For a given number N(0<N<=100) find out the minimum positive integer X divisible by N, where the sum of digits of X is equal to N, and X is not equal to N.
public static int getSmallestNumber(int input1) {
int res =0;
for(int i=2;i<10000;i++) {
if(getSum(input1*i) == input1) {
res = input1*i;
break;
}
}
return res;
}
static int getSum(int n) {
int sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = n/10;
}
return sum;
}
Example: Input: 9
output: 18
Explanation: 1+8 is 9 and 18 is divisible by 9
Example2: Input: 10
output: 190
Explanation: 1+9+0 is 10 and 190 is divisible by 10
My solution works for smaller integer but breaks for larger numbers like 98, 99 or 100.
we need to know that if we multiply multiple of 7, 35(3+5=8) with 7 again then obviously a larger number 42(4+2=6) comes but have lower sum of digits so the change in digits with each multiplication seems uncertain. So, if we can find a way to generalize this change in digits of upcoming next to next larger multiples of our number then a time efficient solution can be designed.
I think you need to add the actual check on the condition (i % input1 == 0) to check the divisibility.
public static int getSmallestNumber (int input1) {
int res = 0;
for (int i = input1; i < 10000; i++) {
if (getSum(i) == input1 && i % input1 == 0) {
res = i;
break;
}
}
return res;
}
static int getSum (int n) {
int sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = n/10;
}
return sum;
}
I will get to the point quickly. Basically smith numbers are: Composite number the sum of whose digits is the sum of the digits of its prime factors (excluding 1). (The primes are excluded since they trivially satisfy this condition). One example of a Smith number is the beast number 666=2·3·3·37, since 6+6+6=2+3+3+(3+7)=18.
what i've tried:
In a for loop first i get the sum of the current number's(i) digits
In same loop i try to get the sum of the number's prime factors digits.
I've made another method to check if current number that is going to proccessed in for loop is prime or not,if its prime it will be excluded
But my code is seems to not working can you guys help out?
public static void main(String[] args) {
smithInrange(1, 50);
}
public static void smithInrange(int start_val, int end_val) {
for (int i = start_val; i < end_val; i++) {
if(!isPrime(i)) { //since we banned prime numbers from this process i don't include them
int for_digit_sum = i, digit = 0, digit_sum = 0, for_factor_purpose = i, smith_sum = 0;
int first = 0, second = 0, last = 0;
// System.out.println("current number is" + i);
while (for_digit_sum > 0) { // in this while loop i get the sum of current number's digits
digit = for_digit_sum % 10;
digit_sum += digit;
for_digit_sum /= 10;
}
// System.out.println("digit sum is"+digit_sum);
while (for_factor_purpose % 2 == 0) { // i divide the current number to 2 until it became an odd number
first += 2;
for_factor_purpose /= 2;
}
// System.out.println("the first sum is " + first);
for (int j = 3; j < Math.sqrt(for_factor_purpose); j += 2) {
while (for_factor_purpose % j == 0) { // this while loop is for getting the digit sum of every prime
// factor that j has
int inner_digit = 0, inner_temp = j, inner_digit_sum = 0;
while (inner_temp > 0) {
inner_digit = inner_temp % 10;
second += inner_digit;
inner_temp /= 10;
}
// System.out.println("the second sum is " + second);
for_factor_purpose /= j;
}
}
int last_temp = for_factor_purpose, last_digit = 0, last_digit_sum = 0;
if (for_factor_purpose > 2) {
while (last_temp > 0) {
last_digit = last_temp % 10;
last += last_digit;
last_temp /= 10;
}
// System.out.println("last is " + last);
}
smith_sum = first + second + last;
// System.out.println("smith num is "+ smith_sum);
// System.out.println(smith_sum);
if (smith_sum == digit_sum) {
System.out.println("the num founded is" + i);
}
}
}
}
public static boolean isPrime(int i) {
int sqrt = (int) Math.sqrt(i) + 1;
for (int k = 2; k < sqrt; k++) {
if (i % k == 0) {
// number is perfectly divisible - no prime
return false;
}
}
return true;
}
the output is:
the num founded is4
the num founded is9
the num founded is22
the num founded is25
the num founded is27
the num founded is49
how ever the smith number between this range(1 and 50) are:
4, 22 and 27
edit:I_ve found the problem which is :
Math.sqrt(for_factor_purpose) it seems i should add 1 to it to eliminate square numbers. Thanks to you guys i've see sthe solution on other perspectives.
Keep coding!
Main loop for printing Smith numbers.
for (int i = 3; i < 10000; i++) {
if (isSmith(i)) {
System.out.println(i + " is a Smith number.");
}
}
The test method to determine if the supplied number is a Smith number. The list of primes is only increased if the last prime is smaller in magnitude than the number under test.
static boolean isSmith(int v) {
int sum = 0;
int save = v;
int lastPrime = primes.get(primes.size() - 1);
if (lastPrime < v) {
genPrimes(v);
}
outer:
for (int p : primes) {
while (save > 1) {
if (save % p != 0) {
continue outer;
}
sum += sumOfDigits(p);
save /= p;
}
break;
}
return sum == sumOfDigits(v) && !primes.contains(v);
}
Helper method to sum the digits of a number.
static int sumOfDigits(int i) {
return String.valueOf(i).chars().map(c -> c - '0').sum();
}
And the prime generator. It uses the list as it is created to determine if a given
number is a prime.
static List<Integer> primes = new ArrayList<>(List.of(2, 3));
static void genPrimes(int max) {
int next = primes.get(primes.size() - 1);
outer:
while (next <= max) {
next += 2;
for (int p : primes) {
if (next % p == 0) {
continue outer;
}
if (p * p > next) {
break;
}
}
primes.add(next);
}
}
}
I do not want to spoil the answer finding, but just some simpler code snippets,
making everything simpler, and more readable.
public boolean isSmith(int a) {
if (a < 2) return false;
int factor = findDivisor(a);
if (factor == a) return false;
int sum = digitSum(a);
// loop:
a /= factor;
sum -= digitSum(factor);
...
}
boolean isPrime(int a){
for(int i = 2; i*i <= a; i++) {
if (a % i == 0) {
return false;
}
}
return true;
}
int findDivisor(int a){
for(int i = 2; i*i <= a; i++) {
if (a % i == 0) {
return i;
}
}
return a;
}
int digitSum(int a) {
if (a < 10) {
return a;
}
int digit = a % 10;
int rest = a / 10;
return digit + digitSum(rest);
}
As you see integer division 23 / 10 == 2, and modulo (remainder) %: 23 % 10 == 3 can simplify things.
Instead of isPrime, finding factor(s) is more logical. In fact the best solution is not using findDivisor, but immediately find all factors
int factorsSum = 0;
int factorsCount = 0;
for(int i = 2; i*i <= a; i++) {
while (a % i == 0) {
factorsSum += digitSum(i);
a /= i;
factorsCount++;
}
}
// The remaining factor >= sqrt(original a) must be a prime.
// (It cannot contain smaller factors.)
factorsSum += digitSum(a);
factorsCount++;
Here is the code. If you need further help, please let me know. The code is pretty self explanatory and a decent bit was taken from your code but if you need me to explain it let me know.
In short, I created methods to check if a number is a smith number and then checked each int in the range.
import java.util.*;
public class MyClass {
public static void main(String args[]) {
System.out.println(smithInRange)
}
public int factor;
public boolean smithInRange(int a, int b){
for (int i=Math.min(a,b);i<=Math.max(a,b);i++) if(isSmith(i)) return true;
return false;
}
public boolean isSmith(int a){
if(a<2) return false;
if(isPrime(a)) return false;
int digits=0;
int factors=0;
String x=a+¨" ";
for(int i=0;i<x.length()-1;i++) digits+= Integer.parseInt(x.substring(i,i+1));
ArrayList<Integer> pF = new ArrayList<Integer>();
pF.add(a);
while(!aIsPrime(pF)){
int num = pF.get(pF.size-1)
pF.remove(pF.size()-1);
pF.add(factor);
pF.add(num/factor)
}
for(int i: pF){
if((factors+"").length()==1)factors+= i;
else{
String ss= i+" ";
int nums=0;
for(int j=0;j<ss.length()-1;j++){
nums+=Integer.parseInt(ss.substring(j,j+1));
}
}
}
return (factors==digits);
}
public boolean isPrime(int a){
for(int i=2;i<=(int)Math.sqrt(a),i++){
String s = (double)a/(double)i+"";
if(s.substring(s.length()-2).equals(".0")){
return false;
factor = i;
}
}
return true;
}
public boolean aIsPrime(ArrayList<int> a){
for(int i: a) if (!isPrime(a)) return false;
return true;
}
}
In our directions, we have to get a 16 digit number, then sum all the digits in the odd place from right to left. After that, we have to sum all the even place digits from right to left, double the sum, then take module 9. When I try to run my code, I keep getting "Invalid", even if it is with a valid credit card number.
public static boolean validateCreditCard(long number) {
double cardSum = 0;
for (int i = 0; i < 16; i++) {
long cardnumber = (long) Math.pow(10, i);
double oddPlaceSum = 0;
double evenPlaceSum = 0;
if (i % 2 != 0) {
oddPlaceSum += ((int)(number % cardnumber / (Math.pow(10, i))));
} else { // so if i%2 ==0
evenPlaceSum += ((int)(number % cardnumber / (Math.pow(10, i)) * 2 % 9));
}
cardSum += evenPlaceSum + oddPlaceSum;
}
if (cardSum % 10 == 0) {
return true;
System.out.println("Valid");
} else {
return false;
System.out.println("Invalid");
}
}
Try this instead :
Convert the 16 digit number into a String using Long.toString(number).
Iterate through the String character by character and keep track of even and odd indexes.
Convert each char to an Integer using Integer.valueOf() thereby adding them incrementally.
Voila, you got your evenSum and oddSum. Next steps should be trivial.
public static boolean validateCreditCard(long number){
String x = Long.toString(number);
int evenSum = 0;
int oddSum = 0;
for(int i=0; i<x.length; i=i+2) {
oddSum += Integer.valueOf(s[i]);
evenSum += Integer.valueOf(s[i+1]);
}
//Do the next steps with odd and even sums.
Also, do handle IndexOutOfBoundsException as appropriate.
You can do it in a single while loop as digits are fixed, like this:
int digit,evensum,oddsum;
int i=16;
while(i > 0){
digit=number%10;
if(i%2 == 0)
evensum+=digit;
else
oddsum+=digit;
i--;
digit/=10;
}
Try this instead
using Recusion find sum of even placed of digit and sum of odd placed of digit.
class Recursion {
static int count = 0;
static int even =0;
static int odd =0;
public static int Digits(int num) {
if (num > 0) {
count++;
if(count%2 == 0){
even += num%10;
}
else{
odd += num%10;
}
Digits(num / 10);
}
return even;
// for odd
// return odd;
}
public static void main(String[] args) {
int num = 31593;
int res = Digits(num);
System.out.println("Total digits are: " + res);
}
}
Help me How to print both even and odd sum together?
I have used below programs to find first n prime numbers (in below program it is from 2 to n). Can we write a program with single for loop? I also tried recursive approach but it is not working for me.
public static void main(String[] args) {
// Prime numbers in a range
int range = 15;
int num = 1;
int count = 0;
boolean prime = true;
while (count < range) {
num = num + 1;
prime = true;
for (int i = 2; i <= num / 2; i++) {
if (num % i == 0) {
prime = false;
break;
}
}
if (prime) {
count++;
System.out.println(num);
}
}
}
i dont think you can reduce it to one loop. But you can improve your code as Luca mentioned it.
public class PrimeFinder {
private final List<Integer> primes;
private final int primeCapacity;
public PrimeFinder(final int primeCapacity) {
if (primeCapacity < 3) {
throw new IllegalArgumentException("Tkat is way to easy.");
}
this.primeCapacity = primeCapacity;
primes = new ArrayList<>(primeCapacity);
primes.add(2);
}
public void find() {
final Index currentNumber = new Index();
while (primes.size() < primeCapacity) {
if (!primes.stream().parallel().anyMatch(prime -> (currentNumber.value % prime) == 0)) {
primes.add(currentNumber.incremet());
} else {
currentNumber.incremet();
}
}
}
public List<Integer> getPrimes() {
return primes;
}
private class Index {
public int value = 3;
public int incremet() {
return value++;
}
}
public static void main(String[] args) {
PrimeFinder primeFinder = new PrimeFinder(100000);
final long start = System.currentTimeMillis();
primeFinder.find();
final long finish = System.currentTimeMillis();
System.out.println("Score after " + (finish - start) + " milis.");
primeFinder.getPrimes().stream().forEach((prime) -> {
System.out.println(prime);
});
}
}
main rule here is simple, if given number isnt clearly divided by any prime number that you already have found, then it is prime number.
P.S. dont forget that primes.strem().... is loop also, so it is not a one loop code.
P.S.S. you can reduce this much further.
to understand the complexity of your algorithm you don't have to count the number of inner loops but the number of times you are iterating over your elements. To improve the performance of your algorithm you need to investigate if there are some iterations that could be unnecessary.
In your case when you do
for (int i = 2; i <= num / 2; i++) you are testing your num against values that are not necessary.. ex: if a number is divisible by 4 it will be by 2 too.
when you do for (int i = 2; i <= num / 2; i++) with num = 11
i will assume the values 2,3,4,5. 4 here is a not interesting number and represent an iteration that could be avoided.
anyway according to wikipedia the sieve of Eratosthenes is one of the most efficient ways to find all of the smaller primes.
public class PrimeSieve {
public static void main(String[] args) {
int N = Integer.parseInt(args[0]);
// initially assume all integers are prime
boolean[] isPrime = new boolean[N + 1];
for (int i = 2; i <= N; i++) {
isPrime[i] = true;
}
// mark non-primes <= N using Sieve of Eratosthenes
for (int i = 2; i*i <= N; i++) {
// if i is prime, then mark multiples of i as nonprime
// suffices to consider mutiples i, i+1, ..., N/i
if (isPrime[i]) {
for (int j = i; i*j <= N; j++) {
isPrime[i*j] = false;
}
}
}
// count primes
int primes = 0;
for (int i = 2; i <= N; i++) {
if (isPrime[i]) primes++;
}
System.out.println("The number of primes <= " + N + " is " + primes);
}
}
I don't know any way use a single loop for your question, but I do see two places where you can reduce the complexity:
Cache the prime numbers you found and use these prime numbers to decide if a number is prime or not. For example, to decide if 11 is a prime number, you just need to divide it by 2,3,5,7 instead of 2,3,4,5,6,...10
You don't need to check till num/2, you only need to check till the square root of num. For example, for 10, you only need to check 2,3 instead of 2,3,4,5. Because if number n is not prime, then n = a * b where either a or b is smaller than the square root x of n). If a is the smaller one, knowing that n can be divided by a is enough to decide that n is not prime.
So combining 1 & 2, you can improve the efficiency of your loops:
public static void main(String[] args) {
// Prime numbers in a range
int range = 15;
int num = 1;
int count = 0;
boolean prime = true;
ArrayList<Integer> primes = new ArrayList<>(range);
while (num < range) {
num = num + 1;
prime = true;
int numSquareRoot = (int) Math.floor(Math.pow(num, 0.5));
for (Integer smallPrimes : primes) {// only need to divide by the primes smaller than num
if (numSquareRoot > numSquareRoot) {// only need to check till the square root of num
break;
}
if (num % smallPrimes == 0) {
prime = false;
break;
}
}
if (prime) {
System.out.println(num);
primes.add(num);// cache the primes
}
}
}
I have something like this down:
int f = 120;
for(int ff = 1; ff <= f; ff++){
while (f % ff != 0){
}
Is there anything wrong with my loop to find factors? I'm really confused as to the workings of for and while statements, so chances are they are completely wrong.
After this, how would I go about assigning variables to said factors?
The following code will return a list of all factors of a given number:
public ArrayList<Integer> findFactors(int num) {
ArrayList<Integer> factors = new ArrayList<Integer>();
// Skip two if the number is odd
int incrementer = num % 2 == 0 ? 1 : 2;
for (int i = 1; i <= Math.sqrt(num); i += incrementer) {
// If there is no remainder, then the number is a factor.
if (num % i == 0) {
factors.add(i);
// Skip duplicates
if (i != num / i) {
factors.add(num / i);
}
}
}
// Sort the list of factors
Collections.sort(factors);
return factors;
}
This answer improves Sharad Dargan's answer in two ways:
Based on an idea used in this answer, you can speed up the solution by determining the value to increment by, based on whether the number is even or odd.
Add the following line of code before the for loop:
int incrementer = num % 2 == 0 ? 1 : 2;
Then change the last part of the loop to:
i += incrementer
If the number is odd, it then will skip all even numbers, rather than always incrementing by one no matter what.
Sharad stores the upper limit value in a variable and then uses that variable in the for loop:
int upperlimit = (int)(Math.sqrt(a));
...
for(int i = 1; i <= upperlimit; i+= 1)
Instead, place Math.sqrt(num) directly in the for loop and skip the upper limit variable:
for (int i = 1; i <= Math.sqrt(num); i += incrementer) {
This will allow you to skip the casting part of the code, creating cleaner code.
Some JUnit test cases you can then use:
#Test
public void test12() {
FindFactors find = new FindFactors();
int num = 12;
List<Integer> factors = Arrays.asList(1, 2, 3, 4, 6, 12);
assertEquals(factors, find.findFactors(num));
}
#Test
public void test1000000() {
FindFactors find = new FindFactors();
int num = 1000000;
List<Integer> factors = Arrays.asList(1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 160, 200,
250, 320, 400, 500, 625, 800, 1000, 1250, 1600, 2000, 2500, 3125, 4000, 5000, 6250, 8000, 10000, 12500,
15625, 20000, 25000, 31250, 40000, 50000, 62500, 100000, 125000, 200000, 250000, 500000, 1000000);
assertEquals(factors, find.findFactors(num));
}
#Test
public void test1() {
FindFactors find = new FindFactors();
int num = 1;
List<Integer> factors = Arrays.asList(1);
assertEquals(factors, find.findFactors(num));
}
#Test
public void test0() {
FindFactors find = new FindFactors();
int num = 0;
List<Integer> factors = new ArrayList<Integer>();
assertEquals(factors, find.findFactors(num));
}
Here is how to get all factors of the given number.
public class Factors {
public static void main(String[] args){
int n = 420;
for(int i=2; i<=n; i++){
while(n%i==0){
System.out.println(i + "| " + n);
System.out.println(" -----");
n = n/i;
}
}
}
}
Output:
2| 420
-----
2| 210
-----
3| 105
-----
5| 35
-----
7| 7
-----
public class Solution {
public ArrayList<Integer> allFactors(int a) {
int upperlimit = (int)(Math.sqrt(a));
ArrayList<Integer> factors = new ArrayList<Integer>();
for(int i=1;i <= upperlimit; i+= 1){
if(a%i == 0){
factors.add(i);
if(i != a/i){
factors.add(a/i);
}
}
}
Collections.sort(factors);
return factors;
}
}
The above solution simply works like calculating prime factors.
The difference being for every prime factor we keep calculating the other part of the product i.e the reqd number.
In order to find the factors of a given number, you only need to check upto the square root of the given number.
For example, in order to find the factors of 6, you only need to check till 2.45 (√6). The factors of 6 will be 1 and 2, and their converse numbers, i.e. 3 and 6.
I have made a program that determines the factors of a given number and displays them. Here is the necessary code:
Scanner input = new Scanner(System.in);
System.out.print("Enter integer: ");
long num = input.nextLong();
for(long i = 1; i <= Math.sqrt(num); i++) {
if(num % i == 0) {
System.out.println(i);
if(i != num/i) {
System.out.println(num/i);
}
}
}
You just need this program to find the factors of a given number. However, if you want to take it a step further and display the factors arranged in ascending order, then the necessary code is as follows:
Scanner input = new Scanner(System.in);
System.out.print("Enter integer: ");
long num = input.nextLong();
ArrayList<Long> list1 = new ArrayList<>(), list2 = new ArrayList<>();
long currentTime = System.currentTimeMillis();
for(long i = 1; i <= Math.sqrt(num); i++) {
if(num % i == 0) {
list1.add(i);
if(i != num/i) {
list2.add(num/i);
}
}
}
int n1 = list1.size() - 1;
int n2 = list2.size() - 1;
for(int i = 0; i <= n1; i++) {
System.out.println(list1.get(i));
}
for(int i = n2; i >= 0; i--) {
System.out.println(list2.get(i));
}
What this does: This program stores the factors of the number upto the number's square root in one list (list1), and the converse of these numbers in another list (list2). It then prints the elements of both lists (as shown).
There's nothing wrong with your for loop, but a while loop is the wrong thing to be using here.
The logic of your for loop is:
Set ff to 1.
Keep going while ff <= f.
After you've done everything in the for loop, add 1 to ff.
This looks like it is exactly as you want.
The while loop isn't right, though. It will continue to do whatever code you write there for as long as ff is a factor of f, so unless you change them in the while code, you'll get an infinite loop. However, changing that to an if statement will give you what you want.
Since you're checking for factors, you don't actually need to check all possibilities up to f - only up to the square root of f. Whenever you find that ff is a factor, output both ff and f/ff as factors, unless f is a sqare number.
public static void printFactors(int number) {
if (number < 1 )
System.out.println("Invalid Value");
for (int i = 1 ; i <= number ; ++i) {
if ( number % i == 0)
System.out.println(i);
}
}
}
It looks like you are not going to do something with either f or ff in your while loop? If so, the expression f%ff != 0 is either false (and then it will go to the next in the for loop), or it is true, and it will end up in an infinite loop.
Are you sure you need the while like this?
Slightly modified solution: You can first check if variable x is divisible by variable y. If yes, we will count 1 and will repeat this process. For the loop counter, x/y is used and you should check x>0 to avoid repetition when x becomes zero but loop is not finished yet.
public class Factor {
public static void main(String[] args) {
int x = 48;
int x1 = x;
int y = 2;
int k = x / y;
int j = 0;
for (int i = 1; i < k; i++) {
if ((x % y) == 0 && x > 0)
j++;
x = x / 2;
}
System.out.println(+x1 + " is a factor of " + y + " for " + j
+ " times.");
}
}
I got all the factors just fine with this (I just modified the algorithm in the question).
int num1 = 120;
for(int num2=1;num2<=num1;num2++)
{
if (num1%num2 != 0)
System.out.println(num2);
}
import java.util.Scanner;
public class Factors
{
Scanner scn=new Scanner(System.in);
int num=scn.nextInt();
public void findFactor()
{
System.out.println("Factors are");
System.out.println("1");
for(int i=2;i<=num;i++)
{
if(num%i==0)
{
num=num/i;
System.out.println(i);
i=2;
}
}
}
public static void main(String[] args)
{
while(1==1)
{
System.out.println("Enter a Number");
Factors fct=new Factors();
fct.findFactor();
}
}
}
Utilizing Streams introduced in Java 8, the following will print the factors for a given number.
int input = 1500;
IntStream.rangeClosed(1, input)
.filter(e -> input % e == 0)
.forEach(System.out::println);
This is how you write it yourself like a boss. Needs to add if statements to handle one and two, but besides that; this method is as sexy as it gets
public static void primerize(int n){
boolean reduced = false;
while(n > 2){
if(n%2 == 0){
System.out.println(2 + "," + n/2);
n /= 2;
}
else{
int i = isPrime(n);
if(i == n && reduced == false){
System.out.println(1 + "," + n);
n /= n;
}
else if(i == n){
n/= n;
}
else{
System.out.println(i + "," + n/i);
n = i;
reduced = true;
}
}
}}
public static int isPrime(int n){
for(int i = (n/3); i > 0; i--){
if(i == 1){
return n;
}
else if(n%i == 0){
return i;
}
}
return 0;}
This code will give you the factors.
ArrayList<Integer> arr = new ArrayList<>();
int x=48;
int y=1;
while(x!=1)
{
if(x%y==0)
{
x=x/y;
arr.add(y);
if(y==1)
{
y++;
}
}
else
{
y+=1;
}
}
System.out.println(arr);
Easiest way using recursive function
public static int factorial(int n){
if(n!=1)
return n*factorial(n-1);
return 1;
}