I have been trying to solve a Java exercise on a Codility web page.
Below is the link to the mentioned exercise and my solution.
https://codility.com/demo/results/demoH5GMV3-PV8
Can anyone tell what can I correct in my code in order to improve the score?
Just in case here is the task description:
A small frog wants to get to the other side of a river. The frog is currently located at position 0, and wants to get to position X. Leaves fall from a tree onto the surface of the river.
You are given a non-empty zero-indexed array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in minutes.
The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only when leaves appear at every position across the river from 1 to X.
For example, you are given integer X = 5 and array A such that:
A[0] = 1
A[1] = 3
A[2] = 1
A[3] = 4
A[4] = 2
A[5] = 3
A[6] = 5
A[7] = 4
In minute 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the river.
Write a function:
class Solution { public int solution(int X, int[] A); }
that, given a non-empty zero-indexed array A consisting of N integers and integer X, returns the earliest time when the frog can jump to the other side of the river.
If the frog is never able to jump to the other side of the river, the function should return −1.
For example, given X = 5 and array A such that:
A[0] = 1
A[1] = 3
A[2] = 1
A[3] = 4
A[4] = 2
A[5] = 3
A[6] = 5
A[7] = 4
the function should return 6, as explained above. Assume that:
N and X are integers within the range [1..100,000];
each element of array A is an integer within the range [1..X].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(X), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
And here is my solution:
import java.util.ArrayList;
import java.util.List;
class Solution {
public int solution(int X, int[] A) {
int list[] = A;
int sum = 0;
int searchedValue = X;
List<Integer> arrayList = new ArrayList<Integer>();
for (int iii = 0; iii < list.length; iii++) {
if (list[iii] <= searchedValue && !arrayList.contains(list[iii])) {
sum += list[iii];
arrayList.add(list[iii]);
}
if (list[iii] == searchedValue) {
if (sum == searchedValue * (searchedValue + 1) / 2) {
return iii;
}
}
}
return -1;
}
}
You are using arrayList.contains inside a loop, which will traverse the whole list unnecessarily.
Here is my solution (I wrote it some time ago, but I believe it scores 100/100):
public int frog(int X, int[] A) {
int steps = X;
boolean[] bitmap = new boolean[steps+1];
for(int i = 0; i < A.length; i++){
if(!bitmap[A[i]]){
bitmap[A[i]] = true;
steps--;
if(steps == 0) return i;
}
}
return -1;
}
Here is my solution. It got me 100/100:
public int solution(int X, int[] A)
{
int[] B = A.Distinct().ToArray();
return (B.Length != X) ? -1 : Array.IndexOf<int>(A, B[B.Length - 1]);
}
100/100
public static int solution (int X, int[] A){
int[]counter = new int[X+1];
int ans = -1;
int x = 0;
for (int i=0; i<A.length; i++){
if (counter[A[i]] == 0){
counter[A[i]] = A[i];
x += 1;
if (x == X){
return i;
}
}
}
return ans;
}
A Java solution using Sets (Collections Framework) Got a 100%
import java.util.Set;
import java.util.TreeSet;
public class Froggy {
public static int solution(int X, int[] A){
int steps=-1;
Set<Integer> values = new TreeSet<Integer>();
for(int i=0; i<A.length;i++){
if(A[i]<=X){
values.add(A[i]);
}
if(values.size()==X){
steps=i;
break;
}
}
return steps;
}
Better approach would be to use Set, because it only adds unique values to the list. Just add values to the Set and decrement X every time a new value is added, (Set#add() returns true if value is added, false otherwise);
have a look,
public static int solution(int X, int[] A) {
Set<Integer> values = new HashSet<Integer>();
for (int i = 0; i < A.length; i++) {
if (values.add(A[i])) X--;
if (X == 0) return i;
}
return -1;
}
do not forget to import,
import java.util.HashSet;
import java.util.Set;
Here's my solution, scored 100/100:
import java.util.HashSet;
class Solution {
public int solution(int X, int[] A) {
HashSet<Integer> hset = new HashSet<Integer>();
for (int i = 0 ; i < A.length; i++) {
if (A[i] <= X)
hset.add(A[i]);
if (hset.size() == X)
return i;
}
return -1;
}
}
Simple solution 100%
public int solution(final int X, final int[] A) {
Set<Integer> emptyPosition = new HashSet<Integer>();
for (int i = 1; i <= X; i++) {
emptyPosition.add(i);
}
// Once all the numbers are covered for position, that would be the
// moment when the frog will jump
for (int i = 0; i < A.length; i++) {
emptyPosition.remove(A[i]);
if (emptyPosition.size() == 0) {
return i;
}
}
return -1;
}
Here's my solution.
It isn't perfect, but it's good enough to score 100/100.
(I think that it shouldn't have passed a test with a big A and small X)
Anyway, it fills a new counter array with each leaf that falls
counter has the size of X because I don't care for leafs that fall farther than X, therefore the try-catch block.
AFTER X leafs fell (because it's the minimum amount of leafs) I begin checking whether I have a complete way - I'm checking that every int in count is greater than 0.
If so, I return i, else I break and try again.
public static int solution(int X, int[] A){
int[] count = new int[X];
for (int i = 0; i < A.length; i++){
try{
count[A[i]-1]++;
} catch (ArrayIndexOutOfBoundsException e){ }
if (i >= X - 1){
for (int j = 0; j< count.length; j++){
if (count[j] == 0){
break;
}
if (j == count.length - 1){
return i;
}
}
}
}
return -1;
}
Here's my solution with 100 / 100.
public int solution(int X, int[] A) {
int len = A.length;
if (X > len) {
return -1;
}
int[] isFilled = new int[X];
int jumped = 0;
Arrays.fill(isFilled, 0);
for (int i = 0; i < len; i++) {
int x = A[i];
if (x <= X) {
if (isFilled[x - 1] == 0) {
isFilled[x - 1] = 1;
jumped += 1;
if (jumped == X) {
return i;
}
}
}
}
return -1;
}
Here's what I have in C#. It can probably still be refactored.
We throw away numbers greater than X, which is where we want to stop, and then we add numbers to an array if they haven't already been added.
When the count of the list has reached the expected number, X, then return the result. 100%
var tempArray = new int[X+1];
var totalNumbers = 0;
for (int i = 0; i < A.Length; i++)
{
if (A[i] > X || tempArray.ElementAt(A[i]) != 0)
continue;
tempArray[A[i]] = A[i];
totalNumbers++;
if (totalNumbers == X)
return i;
}
return -1;
below is my solution. I basically created a set which allows uniques only and then go through the array and add every element to set and keep a counter to get the sum of the set and then using the sum formula of consecutive numbers then I got 100% . Note : if you add up the set using java 8 stream api the solution is becoming quadratic and you get %56 .
public static int solution2(int X, int[] A) {
long sum = X * (X + 1) / 2;
Set<Integer> set = new HashSet<Integer>();
int setSum = 0;
for (int i = 0; i < A.length; i++) {
if (set.add(A[i]))
setSum += A[i];
if (setSum == sum) {
return i;
}
}
return -1;
}
My JavaScript solution that got 100 across the board. Since the numbers are assumed to be in the range of the river width, simply storing booleans in a temporary array that can be checked against duplicates will do. Then, once you have amassed as many numbers as the quantity X, you know you have all the leaves necessary to cross.
function solution(X, A) {
covered = 0;
tempArray = [];
for (let i = 0; i < A.length; i++) {
if (!tempArray[A[i]]) {
tempArray[A[i]] = true;
covered++
if(covered === X) return i;
}
}
return -1;
}
Here is my answer in Python:
def solution(X, A):
# write your code in Python 3.6
values = set()
for i in range (len(A)):
if A[i]<=X :
values.add(A[i])
if len(values)==X:
return i
return -1
Just tried this problem as well and here is my solution. Basically, I just declared an array whose size is equal to position X. Then, I declared a counter to monitor if the necessary leaves have fallen at the particular spots. The loop exits when these leaves have been met and if not, returns -1 as instructed.
class Solution {
public int solution(int X, int[] A) {
int size = A.length;
int[] check = new int[X];
int cmp = 0;
int time = -1;
for (int x = 0; x < size; x++) {
int temp = A[x];
if (temp <= X) {
if (check[temp-1] > 0) {
continue;
}
check[temp - 1]++;
cmp++;
}
if ( cmp == X) {
time = x;
break;
}
}
return time;
}
}
It got a 100/100 on the evaluation but I'm not too sure of its performance. I am still a beginner when it comes to programming so if anybody can critique the code, I would be grateful.
Maybe it is not perfect but its straightforward. Just made a counter Array to track the needed "leaves" and verified on each iteration if the path was complete. Got me 100/100 and O(N).
public static int frogRiver(int X, int[] A)
{
int leaves = A.Length;
int[] counter = new int[X + 1];
int stepsAvailForTravel = 0;
for(int i = 0; i < leaves; i++)
{
//we won't get to that leaf anyway so we shouldnt count it,
if (A[i] > X)
{
continue;
}
else
{
//first hit!, keep a count of the available leaves to jump
if (counter[A[i]] == 0)
stepsAvailForTravel++;
counter[A[i]]++;
}
//We did it!!
if (stepsAvailForTravel == X)
{
return i;
}
}
return -1;
}
This is my solution. I think it's very simple. It gets 100/100 on codibility.
set.contains() let me eliminate duplicate position from table.
The result of first loop get us expected sum. In the second loop we get sum of input values.
class Solution {
public int solution(int X, int[] A) {
Set<Integer> set = new HashSet<Integer>();
int sum1 = 0, sum2 = 0;
for (int i = 0; i <= X; i++){
sum1 += i;
}
for (int i = 0; i < A.length; i++){
if (set.contains(A[i])) continue;
set.add(A[i]);
sum2 += A[i];
if (sum1 == sum2) return i;
}
return -1;
}
}
Your algorithm is perfect except below code
Your code returns value only if list[iii] matches with searchedValue.
The algorithm must be corrected in such a way that, it returns the value if sum == n * ( n + 1) / 2.
import java.util.ArrayList;
import java.util.List;
class Solution {
public int solution(int X, int[] A) {
int list[] = A;
int sum = 0;
int searchedValue = X;
int sumV = searchedValue * (searchedValue + 1) / 2;
List<Integer> arrayList = new ArrayList<Integer>();
for (int iii = 0; iii < list.length; iii++) {
if (list[iii] <= searchedValue && !arrayList.contains(list[iii])) {
sum += list[iii];
if (sum == sumV) {
return iii;
}
arrayList.add(list[iii]);
}
}
return -1;
}
}
I think you need to check the performance as well. I just ensured the output only
This solution I've posted today gave 100% on codility, but respectivly #rafalio 's answer it requires K times less memory
public class Solution {
private static final int ARRAY_SIZE_LOWER = 1;
private static final int ARRAY_SIZE_UPPER = 100000;
private static final int NUMBER_LOWER = ARRAY_SIZE_LOWER;
private static final int NUMBER_UPPER = ARRAY_SIZE_UPPER;
public static class Set {
final long[] buckets;
public Set(int size) {
this.buckets = new long[(size % 64 == 0 ? (size/64) : (size/64) + 1)];
}
/**
* number should be greater than zero
* #param number
*/
public void put(int number) {
buckets[getBucketindex(number)] |= getFlag(number);
}
public boolean contains(int number) {
long flag = getFlag(number);
// check if flag is stored
return (buckets[getBucketindex(number)] & flag) == flag;
}
private int getBucketindex(int number) {
if (number <= 64) {
return 0;
} else if (number <= 128) {
return 1;
} else if (number <= 192) {
return 2;
} else if (number <= 256) {
return 3;
} else if (number <= 320) {
return 4;
} else if (number <= 384) {
return 5;
} else
return (number % 64 == 0 ? (number/64) : (number/64) + 1) - 1;
}
private long getFlag(int number) {
if (number <= 64) {
return 1L << number;
} else
return 1L << (number % 64);
}
}
public static final int solution(final int X, final int[] A) {
if (A.length < ARRAY_SIZE_LOWER || A.length > ARRAY_SIZE_UPPER) {
throw new RuntimeException("Array size out of bounds");
}
Set set = new Set(X);
int ai;
int counter = X;
final int NUMBER_REAL_UPPER = min(NUMBER_UPPER, X);
for (int i = 0 ; i < A.length; i++) {
if ((ai = A[i]) < NUMBER_LOWER || ai > NUMBER_REAL_UPPER) {
throw new RuntimeException("Number out of bounds");
} else if (ai <= X && !set.contains(ai)) {
counter--;
if (counter == 0) {
return i;
}
set.put(ai);
}
}
return -1;
}
private static int min(int x, int y) {
return (x < y ? x : y);
}
}
This is my solution it got me 100/100 and O(N).
public int solution(int X, int[] A) {
Map<Integer, Integer> leaves = new HashMap<>();
for (int i = A.length - 1; i >= 0 ; i--)
{
leaves.put(A[i] - 1, i);
}
return leaves.size() != X ? -1 : Collections.max(leaves.values());
}
This is my solution
public func FrogRiverOne(_ X : Int, _ A : inout [Int]) -> Int {
var B = [Int](repeating: 0, count: X+1)
for i in 0..<A.count {
if B[A[i]] == 0 {
B[A[i]] = i+1
}
}
var time = 0
for i in 1...X {
if( B[i] == 0 ) {
return -1
} else {
time = max(time, B[i])
}
}
return time-1
}
A = [1,2,1,4,2,3,5,4]
print("FrogRiverOne: ", FrogRiverOne(5, &A))
Actually I re-wrote this exercise without seeing my last answer and came up with another solution 100/100 and O(N).
public int solution(int X, int[] A) {
Set<Integer> leaves = new HashSet<>();
for(int i=0; i < A.length; i++) {
leaves.add(A[i]);
if (leaves.contains(X) && leaves.size() == X) return i;
}
return -1;
}
I like this one better because it is even simpler.
This one works good on codality 100% out of 100%. It's very similar to the marker array above but uses a map:
public int solution(int X, int[] A) {
int index = -1;
Map<Integer, Integer> map = new HashMap();
for (int i = 0; i < A.length; i++) {
if (!map.containsKey(A[i])) {
map.put(A[i], A[i]);
X--;
if (X == 0) {index = i;break;}
}
}
return index;
}
%100 with js
function solution(X, A) {
let leafSet = new Set();
for (let i = 0; i < A.length; i += 1) {
if(A[i] <= 0)
continue;
if (A[i] <= X )
leafSet.add(A[i]);
if (leafSet.size == X)
return i;
}
return -1;
}
With JavaScript following solution got 100/100.
Detected time complexity: O(N)
function solution(X, A) {
let leaves = new Set();
for (let i = 0; i < A.length; i++) {
if (A[i] <= X) {
leaves.add(A[i])
if (leaves.size == X) {
return i;
}
}
}
return -1;
}
100% Solution using Javascript.
function solution(X, A) {
if (A.length === 0) return -1
if (A.length < X) return -1
let steps = X
const leaves = {}
for (let i = 0; i < A.length; i++) {
if (!leaves[A[i]]) {
leaves[A[i]] = true
steps--
}
if (steps === 0) {
return i
}
}
return -1
}
C# Solution with 100% score:
using System;
using System.Collections.Generic;
class Solution {
public int solution(int X, int[] A) {
// go through the array
// fill a hashset, until the size of hashset is X
var set = new HashSet<int>();
int i = 0;
foreach (var a in A)
{
if (a <= X)
{
set.Add(a);
}
if (set.Count == X)
{
return i;
}
i++;
}
return -1;
}
}
https://app.codility.com/demo/results/trainingXE7QFJ-TZ7/
I have a very simple solution (100% / 100%) using HashSet. Lots of people check unnecessarily whether the Value is less than or equal to X. This task cannot be otherwise.
public static int solution(int X, int[] A) {
Set<Integer> availableFields = new HashSet<>();
for (int i = 0; i < A.length; i++) {
availableFields.add(A[i]);
if (availableFields.size() == X){
return i;
}
}
return -1;
}
public static int solutions(int X, int[] A) {
Set<Integer> values = new HashSet<Integer>();
for (int i = 0; i < A.length; i++) {
if (values.add(A[i])) {
X--;
}
if (X == 0) {
return i;
}
}
return -1;
}
This is my solution. It uses 3 loops but is constant time and gets 100/100 on codibility.
class FrogLeap
{
internal int solution(int X, int[] A)
{
int result = -1;
long max = -1;
var B = new int[X + 1];
//initialize all entries in B array with -1
for (int i = 0; i <= X; i++)
{
B[i] = -1;
}
//Go through A and update B with the location where that value appeared
for (int i = 0; i < A.Length; i++)
{
if( B[A[i]] ==-1)//only update if still -1
B[A[i]] = i;
}
//start from 1 because 0 is not valid
for (int i = 1; i <= X; i++)
{
if (B[i] == -1)
return -1;
//The maxValue here is the earliest time we can jump over
if (max < B[i])
max = B[i];
}
result = (int)max;
return result;
}
}
Short and sweet C++ code. Gets perfect 100%... Drum roll ...
#include <set>
int solution(int X, vector<int> &A) {
set<int> final;
for(unsigned int i =0; i< A.size(); i++){
final.insert(A[i]);
if(final.size() == X) return i;
}
return -1;
}
Basically, I would like to know if there is a shorter way of printing out the Fibonacci numbers from 0 - 100.
What I have done is probably very basic, but here is the code:
public static void main(String[] args) {
int number[] = new int[100];
number[0] = 0;
number[1] = 1;
int sum1 = number[0] + number[0];
int sum2 = sum1 + number[1];
int sum3 = sum1 + sum2;
int sum4 = sum2 + sum3;
int sum5 = sum3 + sum4;
int sum6 = sum4 + sum5;
System.out.println(sum1);
System.out.println(sum2);
System.out.println(sum3);
System.out.println(sum4);
System.out.println(sum5);
System.out.println(sum6);
}
And I would be doing this up until 100. But I'm sure there is a quicker way of doing this. How?
You can use a loop. This example uses BigInteger as the number quickly become too large for long.
BigInteger a = BigInteger.ZERO, b = BigInteger.ONE;
System.out.println(1);
for (int i = 0; i < 100000; i++) {
BigInteger c = a.add(b);
System.out.println(c);
a = b;
b = c;
}
prints after a couple of seconds finally
420269270299515438 ... many, many digits deleted ... 9669707537501
Note: you don't need to memorize all the previous values, just the last two.
Standard way to do this is by recursion:
Have a look at this simple snippet, which returns the fibonacci number at position a:
public static long fib(int a){
if (a==1||a==2) return 1;
else return fib(a-1)+fib(a-2);
}
google search ftw!... wasn't even a minute searching
int[] fibonacci = new int[25+1];
fibonacci[1] = 1;
fibonacci[2] = 1;
for ( int i = 3; i < fibonacci.length; i++ )
{
fibonacci[i] = fibonacci[i-2] + fibonacci[i-1];
}
Print number < 100 with n=11
public int getFib(int n){
if(n==0) return 0;
else if(n==1) return 1;
else{
int temp=getFib(n-1)+getFib(n-2);
return temp;
}
}
Just in case you don't want to use the recursive way..
Here is the iterative...
public class Fib2 {
public static int fib(int n, int a, int b)
{
if (n==0)
{
System.out.print("1x +");
return a;
}
else
{
System.out.print("2x +");
return fib(n-1,b,a+b);
}
}
public static void main(String arg[])
{
System.out.println(fib(0,1,1));
}
}
Well, 100 isn't one of the Fibonacci numbers, however if we include 144, then this is a nice succinct example:
class F
{
public static void main( final String[] a )
{
int a = 1, b = 1;
for ( ; b < 145; a = b + (b = a) )
System.out.println( b );
}
}
If you have to do this without arrays here is the way to do it. Just change 20 to 100. This is the way I had to do it for an assignment and it is simple to understand for those still learning.
public static void main(String[] args)
{
int number1 = 1;
int number2 = 1;
int count;
int fib;
System.out.print(number1 + " " + number2 + " ");
for(count = 3; count <= 20; count++)
{
fib = number1 + number2;
number1 = number2;
number2 = fib;
System.out.print(fib + " ");
}
}
This question already has answers here:
How to find GCD, LCM on a set of numbers
(13 answers)
Closed 8 years ago.
Here is the question:
"Write a method named gcd that accepts two integers as parameters and returns the greatest common divisor of the two numbers. The greatest common divisor (GCD) of two integers a and b is the largest integer that is a factor of both a and b. The GCD of any number and 1 is 1, and the GCD of any number and 0 is that number.
One efficient way to compute the GCD of two numbers is to use Euclid's algorithm, which states the following:
GCD(A, B) = GCD(B, A % B)
GCD(A, 0) = Absolute value of A"
I'm really confused as to how to solve this problem. I just want some hints and tips as to what I did wrong in the program I have so far. (I have to put in a Scanner, that is my teacher's requirement.)
Don't give me a full code as I kinda want to solve this out myself. Maybe just give me a hint on how I incorporate this formula that you see above. (And if you're wondering why I put in the == 0, it's because I thought that if you have two numbers, say 0 and 90, their GCD would be 0 right??)
Also, my code has to include while loops...I would've preferred if loops...
Thanks in advance! :)
My current program:
public static void main(String[] args) {
Scanner console = new Scanner(System.in);
int a = console.nextInt();
int b = console.nextInt();
gcd (a, b);
}
public static void gcd(int a, int b) {
System.out.print("Type in two numbers and I will print outs its Greatest Common Divisor: ");
int gcdNum1 = console.nextInt();
int gcdNum2 = console.nextInt();
while (gcdNum1 == 0) {
gcdNum1 = 0;
}
while (gcdNum2 > gcdNum1) {
int gcd = gcdNum1 % gcdNum2;
}
System.out.print(gcdNum1 + gcdNum2);
}
}
A recursive method would be:
static int gcd(int a, int b)
{
if(a == 0 || b == 0) return a+b; // base case
return gcd(b,a%b);
}
Using a while loop:
static int gcd(int a, int b)
{
while(a!=0 && b!=0) // until either one of them is 0
{
int c = b;
b = a%b;
a = c;
}
return a+b; // either one is 0, so return the non-zero value
}
When I'm returning a+b, I'm actually returning the non-zero number assuming one of them is 0.
You can also do it in a three line method:
public static int gcd(int x, int y){
return (y == 0) ? x : gcd(y, x % y);
}
Here, if y = 0, x is returned. Otherwise, the gcd method is called again, with different parameter values.
public static int GCD(int x, int y) {
int r;
while (y!=0) {
r = x%y;
x = y;
y = r;
}
return x;
}
import java.util.Scanner;
public class Main {
public static void main(String [] args)
{
Scanner input = new Scanner(System.in);
System.out.println("Please enter the first integer:");
int b = input.nextInt();
System.out.println("Please enter the second integer:");
int d = input.nextInt();
System.out.println("The GCD of " + b + " and " + d + " is " + getGcd(b,d) + ".");
}
public static int getGcd(int b, int d)
{
int gcd = 1;
if(b>d)
{
for(int i = d; i >=1; i--)
{
if(b%i==0 && d%i ==0)
{
return i;
}
}
}
else
{
for(int j = b; j >=1; j--)
{
if(b%j==0 && d% j==0)
{
return j;
}
}
}
return gcd;
}
}
One way to do it is the code below:
int gcd = 0;
while (gcdNum2 !=0 && gcdNum1 != 0 ) {
if(gcdNum1 % gcdNum2 == 0){
gcd = gcdNum2;
}
int aux = gcdNum2;
gcdNum2 = gcdNum1 % gcdNum2;
gcdNum1 = aux;
}
You do not need recursion to do this.
And be careful, it says that when a number is zero, then the GCD is the number that is not zero.
while (gcdNum1 == 0) {
gcdNum1 = 0;
}
You should modify this to fulfill the requirement.
I am not going to tell you how to modify your code entirely, only how to calculate the gcd.
private static void GCD(int a, int b) {
int temp;
// make a greater than b
if (b > a) {
temp = a;
a = b;
b = temp;
}
while (b !=0) {
// gcd of b and a%b
temp = a%b;
// always make a greater than bf
a =b;
b =temp;
}
System.out.println(a);
}
import java.util.Scanner;
class CalculateGCD
{
public static int calGCD(int a, int b)
{
int c=0,d=0;
if(a>b){c=b;}
else{c=a;}
for(int i=c; i>0; i--)
{
if(((a%i)+(b%i))==0)
{
d=i;
break;
}
}
return d;
}
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter the nos whose GCD is to be calculated:");
int a=sc.nextInt();
int b=sc.nextInt();
System.out.println(calGCD(a,b));
}
}
Now, I just started programing about a week ago, so nothing fancy, but I had this as a problem and came up with this, which may be easier for people who are just getting into programing to understand. It uses Euclid's method like in previous examples.
public class GCD {
public static void main(String[] args){
int x = Math.max(Integer.parseInt(args[0]),Integer.parseInt(args[1]));
int y = Math.min(Integer.parseInt(args[0]),Integer.parseInt(args[1]));
for (int r = x % y; r != 0; r = x % y){
x = y;
y = r;
}
System.out.println(y);
}
}