Java - number in expanded form

up vote
11
down vote

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I have given number and want it to return as a String in expanded form. For example

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

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

70 + 00 + 300 + 000 + 4

Here's my code

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.

asked Sep 6 at 11:22

Asia

786

add a commentÂ |Â

up vote
11
down vote

favorite

I have given number and want it to return as a String in expanded form. For example

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

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

70 + 00 + 300 + 000 + 4

Here's my code

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.

asked Sep 6 at 11:22

Asia

786

add a commentÂ |Â

up vote
11
down vote

favorite

I have given number and want it to return as a String in expanded form. For example

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

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

70 + 00 + 300 + 000 + 4

Here's my code

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.

asked Sep 6 at 11:22

Asia

786

I have given number and want it to return as a String in expanded form. For example

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

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

70 + 00 + 300 + 000 + 4

Here's my code

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.

asked Sep 6 at 11:22

Asia

786

asked Sep 6 at 11:22

Asia

786

asked Sep 6 at 11:22

Asia

786

asked Sep 6 at 11:22

Asia

786

add a commentÂ |Â

7 Answers
7

active

oldest

votes

up vote
13
down vote

accepted

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

 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[i] += '0';

This will result in:

70000 + 0 + 300 + 0 + 4

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(",", " +");

Now the output is

70000 + 300 + 4

answered Sep 6 at 11:31

Eran

262k33406491

result.replace(" + 0","") would be easier?
â€“Â Viktor Mellgren
Sep 6 at 12:10

add a commentÂ |Â

up vote
6
down vote

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

edited Sep 6 at 13:03

answered Sep 6 at 11:32

MBo

42.4k22847

add a commentÂ |Â

up vote
3
down vote

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

answered Sep 6 at 15:43

tobias_k

54.4k963100

1

Why the downvote? If because the terms are in reverse order, well, if that is important, the order could easily be fixed.
â€“Â tobias_k
Sep 6 at 16:12

add a commentÂ |Â

up vote
2
down vote

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;
 List<String> l = new ArrayList<String>();

 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;

edited Sep 6 at 18:25

answered Sep 6 at 18:18

Imago

257312

add a commentÂ |Â

up vote
0
down vote

package backup;

import java.util.Arrays;

public class FileOutput 

 public static void main(String args)

 String expForm = expandedForm(70304);
 //System.out.println(expForm);

 

 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[i] += '0';
 
 
 

 result = Arrays.toString(str);
 result = result.substring(1, result.length()-1).replace(",", " +");
 System.out.println(result);

 return result;

Output : 70000 + 0 + 300 + 0 + 4

Solution in most inner loop you need to add '0' to str[i] : str[i] += '0';

Then you need to replace "+ 0" from the resulted output.

answered Sep 6 at 11:33

Raj

28912

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up vote
0
down vote

for(int i = 0; i < str.length; i++) 
 if(Integer.valueOf(str[i]) > 0) 
 for(int j = 0; j < str.length - i - 1; j++) 
 str[i] += '0';

answered Sep 6 at 11:35

Bartek Wichowski

434416

4

You should add explanations to your answers, not only paste code.
â€“Â Nidhoegger
Sep 6 at 17:18

add a commentÂ |Â

up vote
0
down vote

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:

String result = f(70304);

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:

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

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

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

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

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

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.

So in general:

The depth < n if-check is to see when we are done with the recursive calls.

The g(n-remainder, depth*10) will remove the digits we've already output in a previous recursive iteration, and goes to the next 10^k power in the next recursive iteration

The remainder != 0 if-check determines if the number we want to append was not a 0

answered 18 hours ago

Kevin Cruijssen

3,97873680

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7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

up vote
13
down vote

accepted

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

 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[i] += '0';

This will result in:

70000 + 0 + 300 + 0 + 4

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(",", " +");

Now the output is

70000 + 300 + 4

answered Sep 6 at 11:31

Eran

262k33406491

result.replace(" + 0","") would be easier?
â€“Â Viktor Mellgren
Sep 6 at 12:10

add a commentÂ |Â

up vote
13
down vote

accepted

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

 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[i] += '0';

This will result in:

70000 + 0 + 300 + 0 + 4

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(",", " +");

Now the output is

70000 + 300 + 4

answered Sep 6 at 11:31

Eran

262k33406491

result.replace(" + 0","") would be easier?
â€“Â Viktor Mellgren
Sep 6 at 12:10

add a commentÂ |Â

up vote
13
down vote

accepted

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

 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[i] += '0';

This will result in:

70000 + 0 + 300 + 0 + 4

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(",", " +");

Now the output is

70000 + 300 + 4

answered Sep 6 at 11:31

Eran

262k33406491

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

 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[i] += '0';

This will result in:

70000 + 0 + 300 + 0 + 4

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(",", " +");

Now the output is

70000 + 300 + 4

answered Sep 6 at 11:31

Eran

262k33406491

answered Sep 6 at 11:31

Eran

262k33406491

answered Sep 6 at 11:31

Eran

262k33406491

answered Sep 6 at 11:31

Eran

262k33406491

result.replace(" + 0","") would be easier?
â€“Â Viktor Mellgren
Sep 6 at 12:10

add a commentÂ |Â

result.replace(" + 0","") would be easier?
â€“Â Viktor Mellgren
Sep 6 at 12:10

result.replace(" + 0","") would be easier?
â€“Â Viktor Mellgren
Sep 6 at 12:10

add a commentÂ |Â

up vote
6
down vote

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

edited Sep 6 at 13:03

answered Sep 6 at 11:32

MBo

42.4k22847

add a commentÂ |Â

up vote
6
down vote

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

edited Sep 6 at 13:03

answered Sep 6 at 11:32

MBo

42.4k22847

add a commentÂ |Â

up vote
6
down vote

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

edited Sep 6 at 13:03

answered Sep 6 at 11:32

MBo

42.4k22847

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

edited Sep 6 at 13:03

answered Sep 6 at 11:32

MBo

42.4k22847

edited Sep 6 at 13:03

answered Sep 6 at 11:32

MBo

42.4k22847

answered Sep 6 at 11:32

MBo

42.4k22847

answered Sep 6 at 11:32

MBo

42.4k22847

add a commentÂ |Â

up vote
3
down vote

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

answered Sep 6 at 15:43

tobias_k

54.4k963100

1

Why the downvote? If because the terms are in reverse order, well, if that is important, the order could easily be fixed.
â€“Â tobias_k
Sep 6 at 16:12

add a commentÂ |Â

up vote
3
down vote

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

answered Sep 6 at 15:43

tobias_k

54.4k963100

1

Why the downvote? If because the terms are in reverse order, well, if that is important, the order could easily be fixed.
â€“Â tobias_k
Sep 6 at 16:12

add a commentÂ |Â

up vote
3
down vote

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

answered Sep 6 at 15:43

tobias_k

54.4k963100

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

answered Sep 6 at 15:43

tobias_k

54.4k963100

answered Sep 6 at 15:43

tobias_k

54.4k963100

answered Sep 6 at 15:43

tobias_k

54.4k963100

answered Sep 6 at 15:43

tobias_k

54.4k963100

1

Why the downvote? If because the terms are in reverse order, well, if that is important, the order could easily be fixed.
â€“Â tobias_k
Sep 6 at 16:12

add a commentÂ |Â

1

Why the downvote? If because the terms are in reverse order, well, if that is important, the order could easily be fixed.
â€“Â tobias_k
Sep 6 at 16:12

Why the downvote? If because the terms are in reverse order, well, if that is important, the order could easily be fixed.
â€“Â tobias_k
Sep 6 at 16:12

add a commentÂ |Â

up vote
2
down vote

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;
 List<String> l = new ArrayList<String>();

 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;

edited Sep 6 at 18:25

answered Sep 6 at 18:18

Imago

257312

add a commentÂ |Â

up vote
2
down vote

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;
 List<String> l = new ArrayList<String>();

 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;

edited Sep 6 at 18:25

answered Sep 6 at 18:18

Imago

257312

add a commentÂ |Â

up vote
2
down vote

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;
 List<String> l = new ArrayList<String>();

 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;

edited Sep 6 at 18:25

answered Sep 6 at 18:18

Imago

257312

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;
 List<String> l = new ArrayList<String>();

 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;

edited Sep 6 at 18:25

answered Sep 6 at 18:18

Imago

257312

edited Sep 6 at 18:25

answered Sep 6 at 18:18

Imago

257312

answered Sep 6 at 18:18

Imago

257312

answered Sep 6 at 18:18

Imago

257312

add a commentÂ |Â

up vote
0
down vote

package backup;

import java.util.Arrays;

public class FileOutput 

 public static void main(String args)

 String expForm = expandedForm(70304);
 //System.out.println(expForm);

 

 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[i] += '0';
 
 
 

 result = Arrays.toString(str);
 result = result.substring(1, result.length()-1).replace(",", " +");
 System.out.println(result);

 return result;

Output : 70000 + 0 + 300 + 0 + 4

Solution in most inner loop you need to add '0' to str[i] : str[i] += '0';

Then you need to replace "+ 0" from the resulted output.

answered Sep 6 at 11:33

Raj

28912

add a commentÂ |Â

up vote
0
down vote

package backup;

import java.util.Arrays;

public class FileOutput 

 public static void main(String args)

 String expForm = expandedForm(70304);
 //System.out.println(expForm);

 

 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[i] += '0';
 
 
 

 result = Arrays.toString(str);
 result = result.substring(1, result.length()-1).replace(",", " +");
 System.out.println(result);

 return result;

Output : 70000 + 0 + 300 + 0 + 4

Solution in most inner loop you need to add '0' to str[i] : str[i] += '0';

Then you need to replace "+ 0" from the resulted output.

answered Sep 6 at 11:33

Raj

28912

add a commentÂ |Â

up vote
0
down vote

package backup;

import java.util.Arrays;

public class FileOutput 

 public static void main(String args)

 String expForm = expandedForm(70304);
 //System.out.println(expForm);

 

 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[i] += '0';
 
 
 

 result = Arrays.toString(str);
 result = result.substring(1, result.length()-1).replace(",", " +");
 System.out.println(result);

 return result;

Output : 70000 + 0 + 300 + 0 + 4

Solution in most inner loop you need to add '0' to str[i] : str[i] += '0';

Then you need to replace "+ 0" from the resulted output.

answered Sep 6 at 11:33

Raj

28912

package backup;

import java.util.Arrays;

public class FileOutput 

 public static void main(String args)

 String expForm = expandedForm(70304);
 //System.out.println(expForm);

 

 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[i] += '0';
 
 
 

 result = Arrays.toString(str);
 result = result.substring(1, result.length()-1).replace(",", " +");
 System.out.println(result);

 return result;

Output : 70000 + 0 + 300 + 0 + 4

Solution in most inner loop you need to add '0' to str[i] : str[i] += '0';

Then you need to replace "+ 0" from the resulted output.

answered Sep 6 at 11:33

Raj

28912

answered Sep 6 at 11:33

Raj

28912

answered Sep 6 at 11:33

Raj

28912

answered Sep 6 at 11:33

Raj

28912

add a commentÂ |Â

up vote
0
down vote

for(int i = 0; i < str.length; i++) 
 if(Integer.valueOf(str[i]) > 0) 
 for(int j = 0; j < str.length - i - 1; j++) 
 str[i] += '0';

answered Sep 6 at 11:35

Bartek Wichowski

434416

4

You should add explanations to your answers, not only paste code.
â€“Â Nidhoegger
Sep 6 at 17:18

add a commentÂ |Â

up vote
0
down vote

for(int i = 0; i < str.length; i++) 
 if(Integer.valueOf(str[i]) > 0) 
 for(int j = 0; j < str.length - i - 1; j++) 
 str[i] += '0';

answered Sep 6 at 11:35

Bartek Wichowski

434416

4

You should add explanations to your answers, not only paste code.
â€“Â Nidhoegger
Sep 6 at 17:18

add a commentÂ |Â

up vote
0
down vote

for(int i = 0; i < str.length; i++) 
 if(Integer.valueOf(str[i]) > 0) 
 for(int j = 0; j < str.length - i - 1; j++) 
 str[i] += '0';

answered Sep 6 at 11:35

Bartek Wichowski

434416

for(int i = 0; i < str.length; i++) 
 if(Integer.valueOf(str[i]) > 0) 
 for(int j = 0; j < str.length - i - 1; j++) 
 str[i] += '0';

answered Sep 6 at 11:35

Bartek Wichowski

434416

answered Sep 6 at 11:35

Bartek Wichowski

434416

answered Sep 6 at 11:35

Bartek Wichowski

434416

answered Sep 6 at 11:35

Bartek Wichowski

434416

4

You should add explanations to your answers, not only paste code.
â€“Â Nidhoegger
Sep 6 at 17:18

add a commentÂ |Â

4

You should add explanations to your answers, not only paste code.
â€“Â Nidhoegger
Sep 6 at 17:18

You should add explanations to your answers, not only paste code.
â€“Â Nidhoegger
Sep 6 at 17:18

add a commentÂ |Â

up vote
0
down vote

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:

String result = f(70304);

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:

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

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

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

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

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

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.

So in general:

The depth < n if-check is to see when we are done with the recursive calls.

The g(n-remainder, depth*10) will remove the digits we've already output in a previous recursive iteration, and goes to the next 10^k power in the next recursive iteration

The remainder != 0 if-check determines if the number we want to append was not a 0

answered 18 hours ago

Kevin Cruijssen

3,97873680

add a commentÂ |Â

up vote
0
down vote

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:

String result = f(70304);

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:

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

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

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

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

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

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.

So in general:

The depth < n if-check is to see when we are done with the recursive calls.

The g(n-remainder, depth*10) will remove the digits we've already output in a previous recursive iteration, and goes to the next 10^k power in the next recursive iteration

The remainder != 0 if-check determines if the number we want to append was not a 0

answered 18 hours ago

Kevin Cruijssen

3,97873680

add a commentÂ |Â

up vote
0
down vote

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:

String result = f(70304);

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:

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

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

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

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

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

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.

So in general:

The depth < n if-check is to see when we are done with the recursive calls.

The g(n-remainder, depth*10) will remove the digits we've already output in a previous recursive iteration, and goes to the next 10^k power in the next recursive iteration

The remainder != 0 if-check determines if the number we want to append was not a 0

answered 18 hours ago

Kevin Cruijssen

3,97873680

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:

String result = f(70304);

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:

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

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

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

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

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

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.

So in general:

The depth < n if-check is to see when we are done with the recursive calls.

The g(n-remainder, depth*10) will remove the digits we've already output in a previous recursive iteration, and goes to the next 10^k power in the next recursive iteration

The remainder != 0 if-check determines if the number we want to append was not a 0

answered 18 hours ago

Kevin Cruijssen

3,97873680

answered 18 hours ago

Kevin Cruijssen

3,97873680

answered 18 hours ago

Kevin Cruijssen

3,97873680

answered 18 hours ago

Kevin Cruijssen

3,97873680

add a commentÂ |Â

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