We Keep Coding

How to get the first and last digits of a number and combine both

Example:
If i give a number 12345 , i should get an answer like 15243
in the same way if it is 123456 , i should get 162534
i have already tried getting the first value and appending the last value using reverse technique
public class MyFirstJavaProgram {
public static void main(String []args) {
String str = "12345";
String val = str;
char a;
int num=0;
int d=0;
int n;
for(int i=0; i<=str.length()/2; i++) {
a = str.charAt(i);
num = num*10+Character.getNumericValue(a);
if(Integer.parseInt(str)!=0){
d=Integer.parseInt(str)%10;
num = num*10+d;
n=Integer.parseInt(str)/10;
str = Integer.toString(n);
}
}
System.out.println(num);
}
}
i should get the result if they give even number or odd number

Without doing what is presumably homework for you, imagine you have a loop in which there are two integer variables a and b. Variables a and b are string indexes.
You are taking characters from the string at positions a,b,a,b,a,b etc.
BUT the values of a and b need to change for each iteration. If the length of the String is n, a will follow the sequence 0,1,2,3... and b will follow the sequence (n-1),(n-2),(n-3) etc
The loop should continue while a < b.

This is my solution for your exercise:
Method parameter "A" is Integer number you want to parse.
First you create Char Array from given number and then iterate through it. If i%2 == 0 it means that you take number from beginning otherwise from the end
public static int algorithm(int A) {
StringBuilder shuffleNumber = new StringBuilder();
char[] numbersArray = Integer.toString(A).toCharArray();
for (int i = 0; i < numbersArray.length; i++) {
if (i % 2 == 0)
shuffleNumber.append(numbersArray[i / 2]);
else
shuffleNumber.append(numbersArray[numbersArray.length - i / 2 - 1]);
}
return Integer.parseInt(shuffleNumber.toString());
}

If you want a solution without string methods, there is a not so complicated one:
public static void main(String[] args) throws IOException {
String str = "1234567";
int len = str.length();
int num=0;
char a;
for(int i = 0; i < len / 2; i++) {
a = str.charAt(i);
num = num * 10 + Character.getNumericValue(a);
a = str.charAt(len -1 - i);
num = num * 10 + Character.getNumericValue(a);
}
if (len % 2 == 1) {
a = str.charAt(str.length() / 2);
num = num * 10 + Character.getNumericValue(a);
}
System.out.println(num);
}
will print
1726354
Check the last if that takes care the case of odd number of digits in the number.

public class MyFirstJavaProgram {
public static void main(String []args) {
String str = "12345678";
int val = str.length();
char a;
int num=0;
int d=0;
int n;
for(int i=0; i<=str.length()-2; i++)
{
a = str.charAt(i);
num = num*10+Character.getNumericValue(a);
if(Integer.parseInt(str)!=0)
{
d=Integer.parseInt(str)%10;
num = num*10+d;
n=Integer.parseInt(str)/10;
str = Integer.toString( n );
}
}
if(val%2!=0)
{
num = num*10+Integer.parseInt(str)%10;
System.out.println(num);
}
else{System.out.println(num);}
}
}
this is working for my question... Thanks all

Following is my solution -
public static void solution(String s) {
StringBuffer st = new StringBuffer();
for (int i = 0; i < s.length() / 2; i++) {
st.append(s.charAt(i));
st.append(s.charAt(s.length() - 1 - i)); // appending characters from last
}
if (s.length() % 2 != 0) {
st.append(s.charAt(s.length() / 2));
}
System.out.println(Integer.parseInt(st.toString()));
}
my logic is to keep appending first and last character to new string till i < s.length/2.
If string is of odd length , it means only last character is remaining, append it to your resultant string.
Else , no character is left and you have your complete string.

Maximum Integer Value java

I was trying to solve the Maximum Integer Value problem form Geeksforgeeks.
The problem states the following:
Given a string S of digits(0-9), your task is to find the maximum value that can be obtained from the string by putting either '*' or '+' operators in between the digits while traversing from left to right of the string and picking up a single digit at a time.
Input:
The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains one line of input denoting the string.
Output:
For each testcase, print the maximum value obtained.
this is what I did:
class GFG
{
public static void sort(int[] numbers)
{
int n = numbers.length;
for (int i = 1; i < n; ++i)
{
int key = numbers[i];
int j = i - 1;
while (j >= 0 && numbers[j] > key)
{
numbers[j + 1] = numbers[j];
j = j -1 ;
}
numbers[j + 1] = key;
}
System.out.println(numbers.length - 1);
}
public static void main (String[] args)
{
Scanner sc = new Scanner(System.in);
int testCases = sc.nextInt();
int [] maxNum;
for(int i = 0; i< testCases; i++)
{
String numbers = sc.nextLine();
char[] cNumbers = numbers.toCharArray();
maxNum = new int [cNumbers.length];
for(int j = 0; j + 1 < cNumbers.length; j++)
{
int sum = 0;
int mult = 0;
sum = cNumbers[j] + cNumbers[j + 1];
mult = cNumbers[j] * cNumbers[j + 1];
int maxNumber = Math.max(sum, mult);
maxNum[i] = maxNumber;
}
sort(maxNum);
}
}
}
an example of Input:
2
01230
891
My Output:
-1
4
Correct Output:
9
73
What is wrong with my code?!

Just quick glance it would seem if your digit is less than two it should be added. 2 or larger should get multiplied. Not at a PC to test though.

The idea is to put the operators alternatively and choose the maximum results.
import java.util.Scanner;
public class Demo {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int testCases = Integer.parseInt(sc.nextLine());
for (int i = 0; i < testCases; i++) {
String numbers = sc.nextLine();
int max = 0;
for (int j = 0; j + 1 < numbers.length(); j++) {
int next = Integer.parseInt(numbers.substring(j, j+1));
if (max + next > max * next)
max = max + next;
else
max = max * next;
}
System.out.println(max);
}
sc.close();
}
}

After the execution of
int testCases = sc.nextInt();
the buffer contains a new line character. So when executing the line
String numbers = sc.nextLine();
it read '\n' into numbers, so you got -1 as the first output.
Also you need to convert character to Integer before using it any arithmetic operations.
sum = cNumbers[j] + cNumbers[j+1];
mult = cNumbers[j] * cNumbers[j+1];
So the above code will give you wrong results.
I tried the following sample and worked.
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String inputAsString = sc.nextLine();
int testCases = Integer.parseInt(inputAsString);
int maxNumber = 0;
for (int i = 0; i < testCases; i++) {
String numbers = sc.nextLine();
if(!numbers.matches("\\d+")){
System.out.println("Only numeric values are expected.");
continue;
}
char[] cNumbers = numbers.toCharArray();
int sum = 0;
int mult = 1;
for (int j = 0; j < cNumbers.length; j++) {
int nextNumber = Character.getNumericValue(cNumbers[j]);
sum = sum + nextNumber;
mult = mult * nextNumber;
maxNumber = mult > sum ? mult : sum;
sum = maxNumber;
mult = maxNumber;
}
System.out.println(maxNumber);
}
sc.close();
}

I read your description and what you do is wrong. please read question carefully specially the example in reference site.
as mentioned in comments by moilejter you use sc.nextInt() which doesn't read '\n' and make problem. the next sc.nextLine() will read only a empty string and your program throw exception.
Second problem is that you must calculate max continuously and you don't need an int array (you calculate max result of operation between two successive number and save them in an array which is not correspond to max integer value. you only find max between each two digit but not max of operation on all digit).
Third problem is that you use character as numbers which is made incorrect result. (you must convert them to integer)
So there is a code that works for your output:
public class GFG
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int testCases = Integer.valueOf(sc.nextLine());
for (int i = 0; i < testCases; i++)
{
String numbers = sc.nextLine();
char[] cNumbers = numbers.toCharArray();
long maxUntilNow = cNumbers[0] - '0';
for (int j = 1; j < cNumbers.length; j++)
{
int numberOfThisPlace = cNumbers[j] - '0';
maxUntilNow = Math.max(maxUntilNow + numberOfThisPlace,
maxUntilNow * numberOfThisPlace);
}
System.out.println(maxUntilNow);
}
}
}
I hope this is what you want.

As per the problem statement, we need to obtain maximum value from the string by putting either * or + operators in between the digits while traversing from left to right of the string and picking up a single digit at a time.
So, this can be solved in O(n) without using any sorting algorithm.
The simple logic behind the solution is whenever you find "0" or "1" in any of the operands use "+" and the rest of the places use "*".
Here is my solution which got successfully submitted:
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG {
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
int T = Integer.parseInt(scan.nextLine());
while(T-- > 0) {
String str = scan.nextLine();
maxValue(str);
}
}
static void maxValue(String str) {
long maxNumber = 0;
for(int i = 0; i < str.length(); i++) {
int n = Character.getNumericValue(str.charAt(i));
if (maxNumber == 0 || maxNumber == 1 ||
n == 0 || n == 1) {
maxNumber += n;
} else {
maxNumber *= n;
}
}
System.out.println(maxNumber);
}
}

Reverse numerical numbers without use of libraries

import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
for (int i = 0; i < 4; i++)
{
int number = input.nextInt();
int reverse = 0;
while (number != 0)
{
reverse = reverse * 10;
reverse = reverse + number % 10;
number = number / 10;
}
System.out.println(reverse);
}
}
}
All numbers reverse well but I have problem with reversing numbers that end with zero e.g numbers like 10000 instead of reversing result being 00001 it gives the result as 1 which is not what the question wants is there a way to use integers or string will be the best and easier approach? Thank you

Try the reversing of string concept
String s ="10000";
String n= "";
for(int i=0; i<s.length(); i++){
n = s.charAt(i) + n;
}
System.out.println(n);

You can create a variable length, and use System.out.printf to format output.
With %0 + length + d, it will add 0 on the left to make the output have this length.
import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
for (int i = 0; i < 4; i++)
{
int number = input.nextInt();
int reverse = 0;
int length = String.valueOf(number).length();
while (number != 0)
{
reverse = reverse * 10;
reverse = reverse + number % 10;
number = number / 10;
}
System.out.printf("%0" + length + "d", reverse);
}
}
}

You could also solve it with a StringBuilder. It already has a method called reverse:
public String reverseInput(int number) {
String s = Integer.toString(number);
return new StringBuilder(s).reverse().toString();
}
You can't have a Integer beginning with 0 except 0 itself. So you will have to work with Strings or other datastructures.

How to find sum of first three digits

I'm new in Java and I try to find the sum of first three digits in int number.
I have a number 123456 I want to find the sum of first three digits 123 and then find sum of last three digits 456 and then compare it. I can't understand how to do this. I code next:
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter the number:");
int number = scanner.nextInt();
int length = (int) Math.ceil(Math.log10(number));
for (int i = 1; i <= 3; i++) {
System.out.println(i);
}
System.out.println(length);
System.out.println(number);
}
I think that I shoud to use for loop, but I'm not sure. I know how to count the length of a number. But how to how to find the sum of the first three numbers and last three?

Assuming your length is constant (6):
String first = (""+number).substring(0, 3);
String second = (""+number).substring(3, 6);
int firstSum = 0;
int secondSum = 0;
for (int x = 0; x < first.length; x++)
firstSum += Integer.parseInt(first.charAt(x)+"");
for (int x = 0; x < second.length; x++)
secondSum += Integer.parseInt(second.charAt(x)+"");
System.out.println("Sum of first 3: " + firstSum);
System.out.println("Sum of second 3: " + secondSum);
Read the String API- it has a lot of useful methods. For small programs like this, it's usually pretty easy to convert to String, use String methods, then parse back to int.
EDIT: Andy Turner and Jeremy Kato informed me about Character.getNumericValue(). Thus, the statement:
firstSum += Integer.parseInt(first.charAt(x)+"");
Could be changed to:
firstSum += Character.getNumericValue(first.charAt(x));
Likewise for secondSum.

First, probably should correct your spelling of length - though maybe not, because you don't really need that variable to do this.
What your code should do to make this easier for you is to not convert the input into an int immediately. You'll first want to split the input string into two pieces, then take the two numbers and sum them up with a loop.
Here's my idea of what I'd do:
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter the number:");
String input = scanner.nextLine();
String firstHalf = input.substring(0, 3); // note that depending on how
// many digits you want to change, these numbers must change as well.
String secondHalf = input.substring(3);
int sum1 = 0;
int sum2 = 0;
// don't forget: start at 0, not 1!
for(int i = 0; i<=3; i++){
sum1 += Character.getNumericValue(firstHalf.charAt(i));
sum2 += Character.getNumericValue(secondHalf.charAt(i));
// this call to the Character class converts a character to an int.
}
System.out.println(sum1);
System.out.println(sum2);
}
Just note that this code is made to work with your input being 6 characters, and it assumes they're going to be numbers. It'll crash if you enter too few characters in or if you put letters in, but you can always worry about that later.

Here's a straightforward way of doing it, making the assumption that your integer has an even number of digits...
public class Test {
public static void main(String[] args) {
// Makes the assumption that the input integer contains an even number of digits.
int x = 123456;
String s = Integer.toString(x);
int[] count = new int[2];
for (int i = 0; i < s.length(); i++) {
int digit = Integer.parseInt(s.substring(i, i + 1));
int countIndex = (int)Math.floor(i * 2 / s.length());
count[countIndex] += digit;
}
int half = s.length() / 2;
System.out.println("First Sum of " + s.substring(0, half) + ": " + count[0]);
System.out.println("Second Sum of " + s.substring(half) + ": " + count[1]);
}
}
Alternate Method
Here's another way of doing it without using the arrays above, and actually simplified the logic by abstracting the sum to a method.
public class Test {
public static void main(String[] args) {
// Makes the assumption that the input integer contains an even number of digits.
int x = 123456;
String s = Integer.toString(x);
int half = s.length() / 2;
String firstHalf = s.substring(0, half);
String secondHalf = s.substring(half);
System.out.println(firstHalf);
System.out.println(secondHalf);
int sum1 = sumString(firstHalf);
int sum2 = sumString(secondHalf);
System.out.println("Sum 1: " + sum1);
System.out.println("Sum 2: " + sum2);
}
private static int sumString(String s) {
int sum = 0;
for (int i = 0; i < s.length(); i++) {
sum += Integer.parseInt(s.substring(i, i + 1));
}
return sum;
}
}

Here is an alternative solution which doesn't use Strings or arrays:
public static void main(String args[]){
int num=123456;
int second = 0;
int first = 0;
second = addLastThree(num);
num /= 1000;
first = addLastThree(num);
System.out.println("first -- > " + first);
System.out.println("second -- > " + second);
}
static int addLastThree(int num){
int sum = 0;
for (int i =0; i<3; i++){
sum += (num/Math.pow(10,i)) % 10;
}
return sum;
}

you can try this code and every line was explained in comments.
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter the number:");
int number = scanner.nextInt();
int n=number; //duplicating the value for the reference.
int l = (int) Math.log10(number) + 1; //Length of the number
int half = l/2; //to find half length of the number to sum first and last digits seperately.
int last_Sum = 0; //to store sum of last digits
int first_Sum = 0; //to store sum of first digits
for(int i=0; i<half && number>0; i++){ //to sum last digits
last_Sum = last_Sum + number%10;
number = number/10;
}for(int i=0; i<half && number>0; i++){ //to sum first digits
first_Sum = first_Sum + number%10;
number = number/10;
}
System.out.println("Length of the number "+l);
System.out.println("Number is "+n);
System.out.println("Sum of last digits "+last_Sum);
System.out.println("Sum of first digits "+first_Sum);}}

How to reverse a string of numbers in a nested loop

This is a homework question so I would like help, not an answer.
I'm trying to create 2 triangles out of numbers based on a number entered by the user.
"Enter a number between 2-9: "3"
1
12
123
1
21
321
IE2:
"Enter a number between 2-9: "5"
1
12
123
1234
12345
1
21
321
4321
54321
I have been able to get the first triangle complete. But when I add my nested loop it messes up my first triangle with the numbers developed from the nested loop. It also puts all the numbers in a straight vertical line. I've tried variations for different nest loops and even tried messing with a StringBuilder, but was still unsuccessful.
Here's what I have in code so far:
import java.util.Scanner;
public class NestedLoops
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter a Number between 2-9: ");
int width = input.nextInt();
String r = "";
for (int i = 1; i <= width; i++)
{
r = r + i;
System.out.println(r);
}
}
}
Again, I'm looking for help/understanding and not just an answer.

There are two aspects the 2nd part of the question.
You need to generate strings with the numbers in the reverse order:
You could do this by adding the numbers at the other end.
You could do this by reversing the strings.
You need to arrange that there are spaces to the left.
You could do this by adding the required number of spaces to the left end of the string.
You could do this by using the System.out.format(...) with a template that right aligns the string in a field with the required number of characters. (OK, that's a bit too obscure ...)
Or, you can build the string in a character array or string builder rather than using string concatenation.
The "trick" is to figure out what strategy you are going to use ... before you start cutting code.

try
int width = 5;
// for all lines; number of lines = width
for (int line = 1; line <= width; line++) {
// print numbers from 1 to current line number
for (int n = 1; n <= line; n++) {
System.out.print(n);
}
// end of line
System.out.println();
}
// add empty line between triangles
System.out.println();
// for all lines; number of lines = width
for (int line = 1; line <= width; line++) {
// printing padding spaces, number of spaces = with - line number
int nSpaces = width - line;
for (int i = 0; i < nSpaces; i++) {
System.out.print(" ");
}
// print numbers from number of current line to 1
for (int n = line; n >= 1; n--) {
System.out.print(n);
}
// end of line
System.out.println();
}

Can you just add another loop after your first loop like
String r = "";
String space = "";
for (int i = width; i >= 1; i--)
{
r = r + i;
System.out.println(r);
}
Try it. not yet tested

You need to use a queue.
eg. http://docs.oracle.com/javase/1.5.0/docs/api/java/util/LinkedList.html
Enque the numbers till you reach the max, and then start dequing them.
And while you dequeue, you need to apply the reverse
Queue<String> q = new LinkedList<String>();
for (int i = 1; i <= width; i++)
{
r = r + i;
q.add(r);
System.out.println(r);
}
while(!q.isEmpty()){
String j = q.remove();
//reverse j
System.out.println(reverse(j));
}
I leave the reversing part for you to do :)

public static void main(String[] args)
{
int n = 5;
for(int i=1; i<=n; i++)
{
for (int j=(n*2), k=n; j>1; j--)
{
if (k <= i)
{
System.out.print(k);
}
else
{
System.out.print('*');
}
k += (j)-1 > n ? -1 : 1;
}
System.out.println();
}
}

Just tried to implement in scala. Ignore if you don't like it..:-)
class Triangle extends App
{
val width = Console.readInt()
if (width < 2 || width > 9)
{
throw new RuntimeException()
}
var i, j = 1;
for (i <- 1 to width)
{
for (j <- 1 to i)
{
print(j)
}
print("\n")
}
for (i <- 1 to width)
{
for (dummy <- 1 to width-i)
{
print(" ")
}
for (j <- i to 1 by -1)
{
print(j)
}
print("\n")
}
}