CseWay

Fibonacci-prime number : Given a number N, you need to find if N is fibonacci-prime number or not. A fibonacci-prime is any number that is both a prime and a fibonacci number.

Given a non-negative integer N, find the largest number that is less than or equal to N with monotone increasing digits.

Given a string, sort it in decreasing order based on the frequency of characters.

Find the smallest positive number missing from an unsorted array .

1. Given a number n, write a function that returns count of numbers from 1 to n that don’t contain digit 3 in their decimal representation.

Validating ISBN: Write a C program to define a structure Book with the attributes, book title,
author, subject, isbn number, Based on the given input values display the books information if the
given ISBN is valid.
Following are the rules to verify whether the given ISBN is a valid or not.
The 2001 edition of the official manual of the International ISBN Agency says that the ISBN-10
check digit – which is the last digit of the ten-digit ISBN – must range from 0 to 10 (the symbol X is
used for 10), and must be such that the sum of all the ten digits, each multiplied by its (integer)
weight, descending from 10 to 1, is a multiple of 11.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>

_Bool isValidISBN(long);
int getLength(char*);

int main(int argc, char *argv[]) {
 struct Book {
  char *bookTitle;
  char *bookAuthor;
  char *subject;
  long isbn;
 };

 struct Book book[5];
 int n;
 char *s = (char *)malloc(sizeof(char)*100);
 int length = getLength(s);
 scanf("%d", &n);
 for(int i=0; i<n; i++) {
  scanf("%s", s);
  fflush(stdin);
  int length = getLength(s);
  book[i].bookTitle = (char *)malloc(sizeof(char) * length);
  strcpy(book[i].bookTitle, s);
  scanf("%s", s);
  fflush(stdin);
  length = getLength(s);
  book[i].bookAuthor = (char *)malloc(sizeof(char) * length);
  strcpy(book[i].bookAuthor, s);
  scanf("%s", s);
  length = getLength(s);
  fflush(stdin);
  book[i].subject = (char *)malloc(sizeof(char) * length);
  strcpy(book[i].subject, s);
  scanf("%ld", &book[i].isbn);
  if(isValidISBN(book[i].isbn)==true) {
   printf("{Book Title : %s; ", book[i].bookTitle);
   printf("Book Author : %s; ", book[i].bookAuthor);
   printf("Subject : %s; ", book[i].subject);
   printf("ISBN : %ld}\n", book[i].isbn);
  }
 }
}

int getLength(char *s) {
 int length = 0;
 while(s[length]!='\0') {
  length++;
 }
 return length;
}

_Bool isValidISBN(long isbn) {
 long sum = 0;
 int i = 1;
 while(isbn>0) {
  sum = sum + (isbn)%10 * i;
  isbn = isbn / 10;
  i++;
 }
 if(sum%11==0)
  return true;
 return false;
}

By listing the first six prime numbers: and , we can see that the prime is .
What is the prime number?

a = [0]*200008
for i in range(2,200000):
    if (a[i] == 1):
        continue
    for j in range(2*i,200000,i):
        a[j] = 1
primes = []
for i in range(2,200000):
    if a[i] == 0:
        primes.append(i)
p = int(input())
for _ in range(p):
 t = int(input())
 print(primes[t-1])

Sum of Factorial of the digits of the given number: Write a python program to compute sum of
the factorials of each digit of a given integer. Write a separate function for calculating the
factorial of a number.

def fact(n):
    result = 1
    for i in range(n):
        result *= i + 1
    return result

def sum_of_fact(n):
    result = 0
    if(n <= 0):
     return 1
    else:
     while n > 0:
         result += fact(n % 10)
         n = n // 10
     return result

print(sum_of_fact(int(input())))

import java.util.Scanner;

public class Solution {

 public static long factorialN(long n) {
  if (n <= 0)
   return 1;
  else 
   return n * factorialN(n-1);
 }

 public static void main(String[] args) {
  Scanner scan = new Scanner(System.in);
  long n = scan.nextLong();
  int sum = 0;
  if(n <= 0) {
   System.out.println("1");
  } else {
   while(n != 0) {
    sum = sum + (int)factorialN(n%10);
    n = n/10;
   }
   System.out.println(sum);
  }
 }
}

C
#include <stdio.h>

long long int factorialN(long long int n) {
 if (n <= 0)
  return 1;
 else 
  return n * factorialN(n-1); 
}

int main(int argc, char const *argv[]) {
 long long int n = 0;
 long long int sum = 0;
 scanf("%llu",&n);
 if(n <= 0) {
  printf("1");
 } else {
  while(n != 0) {
   sum = sum + factorialN(n%10);
   n = n/10;
  }
  printf("%llu\n",sum);
 }
 return 0;
}

Moving all Zeros to the Left Side of the Array: Write a program to move all the zeros on the
left side of the array while maintaining the relative order of non-zero elements. You should not
use another array.

Solution :
C

/*
    Float all zeros to the left of an array
*/

#include <stdio.h>
#include <stdlib.h>

void float_zeros(int *arr, int len);
void print_array(int *arr, int len);

int main(void) {
    int n = 0, *arr = NULL, i = 0;
    scanf("%d", &n);

    if(n <= 0) {
        fprintf(stderr, "The array size can't be zero or negative\n");
        return 1;
    }

    arr = (int*)calloc(1, n * sizeof(int));
    for(i = 0; i < n; i++)
        scanf("%d", &arr[i]);

    float_zeros(arr, n);
    print_array(arr, n);

    free(arr);
    return 0;
}

void float_zeros(int *arr, int len) {
    /*
        Find all the 0s in arr, and float them to the left end of arr.
    */
    int i = 0, prev = 0, j = 0, k = 0, temp = 0;
    while(i < len) {
        // Find zeros
        if(*(arr + i) == 0) {
            // Found a zero. Now float it.
            // To do that, we need where to float it to.
            // That would be prev.
            for(j = i; j > prev; j--) {
                k = j - 1;
                temp = *(arr + j);
                *(arr + j) = *(arr + k);
                *(arr + k) = temp;
            }
            prev++;
        }
        i++;
    }
}

void print_array(int *arr, int len) {
    /*
        Prints an int array of length len
    */

    int i = 0;
    while(i < len) {
        printf("%d", *(arr + i));
        if(i != len - 1)
            printf(" ");
        i++;
    }
    printf("\n");
}

JAVA:

/*
    Float zeros
*/

import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        int n = 0;

        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();

        if(n <= 0)
            System.err.println("The array size can't be zero or negative");

        int arr[] = new int[n];
        for(int i = 0; i < n; i++)
            arr[i] = sc.nextInt();

        sc.close();

        // Changing the array elements here.
        int i = 0, prev = 0;
        while(i < n) {
            // Find a zero
            if(arr[i] == 0) {
                // Found zero.
                // Float it
                for(int j = i; j > prev; j--) {
                    int k = j - 1, temp = arr[k];
                    arr[k] = arr[j];
                    arr[j] = temp;
                }
                prev++;
            }
            i++;
        }

        for(i = 0; i < n; i++) {
            System.out.print(arr[i]);
            if(i != n - 1)
                System.out.print(" ");
        }
        System.out.println("");
    }
}

Given an array with N elements, the task is to find the count of factors of a number X which is product of all array elements.
Example:
Input : 2 4 6
Output : 10 2 * 4 * 6 = 48,the factors of 48 are 1, 2,
3, 4, 6, 8, 12, 16, 24, 48 whose count is 10.

Solution:
JAVA
import java.util.*;
import java.lang.*;
import java.io.*;

class Solution {
    static List<Integer> ll = new ArrayList<>();
    static  final int MAX = 101;
    static boolean[] isP = new boolean[MAX];
    static Scanner sc = new Scanner(System.in);
public static void main (String[] args) {
//code
init();
int n = sc.nextInt();

while(n-- > 0) solve();
}

public static void solve(){
    int n = sc.nextInt();
    if(n>0){
    int[] arr = new int[n];
    long ans = 1;
    int flag =0;
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    for(int i = 0; i < n; i++){
        arr[i] = sc.nextInt();
        if(arr[i]>=1){
        flag=1;
        countFact(arr[i], map);
}
    }
    for(int key : map.keySet()){
        ans *= (map.get(key) + 1);
    }
    if(flag==1){
    System.out.println(ans);}
}
   
}
static void countFact(int n, Map<Integer, Integer> map){
    int i = 0;
    while(n != 1){
        while(n % ll.get(i) == 0){
            n = n / ll.get(i);
            map.put(ll.get(i),
                map.getOrDefault(ll.get(i), 0) + 1);
        }
        i++;
    }
}
static void init(){
    Arrays.fill(isP, true);
    isP[0] = isP[1] = false;
    for(int i = 2; i * i < MAX; i++){
        if(isP[i]){
            for(int j = i * i; j < MAX; j += i){
                isP[j] = false;
            }
        }
    }
    for(int i = 2; i < MAX; i++){
        if(isP[i]) ll.add(i);
    }
}
}

Reorganize The Array :

Given an array of elements of length ​ N ​ , ranging from ​ 0 to N-1​ , your task is to write a program that rearranges the elements of the array. All elements may not be present in the array, if element is not present then there will be -1 present in the array. Rearrange the array such that A[i] = i and if i is not present, display -1 at that place.
Examples:
                   Input : arr = {-1, -1, 6, 1, 9, 3, 2, -1, 4, -1}
                   Output : {-1, 1, 2, 3, 4, -1, 6, -1, -1, 9}

Solution :
C
#include <stdio.h>
#include <stdlib.h>
int main() {
int T,i,n;
scanf("%d", &T);
while(T--) {
    scanf("%d", &n);
    int *arr,k;
    arr = (int *)calloc(n+1,sizeof(int));
    for(i = 0; i < n; i++) {
        scanf("%d", &k);
        if(k >= 0) {
            arr[k]++;
        }
    }
    for(i = 0; i < n; i++) {
        if(arr[i] > 0) {
            printf("%d ", i);
        }
        else {
            printf("-1 ");
        }
    }
    printf("\n");
}
return 0;
}

In programming as per the requirement we need to have multiple classes. So when there are multiple classes ,some classes will depend on other classes and some classes doesn't require any other classes help.

Loose Coupling : The name indicates that the class is single and doesn't depend on another class for its functionality. By using this you can isolate the things and make the changes easily because it is not depending on any other class

Tight Coupling : The name indicates that the class is not isolated and depends on multiple classes. This kind of coupling needs to be taken care because any change in one class must be notified by all other classes that are depending on that class.