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Formatting returned Strings?

I'm really new to coding and just got assigned my first coding homework involving methods and returns. I managed to struggle through and end up with this, which I'm pretty proud of, but I'm not quite sure it's right. Along with that, my return statements are all on the same lines instead of formatted how my teacher says they should be ("n is a perfect number", then the line below says "factors: x y z", repeated for each perfect number. Below are the exact instructions plus what it outputs. Anything will help!
Write a method (also known as functions in C++) named isPerfect that takes in one parameter named number, and return a String containing the factors for the number that totals up to the number if the number is a perfect number. If the number is not a perfect number, have the method return a null string (do this with a simple: return null; statement).
Utilize this isPerfect method in a program that prompts the user for a maximum integer, so the program can display all perfect numbers from 2 to the maximum integer
286 is perfect.Factors: 1 2 3 1 2 4 7 14
It should be
6 is perfect
Factors: 1 2 3
28 is perfect
Factors: 1 2 4 7 14
public class NewClass {
public static void main(String[] args) {
Scanner input = new Scanner(System.in) ;
System.out.print("Enter max number: ") ;
int max = input.nextInt() ;
String result = isPerfect(max) ;
System.out.print(result) ;
}
public static String isPerfect(int number) {
String factors = "Factors: " ;
String perfect = " is perfect." ;
for (int test = 1; number >= test; test++) {
int sum = 0 ;
for (int counter = 1; counter <= test/2; counter++) {
if (test % counter == 0) {
sum += counter ;
}
}
if (sum == test) {
perfect = test + perfect ;
for (int counter = 1; counter <= test/2; counter++) {
if (test % counter == 0) {
factors += counter + " " ;
}
}
}
}
return perfect + factors ;
}
}

Couple of things you could do:
Firstly, you do not need two loops to do this. You can run one loop till number and keep checking if it's divisible by the iterating variable. If it is, then add it to a variable called sum.
Example:
.
factors = []; //this can be a new array or string, choice is yours
sum=0;
for(int i=1; i<number; i++){
if(number % i == 0){
sum += i;
add the value i to factors variable.
}
}
after this loop completes, check if sum == number, the if block to return the output with factors, and else block to return the output without factors or factors = null(like in the problem statement)
In your return answer add a newline character between perfect and the factors to make it look like the teacher's output.

You can try the solution below:
public String isPerfect(int number) {
StringBuilder factors = new StringBuilder("Factors: ");
StringBuilder perfect = new StringBuilder(" is perfect.");
int sum = 0;
for (int i = 1; i < number; i++) {
if (number % i == 0) {
sum += i;
factors.append(" " + i);
}
}
if (sum == number) {
return number + "" + perfect.append(" \n" + factors);
}
return number + " is not perfect";
}

Keep separate variables for your template bits for the output and the actual output that you are constructing. So I suggest that you don’t alter factors and perfect and instead declare one more variable:
String result = "";
Now when you’ve found a perfect number, add to the result like this:
result += test + perfect + '\n' + factors;
for (int counter = 1; counter <= test/2; counter++) {
if (test % counter == 0) {
result += counter + " ";
}
}
result += '\n';
I have also inserted some line breaks, '\n'. Then of course return the result from your method:
return result;
With these changes your method returns:
6 is perfect.
Factors: 1 2 3
28 is perfect.
Factors: 1 2 4 7 14
Other tips
While your program gives the correct output, your method doesn’t follow the specs in the assignment. It was supposed to check only one number for perfectness. Only your main program should iterate over numbers to find all perfect numbers up to the max.
You’ve got your condition turned in an unusual way here, which makes it hard for me to read:
for (int test = 1; number >= test; test++) {
Prefer
for (int test = 1; test <= number; test++) {
For building strings piecewise learn to use a StringBuffer or StringBuilder.
Link
Java StringBuilder class on Javapoint Tutorials, with examples.

Find max rating(number) in an array where we cannot skip over 1 or more consecutive number in array

we have an array of ratings, we have to find the max rating in such a away that we cannot skip 1 or more consecutive rating in an arrray
Example-1: {9,-1,-3,-4,-5} output = 9 + -1 + -4 = 4
Explanation: I took 9 the we have to took -1 or -3 we cannot jump to -4 directly as we cannot skip 1 or more consecutive number.
Example-2: {-1,-2,-3,-4,-5} output = -2 + -4 = -6
Example-3: {-3,2,-4,-1,-2,5} output = 2 + -1 + 5 = 6
Example-4: {9,-1,-3,4,5} output = 9 + -1 + 4 + 5 = 17
I tried below code but it is working in case of example: 2,3,4 but not for example 1 similarly failing for other scenario.
static int maximizeRatings(int[] ratings) {
int current = 0;
boolean result = false;
for(int j=0; j<ratings.length;j++){
if(ratings[j]<0){
result = true;
}else{
result = false;
}
}
if(result){
return allnegatine(ratings);
}
for(int i=0; i<ratings.length;i++){
if(i == ratings.length-1){
if(ratings[i] > 0)
current += ratings[i];
}else{
if(ratings[i] >0 && ratings[i+1]>0){
current = ratings[i]+ratings[i+1];
i++;
}
if(ratings[i] > ratings[i+1]){
current += ratings[i];
}else{
current += ratings[i+1];
i++;
}
}
}
return current;
}
private static int allnegatine(int[] ratings) {
int current =0;
for(int i=0; i<ratings.length;i++){
if(ratings.length%2==0){
if(i%2 == 0)
current += ratings[i];
}else{
if(i%2!=0)
current += ratings[i];
}
}
return current;
}
not getting excepted out for some scenarios like example 1 I am getting -6 instead of 4, I am trying to get proper code which will pass all scenarios. Thank you

This is a dynamic programming problem.
Let dp[i] be the max ratings which can be achieved considering only the part of the array that starts at zero, ends at i, and includes ratings[i].
dp[0]=ratings[0]
dp[1]=max(ratings[1],ratings[0]+ratings[1])
dp[i]=max(dp[i-1],dp[i-2])+ratings[i]
Answer: max(dp[n-1],dp[n-2]) where n is the size of the ratings array.
Also you can chose to do away with dp array and maintain 2 variables for dp[i-1] and dp[i-2].

this is typical recursion problem (as long as the input array is reasonably long). You should go thru items and try all possible combinations and then pick the best one.
Because it looks like typical school work I am not sure if I should paste my solution. You should figure it out yourself or at least understand what's going on to be able to implement it yourself next time.
public class RatingService {
public int calculate(List<Integer> input) {
return recursion(input, true, 0);
}
private int recursion(List<Integer> sublist, boolean canSkip, int sum) {
if (sublist.isEmpty()) {
return sum;
}
int skippedSum = Integer.MIN_VALUE;
int notSkippedSum;
Integer integer = sublist.get(0);
if (canSkip) {
skippedSum = recursion(sublist.subList(1, sublist.size()), false, sum);
}
notSkippedSum = recursion(sublist.subList(1, sublist.size()), true, integer + sum);
return skippedSum > notSkippedSum ? skippedSum : notSkippedSum;
}
}

I think you are doing the mistake while checking for all negative numbers in for loop. If the last element in the array is negative then the 'result' variable will be true means that all array is negative but actually its not.
You have to replace the for loop by :
for(int j=0;j<ratings.length;j++){
if(ratings[j]<0){
result=true;
}
else{
result = false;
break;
}
}
It will break the for loop at the index where it founds any positive number i-e: all elements of array are not negative.

How to exit from a method, i.e how can i return from a function in this recursion in java?

How to exit from a method, i.e how can i return from a function in this recursion in java?
public class solution {
public static int countZerosRec(int input){
int n=0;
int k =0;
int count=0;
//Base case
if(n==0)
{
return; // How can i return the method from here, i.e how can i stop the execution of the recursive program now.
}
k=input%10;
count++;
n=input/10;
countZerosRec(n);
int myans=count;
return myans;
}
}
Please help me getting out of this method.
This is a program to count number of zeroes.
Example, 34029030 ans = 3

You can try below approach:
public class MyClass {
public static void main(String args[]) {
System.out.println("total zeroes = " + returnZeroesCount(40300));
}
public static int returnZeroesCount(int input){
if(input == 0)
return 0;
int n = input % 10;
return n == 0 ? 1 + returnZeroesCount(input / 10) : returnZeroesCount(input / 10);
}
}
How it works: Assuming your input > 0, we try to get the last digit of the number by taking the modulus by 10. If it is equal to zero, we add one to the value that we will return. And what will be the value that we would be returning? It will be the number of zeroes present in the remaining number after taking out the last digit of input.
For example, in the below case, 40300: we take out 0 in first step, so we return 1+number of zeroes in 4030. Again, it appears as if we have called our recursive function for the input 4030 now. So, we again return 1+number of zeroes in 403.
In next step, since last number is 3, we simply return 0+total number of zeroes in 40 or simply as total number of zeroes present in 40 and so on.
For ending condition, we check if the input is itself 0. If it is zero then this means that we have exhausted the input number and there are no further numbers to check for. Hence, we return zero in that case. Hope this helps.

If your main focus is to find number of zeroes in a given number , You can use this alternatively:
int numOfZeroes =0;
long example = 670880930;
String zeroCounter = String.valueOf(example);
for(int i=0; i< example.length();i++){
if(zeroCounter.charAt(i) ==0){
numOfZeroes++;
}
}
System.out.print("Num of Zeros are"+ numOfZeroes);` `

Instead of posting a code answer to your question, I'll post a few pointers to get you moving.
As #jrahhali said, as your code is, it'll not get past the return
statement inside the if block(which is an error BTW, because you have an int return
type).
I'd recommend that you move the last two lines to some calling
function(such as a main method). That way all this function will
need to do is do some basic processing and move forward.
You aren't checking k at all. As it is, your count is going to
always increment.
Hope this much is enough for you to figure things out.

int count =0;
private int getZeroCount(int num){
if(num+"".length == 1){
if(num==0){
count++;
}
return count;
}
if(num%10 == 0){
count++;
}
num /= 10;
getZeroCount();
}

Method1 :
public static int countZero1(int input) {
int count = 0;
//The loop takes the remainder for the value of the input, and if it is divided by 10, then its number of digits is 0.
// When the value of the input is less than 0, the cycle ends
while (input >0){
if (input % 10 == 0){
count ++;
}
input /= 10;
}
return count;
}
Method2 :
private static int count = 0;
public static int countZero2(int input) {
//Recursive call function
if (input % 10 == 0){
count ++;
}
input /= 10;
if (input <= 0){
return count;
}else {
return countZero2(input);
}
}

Why is this code giving me a strange output

The method below was written to make a String version of a number, I know there are already methods that do this, such as String.valueOf(), Double.toString(), or even just "" + someNumber.
private static String numToString(double i) {
String revNumber = "";
boolean isNeg = false;
if (i == 0) { //catch zero case
return "0";
}
if (i < 0) {
isNeg = true;
}
i = Math.abs(i);
while (i > 0) { //loop backwards through number, this loop
//finish, otherwise, i would not get any output in 'main()'
revNumber += "" + i % 10; //get the end
i /= 10; //slice end
}
String number = ""; //reversed
for (int k = revNumber.length() - 1; k >= 0; k--) {
number += revNumber.substring(k, k + 1);
}
revNumber = null; //let gc do its work
return isNeg ? "-" + number : number; //result expression to add "-"
//if needed.
}
Although the above method should only be used for ints (32-bit), I made it accept a double (64-bit) argument and I passed a double argument, without a decimal, the output results are the same if I pass an int into the method as well, or with a decimals, etc...
Test:
public static void main(String[] args) {
double test = -134; //Passing double arg
System.out.println(numToString(test)); //results
}
Result: (Maximum memory results for double?):
-323-E5.1223-E33.1123-E43.1023-E43.1913-E43.1813-E43.1713-E43.1613-E43.1513-E43.1413-E43.1313-E43.1213-E43.1113-E43.1013-E43.1903-E43.1803-E999999999999933.1703-E999999999999933.1603-E999999999999933.1503-E999999999999933.1403-E9899999999999933.1303-E999999999999933.1203-E999999999999933.1103-E8899999999999933.1003-E9899999999999933.1992-E999999999999933.1892-E999999999999933.1792-E999999999999933.1692-E999999999999933.1592-E999999999999933.1492-E999999999999933.1392-E999999999999933.1292-E999999999999933.1192-E999999999999933.1092-E999999999999933.1982-E999999999999933.1882-E1999999999999933.1782-E999999999999933.1682-E999999999999933.1582-E999999999999933.1482-E999999999999933.1382-E999999999999933.1282-E1999999999999933.1182-E1999999999999933.1082-E2999999999999933.1972-E2999999999999933.1872-E3999999999999933.1772-E2999999999999933.1672-E1999999999999933.1572-E2999999999999933.1472-E999999999999933.1372-E999999999999933.1272-E1999999999999933.1172-E2999999999999933.1072-E2999999999999933.1962-E1999999999999933.1862-E999999999999933.1762-E999999999999933.1662-E999999999999933.1562-E999999999999933.1462-E999999999999933.1362-E999999999999933.1262-E999999999999933.1162-E999999999999933.1062-E999999999999933.1952-E999999999999933.1852-E999999999999933.1752-E999999999999933.1652-E999999999999933.1552-E999999999999933.1452-E999999999999933.1352-E999999999999933.1252-E999999999999933.1152-E999999999999933.1052-E9899999999999933.1942-E9899999999999933.1842-E999999999999933.1742-E999999999999933.1642-E999999999999933.1542-E999999999999933.1442-E999999999999933.1342-E999999999999933.1242-E999999999999933.1142-E9899999999999933.1042-E999999999999933.1932-E8899999999999933.1832-E9899999999999933.1732-E8899999999999933.1632-E8899999999999933.1532-E8899999999999933.1432-E999999999999933.1332-E9899999999999933.1232-E8899999999999933.1132-E7899999999999933.1032-E8899999999999933.1922-E8899999999999933.1822-E7899999999999933.1722-E6899999999999933.1622-E5899999999999933.1522-E5899999999999933.1422-E6899999999999933.1322-E6899999999999933.1222-E6899999999999933.1122-E7899999999999933.1022-E7899999999999933.1912-E7899999999999933.1812-E8899999999999933.1712-E8899999999999933.1612-E7899999999999933.1512-E7899999999999933.1412-E6899999999999933.1312-E6899999999999933.1212-E7899999999999933.1112-E7899999999999933.1012-E7899999999999933.1902-E7899999999999933.1802-E6899999999999933.1702-E7899999999999933.1602-E7899999999999933.1502-E8899999999999933.1402-E8899999999999933.1302-E8899999999999933.1202-E8899999999999933.1102-E7899999999999933.1002-E7899999999999933.1991-E9899999999999933.1891-E9899999999999933.1791-E8899999999999933.1691-E7899999999999933.1591-E8899999999999933.1491-E8899999999999933.1391-E999999999999933.1291-E9899999999999933.1191-E9899999999999933.1091-E999999999999933.1981-E999999999999933.1881-E999999999999933.1781-E999999999999933.1681-E999999999999933.1581-E1999999999999933.1481-E999999999999933.1381-E999999999999933.1281-E1999999999999933.1181-E2999999999999933.1081-E999999999999933.1971-E1999999999999933.1871-E999999999999933.1771-E2999999999999933.1671-E3999999999999933.1571-E3999999999999933.1471-E2999999999999933.1371-E2999999999999933.1271-E2999999999999933.1171-E999999999999933.1071-E999999999999933.1961-E999999999999933.1861-E999999999999933.1761-E999999999999933.1661-E999999999999933.1561-E999999999999933.1461-E8899999999999933.1361-E8899999999999933.1261-E8899999999999933.1161-E7899999999999933.1061-E7899999999999933.1951-E7899999999999933.1851-E7899999999999933.1751-E7899999999999933.1651-E7899999999999933.1551-E6899999999999933.1451-E6899999999999933.1351-E6899999999999933.1251-E6899999999999933.1151-E6899999999999933.1051-E7899999999999933.1941-E6899999999999933.1841-E7899999999999933.1741-E7899999999999933.1641-E9899999999999933.1541-E999999999999933.1441-E999999999999933.1341-E999999999999933.1241-E999999999999933.1141-E999999999999933.1041-E999999999999933.1931-E9899999999999933.1831-E999999999999933.1731-E999999999999933.1631-E8899999999999933.1531-E9899999999999933.1431-E9899999999999933.1331-E8899999999999933.1231-E8899999999999933.1131-E8899999999999933.1031-E7899999999999933.1921-E7899999999999933.1821-E7899999999999933.1721-E6899999999999933.1621-E7899999999999933.1521-E7899999999999933.1421-E8899999999999933.1321-E7899999999999933.1221-E7899999999999933.1121-E8899999999999933.1021-E8899999999999933.1911-E9899999999999933.1811-E999999999999933.1711-E999999999999933.1611-E999999999999933.1511-E999999999999933.1411-E1999999999999933.1311-E2999999999999933.1211-E3999999999999933.1111-E2999999999999933.1011-E2999999999999933.1901-E3999999999999933.1801-E2999999999999933.1701-E2999999999999933.1601-E3999999999999933.1501-E2999999999999933.1401-E3999999999999933.1301-E3999999999999933.1201-E4999999999999933.1101-E5999999999999933.1001-E4999999999999933.199-E3999999999999933.189-E3999999999999933.179-E3999999999999933.169-E4999999999999933.159-E4999999999999933.149-E4999999999999933.139-E49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This is not because of the complier. It is happening because you are doing
i /= 10; //slice end
So when you do 13.4 after the first run it wont give you 1.34 it will give you something like 1.339999999999999999 which is 1.34.
Check Retain precision with double in Java for more details.
If you just want to reverse the number you can do
private static String numToString(double i) {
String returnString = new StringBuilder(Double.toString(i)).reverse().toString();
return i>=0?returnString:"-"+returnString.substring(0,returnString.length()-1);
}

for (; i > 0; ) { //loop backwards through number
revNumber += "" + i % 10; //get the end
i /= 10; //slice end
}
This loop never finishes until it breaks at a much later time than it should. i % 10 doesn't cut off the end of a double. It works well with an int but not with a double. Hence the 134->13.4->1.34->.134-> etc.... So you get an argumentoutofrange exception or something similar to that. Else the compiler just keeps doing it for the max memory that a double can handle.

Method that determines if a number is 'insipid'

So I have been tasked with writing a program that determines if a number is insipid or not. Meaning, if you take a number:
57 for example, and you add the square of each digit, 5*5 + 7*7, that is now the new number:74. And you keep doing that until you either get 58 which means the number is NOT insipid, or you get 1, meaning the number is. 58 will just repeat a sequence that always ends back on 58.
So I wanted to attempt it with some basic recursion but perhaps I misunderstand the use of recursion here.
Here are the two relevant methods I wrote:
public static boolean insipid(int num){
int dig1 = 0, dig2 = 0, dig3 = 0; // num = 159 for example, dig1 would be 1. Default is 0 in case of a 2 digit number, dig1*dig1 = 0
if(num == 58){ //The number is not insipid
return false;
}
if(num == 1){ // the number is insipid
return true;
}
if (num < 10){
insipid(num * num);
}
if(num>99){
dig1 = (int)(num / 100);
dig2 = (int)((num - 100)/10);
dig3 = num - (((int)(num / 10))*10);
insipid(squaresum(dig1,dig2,dig3));
}
else{
dig2 = (int)(num/10); //the 10s place
dig3 = num - (((int)(num/10)) * 10); // the 1's place
insipid(squaresum(dig1, dig2,dig3)); //dig1 = 0 so I just pass it along with it.
}
}
public static int squaresum(int dig1, int dig2, int dig3){
//Returns the sum of three digits squared.
return (dig1 * dig1) + (dig2 * dig2) + (dig3 + dig3);
}
It gives me an error saying that Insipid() must return a boolean whenever I supply it a number. But I know for any number given it will always either eventually resolve to 58 or 1. So shouldn't true or false always eventually be returned so a boolean is returned and the error is invalid? Obviously this is not the case, but how I see it. Is it the use of the recursion here that is invalid?
Also, if you have any suggestions on how I might clean this up I wouldn't mind harsh criticism, my java is not very good.

If you went with your solution, the only thing missing is the return statements from the recursive calls to incipid.

Update you insipid as below. A method declared with return must return values in all cases. Your method doesn't return any value if it reach second else i.e. if number is anything other than 1 & 58.
public static boolean insipid(int num){
boolean returnValue = false;
if(num == 58){ //The number is not insipid
//do nothing
} if(num == 1){ // the number is insipid
returnValue = true;
}else{
int dig1 = 0, dig2 = 0, dig3 = 0; // num = 159 for example, dig1 would be 1. Default is 0 in case of a 2 digit number, dig1*dig1 = 0
if (num < 10){
returnValue = insipid(num * num);
}
else if(num>99){
dig1 = (int)(num / 100);
dig2 = (int)((num - 100)/10);
dig3 = num - (((int)(num / 10))*10);
returnValue = insipid(squaresum(dig1,dig2,dig3));
}
else{
dig2 = (int)(num/10); //The 10s place
dig3 = num - (((int)(num/10)) * 10); //the 1s place
returnValue = insipid(squaresum(dig1, dig2,dig3)); //dig1 = 0 so I just pass it along with it.
}
}
return returnValue ;
}
Please Note: I just tried to answer your issue regarding return issue ONLY.

In addition to Yogendra Singh's answer, you're calculating digits incorrectly: e.g. with
dig2 = (int)((num -100)/10);
you're assuming that the hundreds digit is 1. Try
dig2 = (int)((num - dig1 * 100)/10);
and similarly for dig3.