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Optimized way to count number of occurrences of a digit in a range of numbers [duplicate]

This question already has answers here:
How to count each digit in a range of integers?
(11 answers)
Closed last year.
I've been trying to find the most optimized way to compute the number of occurrences of each digit from 0 to 9 in a random range of numbers typed in by the user for a random personal project.
Say, the user enters 1 as the lower bound (inclusive) and 20 as the upper bound (inclusive). Output should be like this:
2 12 3 2 2 2 2 2 2 2
User can only enter positive integers.
Now, the below code runs fine for small range of numbers/ small bounds, however, as expected it takes 4 seconds+ on my laptop for large numbers/range.
I've been trying to find a way to make things quicker, I used modulus to get the digits thinking maybe string conversion is to blame, but it didn't increase speed that much. I want to reduce runtime to less than 2 seconds. There must be a way, but what? Here is my original code:
import java.util.Scanner;
public class CountDigitsRandomRange {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
String g = s.nextLine();
while (!g.equals("0 0")) {
String[] n = g.split(" ");
long x = Long.parseLong(n[0]);
long y = Long.parseLong(n[1]);
long zero = 0;
long one = 0;
long two = 0;
long three = 0;
long four = 0;
long five = 0;
long six = 0;
long seven = 0;
long eight = 0;
long nine = 0;
for (long i = x; i <= y; i++) {
String temp = String.valueOf(i);
for (int j = 0; j < temp.length(); j++) {
if (temp.charAt(j) == '0') {
zero++;
}
if (temp.charAt(j) == '1') {
one++;
}
if (temp.charAt(j) == '2') {
two++;
}
if (temp.charAt(j) == '3') {
three++;
}
if (temp.charAt(j) == '4') {
four++;
}
if (temp.charAt(j) == '5') {
five++;
}
if (temp.charAt(j) == '6') {
six++;
}
if (temp.charAt(j) == '7') {
seven++;
}
if (temp.charAt(j) == '8') {
eight++;
}
if (temp.charAt(j) == '9') {
nine++;
}
}
}
System.out.println(zero + " " + one + " " + two + " "+three + " " + four
+ " " + five + " " + six + " " + seven + " " + eight + " " + nine);
g=s.nextLine();
}
}
}
I've seen some solutions online similar to my issue but they're mostly in C/C++, I don't get the syntax.

Here is a simple implementation that uses modulus. If you want a faster code, you will need to find some smart formula that gives you the result without performing the actual computation.
import java.util.Arrays;
public class Counter
{
private static long[] counts = new long[10];
public static void count(long x, long y)
{
Arrays.fill(counts, 0);
for(long val=x; val<=y; val++)
count(val);
}
public static void count(long val)
{
while(val>0)
{
int digit = (int)(val % 10);
counts[digit]++;
val /= 10;
}
}
public static void main(String[] args)
{
count(1, 20);
System.out.println(Arrays.toString(counts));
}
}
Output:
[2, 12, 3, 2, 2, 2, 2, 2, 2, 2]

So here is an issue you might not be aware of #Sammie. In my opinion, you should NOT use the seconds provided by your runner in Java to count time when it comes to making operations more efficient. As far as I am informed, a more objective calculation is to use internal methods of Java which depend on the CPU clock to count time. This way there is less variation between different PC's (although this I don't believe is fully eliminated). Please check my references below:
Clock milis (only use this if you cannot use the solution below)
Nanoseconds
Edit: Here is another stack overflow post discussing this matter. Nanoseconds seem to be preferable.
All you need to do after that is convert into minutes, and you should now be calculating more precisely.

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.

Creating a chain of loops dynamically based on a List

I am practicing Java and I was trying to create a program to calculate the amount of ways an number could be divided using a set number of dividers.
For instance:
100 is the number and the dividers are 50,20,5. What are the possible divisions.
The answer would be:
Amount of 50 : 0, Amount of 20 : 0, Amount of 10 : 10
Amount of 50 : 0, Amount of 20 : 1, Amount of 10 : 8
Amount of 50 : 0, Amount of 20 : 2, Amount of 10 : 6
Amount of 50 : 0, Amount of 20 : 3, Amount of 10 : 4
Amount of 50 : 0, Amount of 20 : 4, Amount of 10 : 2
Amount of 50 : 1, Amount of 20 : 0, Amount of 10 : 5
Amount of 50 : 1, Amount of 20 : 1, Amount of 10 : 3
Amount of 50 : 1, Amount of 20 : 2, Amount of 10 : 1
Amount of 50 : 2
I wrote a code that asks the user an amount and 3 dividers. Now I am trying to figure out if there is a way to dynamically create a code for as many dividers as the user wants. The code is in a way very repetitive and there is a certain pattern to add another divider but I cannot figure out how to implement this dynamic change to the code.
The first code I came up with to do this is the following:
public class Test2 {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Insert the amount:");
int amount = scanner.nextInt();
List<Integer> dividers = new ArrayList<>();
System.out.println("Insert the first divider:");
int tempDivider = scanner.nextInt();
if (!dividers.contains(tempDivider)) {
dividers.add(tempDivider);
}
while (dividers.size()<3) {
System.out.println("Insert the next divider: (" + (3-dividers.size()) + " more to go)");
tempDivider = scanner.nextInt();
if (!dividers.contains(tempDivider)) {
dividers.add(tempDivider);
}
}
dividers.sort(Collections.reverseOrder());
System.out.print("Dividers are: ");
System.out.println(dividers);
int getal1 = dividers.get(0);
int getal2 = dividers.get(1);
int getal3 = dividers.get(2);
int fiftyAmount = amount / getal1;
int fiftyRemainder = amount % getal1;
for (int i = 0; i <= fiftyAmount; i++) {
int currentFiftyAmount = amount - (getal1 * i);
int twentyAmount = currentFiftyAmount / getal2;
int twentyRemainder = currentFiftyAmount % getal2;
if (twentyAmount == 0) {
StringBuilder output = new StringBuilder();
output.append("Amount of " + getal1 + " banknotes: " + i);
if (fiftyRemainder != 0) output.append(", Remainder: " + fiftyRemainder);
System.out.println(output);
} else {
for (int j = 0; j <= twentyAmount; j++) {
int currentTwentyAmount = currentFiftyAmount - (getal2 * j);
int tenAmount = currentTwentyAmount / getal3;
int tenRemainder = currentTwentyAmount % getal3;
if (tenAmount == 0) {
StringBuilder output = new StringBuilder();
output.append("Amount of " + getal1 + " banknotes: " + i + ", Amount of " + getal2 + " banknotes: " + j);
if (tenRemainder != 0) output.append(", Remainder: " + twentyRemainder);
} else {
StringBuilder output = new StringBuilder();
output.append("Amount of " + getal1 + " banknotes: " + i + ", Amount of " + getal2 + " banknotes: " + j +
", Amount of " + getal3 + " banknotes: " + tenAmount);
if (tenRemainder != 0) output.append(", Remainder: " + tenRemainder);
System.out.println(output);
}
}
}
}
}
}
I tried to make this more abstract to figure out a way to automate the creation of the extra dividing loops but I cannot figure it out.
The more abstract version I wrote is the following:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Insert the amount:");
int amount = scanner.nextInt();
List<Integer> dividers = new ArrayList<>();
System.out.println("Insert the first divider:");
dividers.add(scanner.nextInt());
int divider;
while (dividers.size()<2) {
System.out.println("Insert the next divider: (" + (2-dividers.size()) + " more to go)");
divider = scanner.nextInt();
if (!dividers.contains(divider)) {
dividers.add(divider);
}
}
dividers.sort(Collections.reverseOrder());
System.out.print("Dividers are: ");
System.out.println(dividers);
int divided1Amount = amount / dividers.get(0);
int divided1Remainder = amount % dividers.get(0);
for (int i = 0; i <= divided1Amount; i++) {
int currentDivided1Amount = amount - (dividers.get(0) * i);
int divided2Amount = currentDivided1Amount / dividers.get(1);
int divided2Remainder = currentDivided1Amount % dividers.get(1);
if (divided2Amount == 0) {
StringBuilder output = new StringBuilder();
output.append(dividers.get(0) + ":" + i);
if (divided1Remainder != 0) {
output.append(", Remainder: " + divided1Remainder);
}
System.out.println(output);
} else {
StringBuilder output = new StringBuilder();
output.append(dividers.get(0) + ":" + i + "," + dividers.get(1) + ":" + divided2Amount);
if (divided2Remainder != 0) {
output.append(", Remainder: " + divided2Remainder);
}
System.out.println(output);
}
}
}
}
This is also available on GitHub: https://github.com/realm1930/rekendin/blob/master/src/Main.java
Could anybody please enlighten me. I am sorry if I am not clear at my description of the issue.
Thanks

While a fixed number of dividers can be well approached with nested loops, I suggest for the general case to write the solution as a recursive function.
This problem is a good fit for dynamic programming. What I mean by this is that the problem can be broken down into simpler subproblems, and in this way the solution is naturally implemented with recursion. For instance, in your example of expressing 100 as a sum of multiples of 50, 20, and 10, there are three solutions found that all use one 50:
Amount of 50 : 1, Amount of 20 : 0, Amount of 10 : 5
Amount of 50 : 1, Amount of 20 : 1, Amount of 10 : 3
Amount of 50 : 1, Amount of 20 : 2, Amount of 10 : 1
Look at this as solving the subproblem of finding the ways that value 50 can be expressed as multiples of 20 and 10 (that is, 50 is equal to 20*0 + 10*5, 20*1 + 10*3 and 20*2 + 10*1). So you can divide-and-conquer the original problem in this sense.
Let X be the number (e.g. 100) to express, and D1, D2, ... DN the dividers. Here is a possible outline:
If there is just one divider, N = 1, it is easy: there just zero or one solution depending on whether D1 divides X.
Otherwise, possible solutions might have D1 with any multiple from 0, 1, ..., X/D1. So make a loop m1 = 0, 1, ..., X/D1, and recursively solve the subproblem having X' = X - m1*D1 and the remaining dividers D2, ..., DN. This subproblem has one fewer divider, so after enough recursions it reduces to the N = 1 case.
That solves the problem. Note, though, that fully recursing may result in a combinatorially vast number of subproblems to solve. So for efficient solution it is a good idea to store or "memoize" the solutions of previously-solved subproblems in a table so that work isn't repeated.
Other thoughts:
Let Q be the greatest common divisor (GCD) of all the dividers {D1, ..., DN}. If X isn't divisible by Q, then there are no solutions, in which case we can skip the above recursive search entirely. E.g. there is no way to express X = 103 with dividers 50, 20, and 10. This GCD test can also be applied to every subproblem so that some recursive calls can return early.
This problem is a kind of Diophantine equation, more specifically, it is related to the Frobenius coin problem and Frobenius numbers. There is a mathoverflow post discussing it.

My Birthday Problem code is not printing anything

I am an absolute beginner to learning programming and I was given this assignment:
Birthday problem. Suppose that people enter a room one at a time. How people must enter until two share a birthday? Counterintuitively, after 23 people enter the room, there is approximately a 50–50 chance that two share a birthday. This phenomenon is known as the birthday problem or birthday paradox.
Write a program Birthday.java that takes two integer command-line arguments n and trials and performs the following experiment, trials times:
Choose a birthday for the next person, uniformly at random between 0 and n−1.
Have that person enter the room.
If that person shares a birthday with someone else in the room, stop; otherwise repeat.
In each experiment, count the number of people that enter the room. Print a table that summarizes the results (the count i, the number of times that exactly i people enter the room, and the fraction of times that i or fewer people enter the room) for each possible value of i from 1 until the fraction reaches (or exceeds) 50%.
For more information on the assignment
However, my code won't print. I would really appreciate if someone could help me find the problem to my assignment.
public class Birthday {
public static void main(String[] args) {
int n = Integer.parseInt(args[0]); //number of days
int trials = Integer.parseInt(args[1]);
boolean[] birthdays = new boolean[n];
int[] times = new int[n + 2]; //number of times i people entered the room
int r;
for (int t = 1; t <= trials; t++) {
for (int k = 0; k < n; k++) { //reset birthday
birthdays[k] = false;
}
for (int i = 1; i <= n; i++) { //number of times
r = (int) (Math.random() * (n - 1)); //random birthday
if (birthdays[r] = false) {
birthdays[r] = true;
continue;
}
else if (birthdays[r] = true) {
times[i]++; //number of times i people entered the room + 1
break;
}
}
}
int j = 1;
while ((double) times[j] / trials <= 0.5) {
System.out.print(j + "\t" + times[j] + "\t" + ((double) times[j] / trials));
j++;
System.out.println("");
}
}
}

I can spot two errors from your code
As Scary Wombat pointed out, you are miss double equal sign inside of your if statement.
The assignment is asking you to calculate "fraction of times that i or fewer people enter the room", meaning you need to do a summation for the first i indices and divided by trials.
For example, among 1 million trials, the fraction in which first duplicate birthday happens when 4th person enters is
(times[0] + times[1] + times[2] + times[3])/ 1000000
Here is what I got:
1 0 0.0
2 2810 0.00281
3 5428 0.008238
4 8175 0.016413
As you can see the fraction is calculated by adding the first three elements together and then divided by 1000000 (2810 + 5428 + 8175 = 16413) / 1000000 = 0.016413
The way you are calculating the fraction ((double) times[j] / trials) is not correct.

You are not adding the previous counts as shown in the example. To do so, you can create a new variable to store the sums of previous counts. and use it as your while loop condition. For instance, see below..
csum += times[j]; // this adds the previous counts into a cumulative sum.
This cumulative sum is supposed to be the one u use to divide by trials to get your probability. Cheers!

In Java finding numbers that are both a Triangle Number and a Star Number

This is the question I've been assigned:
A so-called “star number”, s, is a number defined by the formula:
s = 6n(n-1) + 1
where n is the index of the star number.
Thus the first six (i.e. for n = 1, 2, 3, 4, 5 and 6) star numbers are: 1, 13, 37,
73, 121, 181
In contrast a so-called “triangle number”, t, is the sum of the numbers from 1 to n: t = 1 + 2 + … + (n-1) + n.
Thus the first six (i.e. for n = 1, 2, 3, 4, 5 and 6) triangle numbers are: 1, 3, 6, 10, 15, 21
Write a Java application that produces a list of all the values of type int that are both star number and triangle numbers.
When solving this problem you MUST write and use at least one function (such as isTriangeNumber() or isStarNumber()
or determineTriangeNumber() or determineStarNumber()). Also you MUST only use the formulas provided here to solve the problem.
tl;dr: Need to output values that are both Star Numbers and Triangle Numbers.
Unfortunately, I can only get the result to output the value '1' in an endless loop, even though I am incrementing by 1 in the while loop.
public class TriangularStars {
public static void main(String[] args) {
int n=1;
int starNumber = starNumber(n);
int triangleNumber = triangleNumber(n);
while ((starNumber<Integer.MAX_VALUE)&&(n<=Integer.MAX_VALUE))
{
if ((starNumber==triangleNumber)&& (starNumber<Integer.MAX_VALUE))
{
System.out.println(starNumber);
}
n++;
}
}
public static int starNumber( int n)
{
int starNumber;
starNumber= (((6*n)*(n-1))+1);
return starNumber;
}
public static int triangleNumber( int n)
{
int triangleNumber;
triangleNumber =+ n;
return triangleNumber;
}
}

Here's a skeleton. Finish the rest yourself:
Questions to ask yourself:
How do I make a Triangle number?
How do I know if something is a Star number?
Why do I only need to proceed until triangle is negative? How can triangle ever be negative?
Good luck!
public class TriangularStars {
private static final double ERROR = 1e-7;
public static void main(String args[]) {
int triangle = 0;
for (int i = 0; triangle >= 0; i++) {
triangle = determineTriangleNumber(i, triangle);
if (isStarNumber(triangle)) {
System.out.println(triangle);
}
}
}
public static boolean isStarNumber(int possibleStar) {
double test = (possibleStar - 1) / 6.;
int reduce = (int) (test + ERROR);
if (Math.abs(test - reduce) > ERROR)
return false;
int sqrt = (int) (Math.sqrt(reduce) + ERROR);
return reduce == sqrt * (sqrt + 1);
}
public static int determineTriangleNumber(int i, int previous) {
return previous + i;
}
}
Output:
1
253
49141
9533161
1849384153

You need to add new calls to starNumber() and triangleNumber() inside the loop. You get the initial values but never re-call them with the updated n values.
As a first cut, I would put those calls immediatly following the n++, so
n++;
starNumber = starNumber(n);
triangleNumber = triangleNumber(n);
}
}

The question here is that "N" neednt be the same for both star and triangle numbers. So you can increase "n" when computing both star and triangle numbers, rather keep on increasing the triangle number as long as its less the current star number. Essentially you need to maintain two variable "n" and "m".

The first problem is that you only call the starNumber() method once, outside the loop. (And the same with triangleNumber().)
A secondary problem is that unless Integer.MAX_VALUE is a star number, your loop will run forever. The reason being that Java numerical operations overflow silently, so if your next star number would be bigger than Integer.MAX_VALUE, the result would just wrap around. You need to use longs to detect if a number is bigger than Integer.MAX_VALUE.
The third problem is that even if you put all the calls into the loop, it would only display star number/triangle number pairs that share the same n value. You need to have two indices in parallel, one for star number and another for triangle numbers and increment one or the other depending on which function returns the smaller number. So something along these lines:
while( starNumber and triangleNumber are both less than or equal to Integer.MAX_VALUE) {
while( starNumber < triangleNumber ) {
generate next starnumber;
}
while( triangleNumber < starNumber ) {
generate next triangle number;
}
if( starNumber == triangleNumber ) {
we've found a matching pair
}
}
And the fourth problem is that your triangleNumber() method is wrong, I wonder how it even compiles.

I think your methodology is flawed. You won't be able to directly make a method of isStarNumber(n) without, inside that method, testing every possible star number. I would take a slightly different approach: pre-computation.
first, find all the triangle numbers:
List<Integer> tris = new ArrayList<Integer>();
for(int i = 2, t = 1; t > 0; i++) { // loop ends after integer overflow
tris.add(t);
t += i; // compute the next triangle value
}
we can do the same for star numbers:
consider the following -
star(n) = 6*n*(n-1) + 1 = 6n^2 - 6n + 1
therefore, by extension
star(n + 1) = 6*(n+1)*n + 1 = 6n^2 + 6n +1
and, star(n + 1) - star(n - 1), with some algebra, is 12n
star(n+1) = star(n) + 12* n
This leads us to the following formula
List<Integer> stars = new ArrayList<Integer>();
for(int i = 1, s = 1; s > 0; i++) {
stars.add(s);
s += (12 * i);
}
The real question is... do we really need to search every number? The answer is no! We only need to search numbers that are actually one or the other. So we could easily use the numbers in the stars (18k of them) and find the ones of those that are also tris!
for(Integer star : stars) {
if(tris.contains(star))
System.out.println("Awesome! " + star + " is both star and tri!");
}
I hope this makes sense to you. For your own sake, don't blindly move these snippets into your code. Instead, learn why it does what it does, ask questions where you're not sure. (Hopefully this isn't due in two hours!)
And good luck with this assignment.
Here's something awesome that will return the first 4 but not the last one. I don't know why the last won't come out. Have fun with this :
class StarAndTri2 {
public static void main(String...args) {
final double q2 = Math.sqrt(2);
out(1);
int a = 1;
for(int i = 1; a > 0; i++) {
a += (12 * i);
if(x((int)(Math.sqrt(a)*q2))==a)out(a);
}
}
static int x(int q) { return (q*(q+1))/2; }
static void out(int i) {System.out.println("found: " + i);}
}