Getting super digit of a number

Given an integer 'z' we can find it's super digit as follow:

If it is a single digit number then it itself is a super digit else the super digit is equal to the super digit of the digit-sum of 'z'.

Example:

1) Input: 5
SuperDigit = 5

2) Input: 1234

SuperDigit(1234) = SuperDigit(1+2+3+4)
                             = SuperDigit(10)
                             = SuperDigit(1+0)
                             = SuperDigit(1)
                             = 1

Input Format:
You are given two numbers n and k. You have to calculate the super digit of p. p is created when the number is concatenated k times. That is, if n= 756  and k=2, then p=756756.
That means you have to calculate super digit for 756756.

The first line contains two space-separated integers n and k.

Java Code:

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

   
    static int digitSum(String n, int k) {
       if(n.length()==1)
           return (Integer.valueOf(n));
        else{
            long sum=0;
            for(int i=0;i<n.length();i++)
            {
                sum=sum+(Long.parseLong(String.valueOf(n.charAt(i))));
               
            }
         
           
            String num=  String.format ("%d", sum);
            return digitSum(num,k);
        }
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String n = in.next();
        int k = in.nextInt();
        long sum=0;
        for(int i=0;i<n.length();i++)
            {
                sum=sum+(Long.parseLong(String.valueOf(n.charAt(i))));
               
            }
        sum=sum*k;
   
        int result = digitSum(Long.toString(sum), k);
        System.out.println(result);
        in.close();
    }
}

Constraints:

1<= n < 10^100000
1<= k< 10^5

 Printing a Series:

 Consider a series of number generated from the following formula

fn+1=fn-1+(fn-2)^2;

and the first two numbers are a0 = 0 and a1 = 1

Find the nth number of the series?

The logic to solve:

1) Generate the series using the above formula until nth term is obtained
2) Print the nth number in the series

Tips: As n can be a larger number so you need to choose a data type capable of handling the larger number.

Solution (Java Code):

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Demo {

    static BigInteger generateSeries(BigInteger t1, BigInteger t2, BigInteger n) {
        int condition=n.compareTo(new BigInteger("0"));
                if(condition==0)
                    return t2;
                else
                {
                        BigInteger one= new BigInteger("1");
                       
                        return generateSeries(t2,(t1.add(t2.multiply(t2))),n.subtract(one));
                   
                }
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
       String a = in.next();
        String b = in.next();
        String c = in.next();
        BigInteger t1= new BigInteger(a);
                BigInteger t2= new BigInteger(b);
                BigInteger n= new BigInteger(c);

        BigInteger term= generateSeries(t1, t2, n.subtract(new BigInteger("2")));
        System.out.println(term);
        in.close();
    }

}

Printing patterns

To print a pattern using '*' and dots ( . ):

code:
#include<stdio.h>
void main()
{
    int i,j,k,m,n;
    printf("enter the value for m and n");
    scanf("%d%d",&n,&m);
        for(i=0;i<((3*n)+1);i++)
        {
            for(j=0;j<3*m+1;j++)
            {
                if(i==0||j==0||i%3==0||j%3==0)
                    printf("*");
                else
                    printf(".");
            }
            printf("\n");
        }

}

input: 3 1
output:

****
*. . *
*. . *
****
*. . *
*. . *
****
*. . *
*. . *
****

input: 3 4

output:

*************

*. . *. . *. . *. . *

*. . *. . *. . *. . *

*************

*. . *. . *. . *. . *

*. . *. . *. . *. . *

*************

*. . *. . *. . *. . *

*. . *. . *. . *. . *

*************

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use of concat() and lower() in sql

Generate the following two result sets:

1)Query an alphabetically ordered list of all names in OCCUPATIONS, immediately followed by the first letter of each profession as a parenthetical (i.e.: enclosed in parentheses). For example: AnActorName(A), ADoctorName(D), AProfessorName(P), and ASingerName(S).

2)Query the number of ocurrences of each occupation in OCCUPATIONS. Sort the occurrences in ascending order, and output them in the following format:

There are total [occupation_count] [occupation]s.

where [occupation_count] is the number of occurrences of an occupation in OCCUPATIONS and [occupation] is the lowercase occupation name. If more than one Occupation has the same [occupation_count], they should be ordered alphabetically.

the table is

occupations( name varchar(45), occupation varchar(25))
QUERY:

select concat(name,'(',left(occupation,1),')')
from occupations
order by name;
select concat('There are total ',count(occupation),' ',lower(occupation),'s','.')
from occupations
group by occupation

order by count(occupation),occupation;

output:
Aamina(D) 
Ashley(P) 
Belvet(P) 
Britney(P) 
Christeen(S) 
Eve(A) 
Jane(S) 
Jennifer(A) 
Jenny(S) 
Julia(D) 
Ketty(A) 
Kristeen(S) 
Maria(P) 
Meera(P) 
Naomi(P) 
Priya(D) 
Priyanka(P) 
Samantha(A) 
There are total 3 doctors. 
There are total 4 actors. 
There are total 4 singers. 

There are total 7 professors. 

Graph colouring using backtracking

GRAPH COLOURING:

Given a graph G (V,E) where V represents the vertices in the graph and E represents the edges in the graph . Given  m different colours to colour the nodes in such a way that no two adjacent nodes are of same colour.

eg:
for the given graph

and given two colours yellow and red

there are two solutions possible:

code:
#include<stdio.h>
int GAM[5][5];
int X[5];
int n,m;
int print()
{
    int i,j;
    printf("THE COLOURS OF EACH NODE IS:");
    printf("for node 1\t2\t3\t4\n");
    for(i=1;i<=n;i++)
    {

        printf("%d",X[i]);
        printf("\n");
    }

}
int issafe( int k,int p)
{
    int c;
    for(c=1;c<k;c++)
    {
        if(X[c]==p && GAM[c][k]==1)
            return 0;
    }
   return 1;
}
void graphcolor(int k,int m)
{
    int p;
    for(p=1;p<=m;p++)// p represents colour no
    {
        if(issafe(k,p))//checking if could be coloured or not
         {
            X[k]=p;
            if(k==n)
            print();
            else
                graphcolor(k+1,m);
         }
    }

}

int main()
{

    printf("enter the number of nodes");
    scanf("%d",&n);
    printf("enter 1 if edge exists else 0");
    int i,j;
    for( i=1;i<=n;i++)
        for(j=1;j<=n;j++)
    {
        printf("edge %d->%d",i,j);
        scanf("%d",&GAM[i][j]);

    }
    printf("enter number of colours available");
    scanf("%d",&m);
    graphcolor(1,m);
    return 0;
}

steps involved ;

for solution 1:(solution two similar to this)

Type of Triangle

Write a query identifying the type of each record in the TRIANGLES table using its three side lengths. Output one of the following statements for each record in the table:

• Not A Triangle: The given values of A, B, and C don't form a triangle.
• Equilateral: It's a triangle with  sides of equal length.
• Isosceles: It's a triangle with  sides of equal length.
• Scalene: It's a triangle with  sides of differing lengths.

Input Format

The TRIANGLES table is described as follows:

Each row in the table denotes the lengths of each of a triangle's three sides.

Sample Input

Sample Output

Isosceles
Equilateral
Scalene
Not A Triangle

Explanation

Values in the tuple  form an Isosceles triangle, because . 
Values in the tuple  form an Equilateral triangle, because . Values in the tuple  form a Scalene triangle, because . 
Values in the tuple  cannot form a triangle because the combined value of sides  and  is not larger than that of side .

Query:
select
case
when (a+b)<=c or (a+c)<=b or (b+c)<=a  then 'Not A Triangle'
when a=b and b=c then 'Equilateral'
when (a=b and b<>c) or (b=c and a<>c) or (a=c and c<>b)  then 'Isosceles'
else    'Scalene'
end
from Triangles;