Java - number in expanded form

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.

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


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


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


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;
}


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 10k power in the next recursive iteration
The remainder != 0 if-check determines if the number we want to append was not a 0


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.

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';
}
}
}