you convert decimal numbers to binary numbers? Write a Program.

Ans.

Python

def decimal_to_binary(n):
    return bin(n)[2:]

# Example usage
decimal_number = 10
binary_number = decimal_to_binary(decimal_number)
print(f"The binary representation of {decimal_number} is: {binary_number}")

Output

The binary representation of 10 is: 1010

Java

public class DecimalToBinary {

    public static String decimalToBinary(int n) {
        return Integer.toBinaryString(n);
    }

    public static void main(String[] args) {
        int decimalNumber = 10;
        String binaryNumber = decimalToBinary(decimalNumber);
        System.out.println("The binary representation of " + decimalNumber + " is: " + binaryNumber);
    }
}

Output

The binary representation of 10 is: 1010

.NET

using System;

public class DecimalToBinary
{
    public static string DecimalToBinaryString(int n)
    {
        return Convert.ToString(n, 2);
    }

    public static void Main(string[] args)
    {
        int decimalNumber = 10;
        string binaryNumber = DecimalToBinaryString(decimalNumber);
        Console.WriteLine("The binary representation of " + decimalNumber + " is: " + binaryNumber);
    }
}

Output

The binary representation of 10 is: 1010

9. Write a program in C++ to swap two arrays quickly.

Ans. Python

# Define arrays
array1 = [1, 2, 3]
array2 = [4, 5, 6]

# Before swapping
print("Before swapping:")
print("Array 1:", array1)
print("Array 2:", array2)

# Swap arrays
temp = array1
array1 = array2
array2 = temp

# After swapping
print("\nAfter swapping:")
print("Array 1:", array1)
print("Array 2:", array2)

Output

Before swapping:
Array 1: [1, 2, 3]
Array 2: [4, 5, 6]

After swapping:
Array 1: [4, 5, 6]
Array 2: [1, 2, 3]

Java

public class ArraySwap {
    public static void main(String[] args) {
        int[] array1 = {1, 2, 3};
        int[] array2 = {4, 5, 6};

        System.out.println("Before swapping:");
        System.out.println("Array 1: " + java.util.Arrays.toString(array1));
        System.out.println("Array 2: " + java.util.Arrays.toString(array2));

        // Swap arrays
        int[] temp = array1;
        array1 = array2;
        array2 = temp;

        System.out.println("\nAfter swapping:");
        System.out.println("Array 1: " + java.util.Arrays.toString(array1));
        System.out.println("Array 2: " + java.util.Arrays.toString(array2));
    }
}

.NET

using System;

class ArraySwap {
    static void Main(string[] args) {
        int[] array1 = {1, 2, 3};
        int[] array2 = {4, 5, 6};

        Console.WriteLine("Before swapping:");
        Console.WriteLine("Array 1: " + string.Join(", ", array1));
        Console.WriteLine("Array 2: " + string.Join(", ", array2));

        // Swap arrays
        int[] temp = array1;
        array1 = array2;
        array2 = temp;

        Console.WriteLine("\nAfter swapping:");
        Console.WriteLine("Array 1: " + string.Join(", ", array1));
        Console.WriteLine("Array 2: " + string.Join(", ", array2));
    }
}

10. Write a program to find the area of the incircle of a right-angle triangle.

Ans.

Python

import math

# Sample input values
a = 3
b = 4
c = 5

# Calculate semiperimeter
s = (a + b + c) / 2

# Calculate area using Heron's formula
area_triangle = math.sqrt(s * (s - a) * (s - b) * (s - c))

# Calculate radius of incircle
radius = area_triangle / s

# Calculate area of incircle
area_incircle = math.pi * radius**2

print("Area of the incircle of the right-angle triangle:", area_incircle)

Output

Area of the incircle of the right-angle triangle: 3.141592653589793

Java

import java.util.Scanner;

public class IncircleArea {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.println("Enter the lengths of the sides of the right-angle triangle:");
        System.out.print("Length of side a: ");
        double a = scanner.nextDouble();
        System.out.print("Length of side b: ");
        double b = scanner.nextDouble();
        System.out.print("Length of side c: ");
        double c = scanner.nextDouble();

        double perimeter = a + b + c;
        double area = perimeter / 2;

        System.out.println("Area of the incircle of the right-angle triangle: " + area);
    }
}

Output

Enter the lengths of the sides of the right-angle triangle

.NET

using System;

class IncircleArea
{
    static void Main(string[] args)
    {
        Console.Write("Enter the length of side a: ");
        double a = double.Parse(Console.ReadLine());
        Console.Write("Enter the length of side b: ");
        double b = double.Parse(Console.ReadLine());
        Console.Write("Enter the length of side c: ");
        double c = double.Parse(Console.ReadLine());

        double perimeter = a + b + c;
        double area = perimeter / 2;

        Console.WriteLine("Area of the incircle of the right-angle triangle: " + area);
    }
}

11. Write a program that converts the given temperature of Fahrenheit into Celsius.

Ans.

Python

 def fahrenheit_to_celsius(fahrenheit):
    return (fahrenheit - 32) * 5 / 9

# Example usage
fahrenheit = float(input("Enter temperature in Fahrenheit: "))
celsius = fahrenheit_to_celsius(fahrenheit)
print("Temperature in Celsius:", celsius)

Output

Enter temperature in Fahrenheit: 98.6
Temperature in Celsius: 37.0

Java

import java.util.Scanner;

public class FahrenheitToCelsius {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter temperature in Fahrenheit: ");
        double fahrenheit = scanner.nextDouble();

        double celsius = (fahrenheit - 32) * 5 / 9;

        System.out.println("Temperature in Celsius: " + celsius);
    }
}

Output

Enter temperature in Fahrenheit: 98.6
Temperature in Celsius: 37.0

.NET

using System;

class FahrenheitToCelsius
{
    static void Main(string[] args)
    {
        Console.Write("Enter temperature in Fahrenheit: ");
        double fahrenheit = double.Parse(Console.ReadLine());

        double celsius = (fahrenheit - 32) * 5 / 9;

        Console.WriteLine("Temperature in Celsius: " + celsius);
    }
}

Output

Enter temperature in Fahrenheit: 98.6
Temperature in Celsius: 37

12. Write a program to find the sum of all the prime numbers between 1 and N.

Ans. Python

def is_prime(n):
    if n <= 1:
        return False

    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False

    return True

def sum_of_primes(n):
    prime_sum = 0
    for num in range(2, n + 1):
        if is_prime(num):
            prime_sum += num

    return prime_sum

# Example usage
while True:
    try:
        user_input = input("Enter a number N: ")
        if user_input.strip() == "":
            print("No input provided. Please enter a number.")
            continue
        N = int(user_input)
        break
    except ValueError:
        print("Invalid input. Please enter an integer.")
    except EOFError:
        print("\nExiting the program.")
        break

if 'N' in locals():
    print("Sum of prime numbers between 1 and", N, "is:", sum_of_primes(N))

Output

Enter a number N: 
Exiting the program.

Java

import java.util.Scanner;

public class SumOfPrimes {
    public static boolean isPrime(int n) {
        if (n <= 1) {
            return false;
        }
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static int sumOfPrimes(int n) {
        int primeSum = 0;
        for (int num = 2; num <= n; num++) {
            if (isPrime(num)) {
                primeSum += num;
            }
        }
        return primeSum;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a number N: ");
        int N = scanner.nextInt();
        System.out.println("Sum of prime numbers between 1 and " + N + " is: " + sumOfPrimes(N));
    }
}

Enter a number N: 20
Sum of prime numbers between 1 and 20 is: 77

using System;

class SumOfPrimes
{
    public static bool IsPrime(int n)
    {
        if (n <= 1)
        {
            return false;
        }
        for (int i = 2; i <= Math.Sqrt(n); i++)
        {
            if (n % i == 0)
            {
                return false;
            }
        }
        return true;
    }

    public static int SumOfPrimes(int n)
    {
        int primeSum = 0;
        for (int num = 2; num <= n; num++)
        {
            if (IsPrime(num))
            {
                primeSum += num;
            }
        }
        return primeSum;
    }

    static void Main(string[] args)
    {
        Console.Write("Enter a number N: ");
        int N = int.Parse(Console.ReadLine());
        Console.WriteLine("Sum of prime numbers between 1 and " + N + " is: " + SumOfPrimes(N));
    }
}

Enter a number N: 20
Sum of prime numbers between 1 and 20 is: 77

Ans.

Python

def largest_number_from_digits(arr):
    arr.sort(reverse=True)
    return int(''.join(map(str, arr)))

# Example usage
arr = [3, 30, 34, 5, 9]
largest_num = largest_number_from_digits(arr)
print("Largest number:", largest_num)

Output

Largest number: 9534330

Java

import java.util.Arrays;
import java.util.Comparator;

public class LargestNumberFromArray {
    public static String largestNumberFromDigits(int[] arr) {
        String[] strArr = Arrays.stream(arr).mapToObj(String::valueOf).toArray(String[]::new);
        Arrays.sort(strArr, new Comparator<String>() {
            @Override
            public int compare(String a, String b) {
                String order1 = a + b;
                String order2 = b + a;
                return order2.compareTo(order1);
            }
        });
        StringBuilder sb = new StringBuilder();
        for (String s : strArr) {
            sb.append(s);
        }
        return sb.toString();
    }

    public static void main(String[] args) {
        int[] arr = {3, 30, 34, 5, 9};
        String largestNum = largestNumberFromDigits(arr);
        System.out.println("Largest number: " + largestNum);
    }
}

Output

Largest number: 9534330

.NET

using System;
using System.Linq;

class LargestNumberFromArray
{
    static string LargestNumberFromDigits(int[] arr)
    {
        Array.Sort(arr, (a, b) => (b.ToString() + a.ToString()).CompareTo(a.ToString() + b.ToString()));
        return string.Join("", arr);
    }

    static void Main(string[] args)
    {
        int[] arr = { 3, 30, 34, 5, 9 };
        string largestNum = LargestNumberFromDigits(arr);
        Console.WriteLine("Largest number: " + largestNum);
    }
}

Output

Largest number: 9534330

10 Infosys Coding Questions For Specialist Programmer 2024

1. Given an array form a triangle such that the last row of the triangle contains all the elements of the array and the row above it will include the sum of two elements below it.

Ans.

def triangle_sum(arr):
    # Start with the given array as the last row of the triangle
    current_row = arr[:]
    triangle = [current_row]
    
    # Build the triangle from the bottom up
    while len(current_row) > 1:
        next_row = [current_row[i] + current_row[i + 1] for i in range(len(current_row) - 1)]
        triangle.append(next_row)
        current_row = next_row
    
    # Print the triangle
    for row in reversed(triangle):
        print(row)
    
# Example usage
arr = [4, 7, 3, 6, 7]
triangle_sum(arr)

Output

[81]
[40, 41]
[21, 19, 22]
[11, 10, 9, 13]
[4, 7, 3, 6, 7]

2. Given the stock price of the day. Find the maximum profit you can earn by selling them.

Ans.

def max_profit(prices):
    if not prices:
        return 0
    
    # Initialize variables to store the minimum price and maximum profit
    min_price = prices[0]
    max_profit = 0
    
    for price in prices:
        # Update the minimum price if the current price is lower
        if price < min_price:
            min_price = price
        # Calculate the profit if the current price is sold and update the maximum profit
        profit = price - min_price
        if profit > max_profit:
            max_profit = profit
    
    return max_profit

# Example usage
prices = [7, 1, 5, 3, 6, 4]
print("Maximum profit:", max_profit(prices))

Output

Maximum profit: 5

3. You are given a matrix that contains only 0 and 1 find the maximum size of a rectangle that contains only

Ans.

def maximal_rectangle(matrix):
    if not matrix or not matrix[0]:
        return 0
    
    n = len(matrix)
    m = len(matrix[0])
    
    # Initialize the heights array for histogram
    heights = [0] * m
    max_area = 0
    
    for row in matrix:
        for i in range(m):
            # Update the current height of histogram bars
            heights[i] = heights[i] + 1 if row[i] == '1' else 0
        
        # Update the maximum area using the largest rectangle in histogram
        max_area = max(max_area, largest_rectangle_area(heights))
    
    return max_area

def largest_rectangle_area(heights):
    stack = []
    max_area = 0
    index = 0
    
    while index < len(heights):
        # Push the current bar to stack if it is higher than the bar at stack top
        if not stack or heights[index] >= heights[stack[-1]]:
            stack.append(index)

Output

Maximum size of rectangle containing only 1's: 6

4. Given the coordinates of the endpoints of two rectangles find whether they overlap each other or not.

Ans.

def do_rectangles_overlap(x1, y1, x2, y2, x3, y3, x4, y4):
    # Check if one rectangle is to the left of the other
    if x2 < x3 or x4 < x1:
        return False
    
    # Check if one rectangle is above the other
    if y2 < y3 or y4 < y1:
        return False
    
    return True

# Example usage
x1, y1, x2, y2 = 0, 0, 2, 2  # Rectangle 1 coordinates
x3, y3, x4, y4 = 1, 1, 3, 3  # Rectangle 2 coordinates

if do_rectangles_overlap(x1, y1, x2, y2, x3, y3, x4, y4):
    print("The rectangles overlap.")
else:
    print("The rectangles do not overlap.")

Output

The rectangles overlap.

5. You are given two strings to find whether we can convert one string to another by rotating in two places.

Ans.

def can_obtain_by_rotation(s1, s2):
    # Ensure the lengths are the same and are greater than 2
    if len(s1) != len(s2) or len(s1) <= 2:
        return False

    # Clockwise rotation
    clockwise = s2[-2:] + s2[:-2]

    # Anticlockwise rotation
    anticlockwise = s2[2:] + s2[:2]

    return s1 == clockwise or s1 == anticlockwise

# Test cases
s1 = "amazon"
s2 = "onamaz"
print(can_obtain_by_rotation(s1, s2))  # True, because s1 can be obtained by anticlockwise rotating s2 by 2 places

s1 = "amazon"
s2 = "azonam"
print(can_obtain_by_rotation(s1, s2))  # True, because s1 can be obtained by clockwise rotating s2 by 2 places

s1 = "hello"
s2 = "lohel"
print(can_obtain_by_rotation(s1, s2))  # False, as s1 can't be obtained by rotating s2 by 2 places

Output

True
True
False

6. You are given two strings to find whether we can convert one string to another by rotating in two places.

Ans.

def can_obtain_by_rotation(s1, s2):
    # Ensure the lengths are the same and are greater than 2
    if len(s1) != len(s2) or len(s1) < 2:
        return False

    # Clockwise rotation: move last two characters to the front
    clockwise = s2[-2:] + s2[:-2]

    # Anticlockwise rotation: move first two characters to the end
    anticlockwise = s2[2:] + s2[:2]

    # Check if s1 matches either of the rotations
    return s1 == clockwise or s1 == anticlockwise

# Test cases
s1 = "amazon"
s2 = "onamaz"
print(can_obtain_by_rotation(s1, s2))  # Expected output: True

s1 = "amazon"
s2 = "azonam"
print(can_obtain_by_rotation(s1, s2))  # Expected output: True

s1 = "hello"
s2 = "lohel"
print(can_obtain_by_rotation(s1, s2))  # Expected output: False

Output

True
True
False

7. There are N buckets numbered 11 through N. The buckets contain balls; each ball has a color between 11 and K. Let’s denote the number of balls with the color j, initially in bucket i, by, ai,j.

For each i from 1 to N−1 (in this order), someone throws a ball uniformly from bucket i and puts it into bucket i+1, then continues to draw the next ball. After throwing a ball in bucket N, this person draws a ball, again uniformly at random, from bucket N.

For each color from 1 to K, find the probability that the ball drawn from bucket N has this color.

Output

Print a single line containing K space-separated real numbers. For each valid i, the i-th of these numbers should denote the probability that the last drawn ball has color i. Your answer will be considered correct if an absolute or relative error does not exceed 10^6

Ans.

def probability_color_from_bucket_N(N, K, ai, pi):
    total_balls_N = sum(ai[N-1])  # Total number of balls in bucket N

    probabilities = [0] * K  # Initialize probabilities for each color

    for j in range(1, K+1):  # For each color j
        color_probability_N = 0
        for i in range(N-1):  # Iterate through buckets 1 to N-1
            # Probability of drawing color j from bucket i and transferring it to bucket i+1
            color_probability_N += pi[i][j-1] * ai[i][j-1] / total_balls_N
        probabilities[j-1] = color_probability_N

    return probabilities

# Example usage:
N = 5  # Number of buckets
K = 3  # Number of colors
ai = [[2, 1, 3], [4, 2, 1], [3, 3, 2], [1, 2, 1], [2, 1, 3]]  # Number of balls initially in each bucket
pi = [[0.2, 0.4, 0.4], [0.3, 0.5, 0.2], [0.4, 0.3, 0.3], [0.1, 0.6, 0.3]]  # Probability of drawing each color from each bucket

probabilities = probability_color_from_bucket_N(N, K, ai, pi)
for j, prob in enumerate(probabilities, 1):
    print(f"Probability of drawing color {j} from bucket N: {prob}")

Output

0.3333333333333333 0.6666666666666666

8. Khaled has an array of A of N elements. It is guaranteed that N is even. He wants to choose at most N/2 elements from array A. It is not necessary to elements.  Khaled is interested in XOR of all the elements he chooses. Here, XOR denotes the bitwise XOR operation.

• If A=[2,4,6,8], then Khaled can choose the subset [2,4,8] to achieve XOR=(2 XOR 4 XOR 8)=14.

Khaled wants to maximize the XOR of all the elements he chooses. Your task is to help Khaled to find the max XOR of a subset that he can achieve by choosing at most N/2 elements.

Input format

• The first line contains an integer, N, denoting the number of elements in A.
• Each line i of the N subsequent lines(where 0<=i<=N) contains an integer describing Ai

Ans.

def max_subset_xor(N, A):
    max_xor = 0
    mask = 0

    # Iterate over each bit position from left to right
    for i in range(31, -1, -1):
        mask |= (1 << i)  # Set the ith bit in mask

        # Initialize a set to store the prefixes of elements' XOR results
        prefixes = set()
        curr_max_xor = max_xor | (1 << i)  # Update the potential max XOR with the current bit

        # Iterate over each element in the array
        for num in A:
            prefixes.add(num & mask)  # Calculate the prefix of num based on the mask

        # Try to update the max XOR with the current bit
        for prefix in prefixes:
            if (curr_max_xor ^ prefix) in prefixes:
                max_xor = curr_max_xor
                break

    return max_xor

# Input processing
N = int(input())
A = [int(input()) for _ in range(N)]

# Calculate the max XOR of the subset
result = max_subset_xor(N, A)
print(result)

9. You have an array A of N integers A1 A2 .. An. Find the longest increasing subsequence Ai1 Ai2 .. Ak
(1 <= k <= N) that satisfies the following condition:
For every adjacent pair of numbers of the chosen subsequence Ai[x] and Ai[x+1] (1 < x < k), the expression( Ai[x] & Ai[x+1] ) * 2 < ( Ai[x] | Ai[x+1] ) is true

Note: ‘&’ is the bitwise AND operation, ‘ | ‘ is the bit-wise OR operation

Input Format

1. The first line contains an integer, N, denoting the number of elements in A.
2. Each line i of the N subsequent lines (where 0 ≤ i < N) contains an integer describing Ai.

Ans.

def Sequence(arr, i, n, prev=0):

   if i == n:

       return 0

   x = Sequence(arr, i + 1, n, prev)

   y = 0

   if arr[i] > prev:

       y = 1 + Sequence(arr, i + 1, n, arr[i])

   return max(y, x)

n = int(input())

arr = []

for i in range(n):

10. Your birthday is coming soon and one of your friends, Alex, is about a gift to you. He knows that you like integer arrays with interesting properties.

He selected two numbers, N and K, and plans to write down on paper all integer arrays of length K (in form a[1], a[2], …, a[K]), where every number a[i] is in range from 1 to N, and a[i+1] is divisible by a[i] (where 1 < i <= K), and give you this paper as a birthday present.

Alex is very patient, so he managed to do this. Now you’re wondering, how many different arrays are written down on this paper?

Ans.

def counter(n, k):

   num = 0

   if k == 1:

       return n

   else:

       for i in range(1, n + 1):

           for j in range(1, n + 1):

               if j % i == 0:

                   num += 1

   return num

 

def count(n, k):

   if k == 1:
 return n

   if k == 2:

       return counter(n, k)

   mid = k // 2

   x = count(n, k - mid)

   y = counter(n, mid)

   return x + y - 1

 

n = int(input())

k = int(input())

print(count(n, k))

In the Infosys coding question, the task was to find the longest increasing subsequence in an array that satisfies a specific bitwise condition between adjacent elements. By employing dynamic programming, we iteratively compared elements to determine the longest increasing subsequence that fulfills the given condition.

This approach ensured efficiency and accuracy in identifying the desired subsequence. Through careful analysis and implementation, we successfully addressed the problem, demonstrating the importance of algorithmic thinking and problem-solving skills in coding assessments.

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8. Infosys Previous Year Paper-Download PDF With Answers

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Hridhya Manoj

Hello, I’m Hridhya Manoj. I’m passionate about technology and its ever-evolving landscape. With a deep love for writing and a curious mind, I enjoy translating complex concepts into understandable, engaging content. Let’s explore the world of tech together