Minuscule Technologies Fresher Interview Questions
This blog explains about Minuscule Technologies Pvt. Ltd Fresher Interview Questions are given below:
1. Write a program to find the second largest number in an array, without sorting.
| // JAVA Code for Find Second largest // element in an array class GFG {
/* Function to print the second largest elements */ public static void print2largest(int arr[], int arr_size) { int i, first, second;
/* There should be atleast two elements */ if (arr_size < 2) { System.out.print(” Invalid Input “); return; }
first = second = Integer.MIN_VALUE; for (i = 0; i < arr_size ; i++) { /* If current element is smaller than first then update both first and second */ if (arr[i] > first) { second = first; first = arr[i]; }
/* If arr[i] is in between first and second then update second */ else if (arr[i] > second && arr[i] != first) second = arr[i]; }
if (second == Integer.MIN_VALUE) System.out.print(“There is no second largest”+ ” element\n”); else System.out.print(“The second largest element”+ ” is “+ second); }
/* Driver program to test above function */ public static void main(String[] args) { int arr[] = {12, 35, 1, 10, 34, 1}; int n = arr.length; print2largest(arr, n); } } |
Output:
The second largest element is 34
2. Write program to find number of line, word and characters available in a paragraph/file.
| // Java program to count the // number of charaters in a file import java.io.*;
public class Test { public static void main(String[] args) throws IOException { File file = new File(“C:\\Users\\Mayank\\Desktop\\1.txt”); FileInputStream fileStream = new FileInputStream(file); InputStreamReader input = new InputStreamReader(fileStream); BufferedReader reader = new BufferedReader(input);
String line;
// Initializing counters int countWord = 0; int sentenceCount = 0; int characterCount = 0; int paragraphCount = 1; int whitespaceCount = 0;
// Reading line by line from the // file until a null is returned while((line = reader.readLine()) != null) { if(line.equals(“”)) { paragraphCount++; } if(!(line.equals(“”))) {
characterCount += line.length();
// \\s+ is the space delimiter in java String[] wordList = line.split(“\\s+”);
countWord += wordList.length; whitespaceCount += countWord -1;
// [!?.:]+ is the sentence delimiter in java String[] sentenceList = line.split(“[!?.:]+”);
sentenceCount += sentenceList.length; } }
System.out.println(“Total word count = ” + countWord); System.out.println(“Total number of sentences = ” + sentenceCount); System.out.println(“Total number of characters = ” + characterCount); System.out.println(“Number of paragraphs = ” + paragraphCount); System.out.println(“Total number of whitespaces = ” + whitespaceCount); } } |
Output:
Total word count = 5
Total number of sentences = 3
Total number of characters = 21
Number of paragraphs = 2
Total number of whitespaces = 7
3. Program to find the summation of integer number (for e.g, input = 8973 8+9+7+3 = 27 = 2+7 = 9 output should be 9)
| // Java Program to find sum of series import java.io.*;
class GFG {
// Function to return sum of // 1/1 + 1/2 + 1/3 + ..+ 1/n static double sum(int n) { double i, s = 0.0; for (i = 1; i <= n; i++) s = s + 1/i; return s; }
// Driven Program public static void main(String args[]) { int n = 5; System.out.printf(“Sum is %f”, sum(n));
} }
// This code is contributed by Nikita Tiwari. |
Output:
2.283333
Type 2
#include<stdio.h>
int main(){
int num,sum=0,r;
printf("Enter a number: ");
scanf("%d",&num);
while(num){
r=num%10;
num=num/10;
sum=sum+r;
}
printf("Sum of digits of number: %d",sum);
return 0;
}Sample output:
Enter a number: 123
Sum of digits of the number: 6
4. Print astriks in pyramid format, input should be number of rows
*
* * *
* * * * *
* * * * * * *| // Java code to demonstrate star pattern public class GeeksForGeeks { // Function to demonstrate printing pattern public static void printTriagle(int n) { // outer loop to handle number of rows // n in this case for (int i=0; i<n; i++) {
// inner loop to handle number spaces // values changing acc. to requirement for (int j=n-i; j>1; j–) { // printing spaces System.out.print(” “); }
// inner loop to handle number of columns // values changing acc. to outer loop for (int j=0; j<=i; j++ ) { // printing stars System.out.print(“* “); }
// ending line after each row System.out.println(); } }
// Driver Function public static void main(String args[]) { int n = 5; printTriagle(n); } } |
Output:
*
* *
* * *
* * * *
* * * * *5. Find a String is a palindrome or not? (for eg “MALAYALAM” – Reverse of string also give same value)
| // Java program to find if given // string is K-Palindrome or not class GFG {
// find if given string is // K-Palindrome or not static int isKPalDP(String str1, String str2, int m, int n) {
// Create a table to store // results of subproblems int dp[][] = new int[m + 1][n + 1];
// Fill dp[][] in bottom up manner for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) {
// If first string is empty, // only option is to remove all // characters of second string if (i == 0) { // Min. operations = j dp[i][j] = j; }
// If second string is empty, // only option is to remove all // characters of first string else if (j == 0) { // Min. operations = i dp[i][j] = i; }
// If last characters are same, // ignore last character and // recur for remaining string else if (str1.charAt(i – 1) == str2.charAt(j – 1)) { dp[i][j] = dp[i – 1][j – 1]; }
// If last character are different, // remove it and find minimum else { // Remove from str1 // Remove from str2 dp[i][j] = 1 + Math.min(dp[i – 1][j], dp[i][j – 1]); } } } return dp[m][n]; }
// Returns true if str is k palindrome. static boolean isKPal(String str, int k) { String revStr = str; revStr = reverse(revStr); int len = str.length();
return (isKPalDP(str, revStr, len, len) <= k * 2); }
static String reverse(String str) { StringBuilder sb = new StringBuilder(str); sb.reverse(); return sb.toString(); }
// Driver program public static void main(String[] args) { String str = “acdcb”; int k = 2; if (isKPal(str, k)) { System.out.println(“Yes”); } else { System.out.println(“No”); } } }
// This code is contributed by // PrinciRaj1992 |
Output :
Yes
6. How to find the first non-repeated character of a given String
| // Java program to find first non-repeating character
class GFG { static final int NO_OF_CHARS = 256; static char count[] = new char[NO_OF_CHARS];
/* calculate count of characters in the passed string */ static void getCharCountArray(String str) { for (int i = 0; i < str.length(); i++) count[str.charAt(i)]++; }
/* The method returns index of first non-repeating character in a string. If all characters are repeating then returns -1 */ static int firstNonRepeating(String str) { getCharCountArray(str); int index = -1, i;
for (i = 0; i < str.length(); i++) { if (count[str.charAt(i)] == 1) { index = i; break; } }
return index; }
// Driver method public static void main (String[] args) { String str = “geeksforgeeks”; int index = firstNonRepeating(str);
System.out.println(index == -1 ? “Either all characters are repeating or string ” + “is empty” : “First non-repeating character is ” + str.charAt(index)); } } |
Output:
First non-repeating character is f
7. In an array 1-100 multiple numbers are duplicates, how do you find it.
| class RepeatElement { void printRepeating(int arr[], int size) { int i, j; System.out.println(“Repeated Elements are :”); for (i = 0; i < size; i++) { for (j = i + 1; j < size; j++) { if (arr[i] == arr[j]) System.out.print(arr[i] + ” “); } } }
public static void main(String[] args) { RepeatElement repeat = new RepeatElement(); int arr[] = {4, 2, 4, 5, 2, 3, 1}; int arr_size = arr.length; repeat.printRepeating(arr, arr_size); } } |
Output :
Repeating elements are 4 2
8. How do you count several vowels and consonants in a given string?
- public class CountVowelConsonant {
- public static void main(String[] args) {
- //Counter variable to store the count of vowels and consonants
- int vCount = 0, cCount = 0;
- //Declare a string
- String str = “This is a really simple sentence”;
- //Converting entire string to lower case to reduce the comparisons
- str = str.toLowerCase();
- for(inti = 0; i < str.length(); i++) {
- //Checks whether a character is a vowel
- if(str.charAt(i) == ‘a’|| str.charAt(i) == ‘e’ || str.charAt(i) == ‘i’ || str.charAt(i) == ‘o’ || str.charAt(i) == ‘u’) {
- //Increments the vowel counter
- vCount++;
- }
- //Checks whether a character is a consonant
- elseif(str.charAt(i) >= ‘a’ && str.charAt(i)<=’z’) {
- //Increments the consonant counter
- cCount++;
- }
- }
- out.println(“Number of vowels: “+ vCount);
- out.println(“Number of consonants: “+ cCount);
- }
- }
Output:
Number of vowels: 10
Number of consonants: 17
9. Write a program to find Fibonacci and factorial using recurssion.Ra
| /* * C Program to print the Fibonacci series using recursion */ #include <stdio.h> #include <conio.h>
int fibonacci(int term); int main(){ int terms, counter; printf(“Enter number of terms in Fibonacci series: “); scanf(“%d”, &terms); /* * Nth term = (N-1)th therm + (N-2)th term; */ printf(“Fibonacci series till %d terms\n”, terms); for(counter = 0; counter < terms; counter++){ printf(“%d “, fibonacci(counter)); } getch(); return 0; } /* * Function to calculate Nth Fibonacci number * fibonacci(N) = fibonacci(N – 1) + fibonacci(N – 2); */ int fibonacci(int term){ /* Exit condition of recursion*/ if(term < 2) return term; return fibonacci(term – 1) + fibonacci(term – 2); } |
Program Output
Enter number of terms in Fibonacci series: 9
Fibonacci series till 9 terms
0 1 1 2 3 5 8 13 21
Final Thoughts:
Preparing for interviews can feel challenging, but having clarity on the types of questions asked can make a huge difference in your confidence and performance. The insights shared in this guide are designed to help you understand real interview patterns and strengthen your technical foundation.
At Payilagam, we focus on empowering learners with practical knowledge and industry-relevant skills, ensuring you are well-equipped to face interviews at top companies. Whether you’re targeting opportunities as a fresher or aiming specifically to crack minuscule technologies interview questions, the concepts and preparation tips discussed here will guide you in the right direction.
Stay committed, keep practising, and let Payilagam support you in your journey toward a successful career!
REFERENCES:
