Write number in expanded form java

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import java.util.Scanner; class Input < public static void main(String[] args) < Scanner input = new Scanner(System.in); System.out.println("Enter your name: "); String inp = input.next(); System.out.println("Hello, " + inp); >>

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apply plugin:'application' mainClassName = 'HelloWorld' run < standardInput = System.in >sourceSets < main < java < srcDir './' >> > repositories < jcenter() >dependencies < // add dependencies here as below implementation group: 'org.apache.commons', name: 'commons-lang3', version: '3.9' >

About Java

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Syntax help

Variables

short x = 999; // -32768 to 32767 int x = 99999; // -2147483648 to 2147483647 long x = 99999999999L; // -9223372036854775808 to 9223372036854775807 float x = 1.2; double x = 99.99d; byte x = 99; // -128 to 127 char x = 'A'; boolean x = true;

Loops

1. If Else:

When ever you want to perform a set of operations based on a condition If-Else is used.

if(conditional-expression) < // code >else < // code >

2. Switch:

Switch is an alternative to If-Else-If ladder and to select one among many blocks of code.

3. For:

For loop is used to iterate a set of statements based on a condition. Usually for loop is preferred when number of iterations is known in advance.

for(Initialization; Condition; Increment/decrement) < //code >

4. While:

While is also used to iterate a set of statements based on a condition. Usually while is preferred when number of iterations are not known in advance.

5. Do-While:

Do-while is also used to iterate a set of statements based on a condition. It is mostly used when you need to execute the statements atleast once.

Classes and Objects

Class is the blueprint of an object, which is also referred as user-defined data type with variables and functions. Object is a basic unit in OOP, and is an instance of the class.

How to create a Class:

class keyword is required to create a class.

Example:

How to create a Object:

How to define methods in a class:

public class Greeting < static void hello() < System.out.println("Hello.. Happy learning!"); >public static void main(String[] args) < hello(); >>

Collections

Collection is a group of objects which can be represented as a single unit. Collections are introduced to bring a unified common interface to all the objects.

Collection Framework was introduced since JDK 1.2 which is used to represent and manage Collections and it contains:

This framework also defines map interfaces and several classes in addition to Collections.

Advantages:

• High performance
• Reduces developer’s effort
• Unified architecture which has common methods for all objects.

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How to Write Number in Expanded Form in Java

You will be given a number and you will need to return it as a string in Expanded Form.

Challenge.expandedForm(12); // Should return "10 + 2" Challenge.expandedForm(42); // Should return "40 + 2" Challenge.expandedForm(70304); // Should return "70000 + 300 + 4" 

NOTE: All numbers will be whole numbers greater than 0.

The solution in Java code#

public class Challenge < public static String expandedForm(int num) < StringBuffer res = new StringBuffer(); int d = 1; while(num >0) < int nextDigit = num % 10; num /= 10; if (nextDigit >0) < res.insert(0, d * nextDigit); res.insert(0, " + "); >d *= 10; > return res.substring(3).toString(); > > 

import java.util.LinkedList; public class Challenge < public static String expandedForm(int num) < LinkedListexpandedList = new LinkedList<>(); int digit; int multiplier = 1; while (num > 0) < digit = (num % 10) * multiplier; if (digit != 0) expandedList.push(Integer.toString(digit)); num /= 10; multiplier *= 10; >return String.join(" + ", expandedList); > > 

import java.util.stream.IntStream; import static java.util.stream.Collectors.joining; public class Challenge < public static String expandedForm(int num) < var ns = ""+num; return IntStream.range(0, ns.length()) .mapToObj(i ->ns.charAt(i) + "0".repeat(ns.length() - 1 - i)) .filter(e -> !e.matches("0+")) .collect(joining(" + ")); > > 

Test cases to validate our solution#

import org.junit.Test; import static org.junit.Assert.assertEquals; import org.junit.runners.JUnit4; public class SolutionTest < @Test public void testSomething() < assertEquals("10 + 2", Challenge.expandedForm(12)); assertEquals("40 + 2", Challenge.expandedForm(42)); assertEquals("70000 + 300 + 4", Challenge.expandedForm(70304)); >> 

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Java — number in expanded form

My function works for first and second case, but with 70304 it gives this:

import java.util.Arrays; public static String expandedForm(int num) < String[] str = Integer.toString(num).split(""); String result = ""; for(int i = 0; i < str.length-1; i++) < if(Integer.valueOf(str[i]) >0) < for(int j = i; j < str.length-1; j++) < str[j] += '0'; >> > result = Arrays.toString(str); result = result.substring(1, result.length()-1).replace(",", " +"); System.out.println(result); return result; > 

I think there’s a problem with the second loop, but can’t figure out why.

Asia

People also ask

Go through the below steps to write the numbers in expanded form: Step 1: Get the standard form of the number. Step 2: Identify the place value of the given number using the place value chart. Step 3: Multiply the given digit by its place value and represent the number in the form of (digit × place value).

6 Answers

You should be adding ‘0’s to str[i] , not str[j] :

You still have to get rid of the 0 digits.

One possible way to get rid of them:

result = result.substring(1, result.length()-1).replace(", 0","").replace(",", " +"); 

answered Oct 14 ’22 23:10

Eran

public class Kata < public static String expandedForm(int num) < String outs = ""; for (int i = 10; i < num; i *= 10) < int rem = num % i; outs = (rem >0) ? " + " + rem + outs : outs; num -= rem; > outs = num + outs; return outs; > > 

ABHISHEK SHUKLA

Pseudocode uses integer arithmetics to extract decimal digits one-by-one (from the right one):

mul = 1 //will contain power of 10 while (num > 0): dig = num % 10 //integer modulo retrieves the last digit if (dig > 0): //filter out zero summands add (dig * mul) to output //like 3 * 100 = 300 num = num / 10 //integer division removes the last decimal digit 6519 => 651 mul = mul * 10 //updates power of 10 for the next digit 

MBo

You could do the same with pure math, using modulo % and integer division / , e.g. using Stream API:

int n = 70304; String res = IntStream .iterate(1, k -> n / k > 0, k -> k * 10) // divisors .map(k -> (n % (k*10) / k ) * k) // get 1s, 10s, 100s, etc. .filter(x -> x > 0) // throw out zeros .mapToObj(Integer::toString) // convert to string .collect(Collectors.joining(" + ")); // join with '+' System.out.println(res); // 4 + 300 + 70000 

tobias_k

There are many variations possible. If the usage of a list is allowed:

public static String expandedForm(int num) < String[] str = Integer.toString(num).split(""); String result; Listl = new ArrayList(); for(int i = 0; i < str.length; i++)< if(Integer.valueOf(str[i]) >0) < String s = str[i]; for(int j = i; j < str.length - 1; j++)< s += '0'; >l.add(s); > > result = l.toString(); result = result.substring(1, result.length() - 1).replace(",", " +"); System.out.println(result); return result; > 

One could also work directly on result:

public static String expandedForm2(int num) < String[] str = Integer.toString(num).split(""); String result = ""; for(int i = 0; i < str.length; i++)< if(Integer.valueOf(str[i]) >0) < result += str[i]; for(int j = i; j < str.length - 1; j++)< result += '0'; >result += " + "; > > result = result.substring(0, result.length() - 3); System.out.println(result); return result; > 

Imago

This is also possible to do recursively. Here an example implementation:

String g(int n, int depth) < // Recursive method with 2 int parameters & String return-type int remainder = n % depth; // The current recursive remainder if(depth < n)< // If we aren't done with the number yet: int nextDepth = depth * 10; // Go to the next depth (of the power of 10) int nextN = n - remainder; // Remove the remainder from the input `n` // Do a recursive call with these next `n` and `depth` String resultRecursiveCall = g(nextN, nextDepth); if(remainder != 0)< // If the remainder was not 0: // Append a " + " and this remainder to the result resultRecursiveCall += " + " + remainder; >return resultRecursiveCall; // And return the result > else < // Else: return Integer.toString(n); // Simply return input `n` as result >> String f(int n) < // Second method so we can accept just integer `n` return g(n, 1); // Which will call the recursive call with parameters `n` and 1 >

The second method is so we can call the method with just a single input n . For example:

Which will result in the String 70000 + 300 + 4 .

Try it online.

To explain a bit more in depth of what this recursive method does, let’s just do a step-by-step for the input 70304 :

1. In the first recursive iteration: n=70304 , depth=1 , remainder=70304%1 = 0 .
   • Since depth < n is truthy, it will do a recursive call with 70304-0 and 1*10
   • And since remainder is 0, it will append nothing more to the result
2. In the second recursive iteration: n=70304 , depth=10 , remainder=70304%10 = 4 .
   • Since depth < n is still truthy, it will do a recursive call with 70304-4 and 10*10
   • And since remainder is 4, it will append a » + » and this 4 to the result
3. In the third recursive iteration: n=70300 , depth=100 , remainder=70300%100 = 0 .
   • Since depth < n is still truthy, it will do a recursive call with 70300-0 and 100*10
   • And since remainder is 0, it will append nothing more to the result
4. In the fourth recursive iteration: n=70300 , depth=1000 , remainder=70300%1000 = 300 .
   • Since depth < n is still truthy, it will do a recursive call with 70300-300 and 1000*10
   • And since remainder is 300, it will append a » + » and this 300 to the result
5. In the fifth recursive iteration: n=70000 , depth=10000 , remainder=70000%10000 = 0 .
   • Since depth < n is still truthy, it will do a recursive call with 70000-0 and 10000*10
   • And since remainder is 0, it will append nothing more to the result
6. In the sixth recursive iteration: n=70000 , depth=100000 , remainder=70000%100000 = 70000 .
   • Since now depth < n is falsey, it won't do any more recursive calls, but instead return the current n (which is 70000 ).

And since these were all recursive calls, we should actually look at it backwards for the result, so it will result in 70000 + 300 + 4 .

answered Oct 15 ’22 01:10

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