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Jump Game Correction

I'm not sure how to improve my solution for the jump game question (https://leetcode.com/problems/jump-game/description/) .It seems logically correct but I encounter a java.lang.Stackoverflow error for large inputs like [nums=[1,1,1,1,1,1,1,........]]
I know that I'm overflowing due to too many recursive calls , how can i optimize it ?
class Solution {
public boolean canJump(int[] nums) {
int []f = new int[1];
f[0] = 0;
return help(nums,0,f);
}
public static boolean help(int[]nums, int c, int[] f) {
if(c==nums.length-1) {
f[0] =1;
return true;
}
if(f[0]==1) return true;
int val = nums[c];
for(int i = 1; i <= val; i++) {
if(f[0]==1) return true;
return help(nums,(c + i), f);
}
return false;
}
}

This can be solved in T:O(N) with greedy algorithm:
class Solution {
public int jump(int[] nums) {
int res=0, l=0, r=0;
while(r<(nums.length-1)){
int farthest=0;
for (int i=l;i<r+1;i++){
farthest=Integer.max(farthest,i+nums[i]);
}
l=r+1;
r=farthest;
res++;
}
return res;
}
}
Explanation
nums = [2,3,1,1,4]
you jump 2 steps. you get the index 2.so you are jumping elements [3,1]. Nov you loop through [3,1] and check which one will take you to the further. In this case it is 3. Then update the variables. Increase res=1
From element 3 which is index-1. you jump 3 steps, jumping [1,1,4]. Now loop through [1,1,4] find the farthest. which is 4. increase res=2. Since out of while loop, return res=2.
It seems to have two nested loops, but if you look carefully, total iteration is N.

Using IsPrime method in Java [closed]

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This is my code to use IsPrime method to determine total number of primes between 0 and 1000 and print the total number of primes at last. Can anyone tell what's wrong with the code.
public static void main(String[] args) {
int z=0;
// z is the variable that holds total number of primes
//n is divisor
//i is dividend
if (isPrime(i)) {
z++;
}
System.out.print(z+"\n");
}
public static boolean isPrime(int n){
{
for(i=0; i<1000; i++)
{
for(n=0; n<i; n++)
if(i%n==0)
return false;
else
return true;
}
}
}
Thanks in advance

Formatting it might help you discover the error.
I see a few things that's wrong with your code:
From what I can see, you have an extra open curly bracket in your isPrime method.
i isn't declared in your main method.
You need to wrap your if(isPrime(i)) statement inside a for loop that goes from 0 to 1000. Like the following:
for (int i = 0; i <= 1000; i++) {
if (isPrime(i))
z++;
}
That way, it will be actually checking all the prime numbers from 0 to 1000
For good coding practices, I would name your z variable to be something like counter so that it's clear what that variable is supposed to be doing. i in for-loop is okay since that's a common way to index through the loop.
You can also use several tactics to optimize your code. You can use Math.sqrt() function, as well as start your for loop from 3 and go up by increment of 2 (since any even number will be dividable by 2) and initialize your counter from 1 since 2 will already be a prime number.

public static boolean isPrime(int n){
int factors = 0;
for(int i = 1; i <= n; i++){
if(n % i == 0) // ensure that you mod n not i
factors++;
}
// if factors count is equals to 2 then it is prime number else it's not prime number
if(factors == 2)
return true;
else
return false;
}
Check this modified code once for your reference.

I think you had a few things turned around.
Why not loop from 1 to 1000 in your primary function and then use the isPrime function to determine if each number is prime.
In the isPrime function, you count from 2 to 1/2 the value of the number and do the divisions to determine if it is prime. Return False if it is divisible.
public static void main(String[] args) {
int z=0;
for (i=1;i<=1000;i++) {
if (isPrime(i))
{
z++;
}
}
System.out.print(z+"\n");
}
public static boolean isPrime(int n){
for(i=2; i<=n/2; i++)
{
if(n%i==0) return false;
}
return true;
}

This will count the number of primes based on this link and your original answer...
public static void main(String[] args) {
int isPrimeCount = 0;
for(i=0; i<1000; i++)
{
if(Check_Prime(i))
{
isPrimeCount++;
}
System.out.println(isPrimeCount);
}
}
private static boolean Check_Prime(int number) {
int i;
for (i = 2; i <= number - 1; i++)
{
if (number % i == 0)
{
return false;
}
}
if (i == number)
{
return true;
}
return false;
}

You need to define the variable "i" before you pass it to the isPrime() method in main(). It seems as though whoever wrote the code did not fully understand what a prime number was. According to wikipedia "A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. " With this in mind you need to make sure that the value you pass to the isPrime() method is greater than 1. After look at your code I made some changes. I made the isPrime() method return false if the input value is <=1. Also, I made the isPrime() method return true if the input value is 2. I made other changes in the code that would make the for statement operate n-1 times because that is all that is needed to find out if the number is prime because all numbers are divisible by themselves. Also the for statement starts at the value 2. Your if-then-else return statements within the for loop is illogical because it will return a value without going through the entire loop. You did not need the inner for loop.
Here's a link on prime numbers
Here is the new code:
public class AreaComparison {
/**
* Starts the program.
*
* #param command line arguments
*/
public static void main(String[] args) {
int z = 0;
int i = 2;
// z is the variable that holds total number of primes
//n is divisor
//i is dividend
if (isPrime(i)) {
z++;
}
System.out.print(z + "\n");
}
public static boolean isPrime(int n) {
if (n <= 1) {
return false;
}
if(n == 2){
return true;
}
for (int i = 2; i < n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
}

Translate Pseudocode and get the result

Can you help translating this Pseudocode into java and get the result
as I tried but didn't get the results and I'm starting to learn Java.
num=2
create list called list
while(true){
empty list
bool = false
for i=1 to num-1
if(num mod i==0){
add i to list
if((i &(bitwise)1)==1)
bool=!bool
}
if sum(list)==num && bool//logical and
return num; //Found the number we seek
num++
}
This is Java translating but I'm not sure as it didn't get the result numbers
/* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.ArrayList;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
//set a number
int num=2;
//make list called list
ArrayList<Integer> list = new ArrayList<Integer>();
//run this while true, but just saying true makes an infinite loop
boolean bool = true;
while(bool){
//clear the list
list.clear();
//set
bool = false;
for (int i= 1; i < num-1; i = i + 1){
//If number modulo i is 0
if((num % i) ==0){
list.add(i);
//If bitwise and of i and 1 is true
//This is doing things with the underlying binary
if((i & 1)==1)
//Reverse the valuse of bool
bool = !bool;
}
int sum = 0;
//for each element in the list
for(int x: list){
//add it to sum
sum += sum;
}
//execute is sum and num are the same and bool is true
if ((sum == num )&& bool){}
return num;
//This does nothing as it is never reached. return ends your execution
num++;
}
}
}
}

This sure looks like some assignment. Anyways, here's the translation of the pseudocode:
public int code ()
{
int num = 2;
List<Integer> list = new ArrayList<> ();
while (true)
{
list.clear ();
boolean bool = false;
for (int i = 1; i < num; ++i)
{
if (num % i == 0)
{
list.add (i);
if ((i & 1) == 1)
{
bool = !bool;
}
}
}
if (sum (list) == num && bool)
{
return num;
}
num++;
}
}
private int sum (List<Integer> list)
{
return list.stream ().mapToInt (i -> i).sum ();
}
The code is sure taking time to print out any result.
Try calling the code method and you'll see.
Apart from that, what number this code is trying to find, you may ask?
Well, through logical deduction, I can tell that this code is trying to find the smallest Perfect Number that has odd number of odd factors. Try thinking about it.

It will never get the result, most probably at least.
As someone already said it is trying to find "the smallest Perfect Number that has odd number of odd factors", and any number that have in its sum odd number of odd numbers is odd. So its searching for odd perfect number which is yet to proven if it exist or not that is why it will never give you result fast enough. It could take million years it could take forever. The question was supposed to be solved by logic.
BTW: Actually I am the one who made this pseudocode, better luck solving the code next weak. (:

How to Check Array for 3 Numbers that Add to a Specific value in JAVA?

Alright guys, I've been stuck on this problem for a while now and have not been able to get past it. This is for Java. I'd appreciate any help at this point. Here are the details: Please note, we must do this in O(n) running time. We are given an array of numbers and must go through it to determine if there are any 3 numbers that sum to a specific number. HOWEVER, we are allowed to reuse any number in the array up to 3 times because we need a total of 3 numbers. We also have to output which 3 numbers gave the sum. Returning true or false.
Below is what I've got:
Do you guys have any suggestions?

You can make a for loop inside of a for loop inside of a for loop. This is for school, so I wont give you the code, but I'll give you the pseudo.
Edit: missed the O(n) part, sorry. This way should work.
public static void main(String[] args)
{
int[] test = {1,8,2,3,11,4};
System.out.println(threeSumTo(test, 6));
}
//check if 3 numbs in an array add up to int x
public static boolean threeSumTo(int[] array, int x)
{
//loop through the array
for (int i = 0; i < array.length; i++) {
boolean result = twoSumTo(array, x - array[i], i);
if (result) {
return result;
}
}
return false;
}
private static boolean twoSumTo(int[] array, int x, int low) {
int high = array.length - 1;
while (low < high) {
if (array[low] + array[high] == x) {
return true;
}
if (array[low] + array[high] > x) {
high--;
} else {
low++;
}
}
return false;
}
}

This seems to be a variation of the 3SUM problem and should obey the same restrictions.
Computing the 3SUM problem in less than O(n^2) is still a unsvoled problem.
Did your teacher ask a trick question or is that some kind of competition?

This is called a 3 sum problem and solving this problem in O(N) is impossible till now. The best you can do is O(N^2).
Check this article out.

Prime Tester for speed

I was given a homework assignment in Java to create classes that find Prime number and etc (you will see in code better).
My code:
class Primes {
public static boolean IsPrime(long num) {
if (num%2==0){
return false;
}
for (int i=3; i*i<=num;i+=2) {
if (num%i==0) {
return false;
}
}
return true;
} // End boolen IsPrime
public static int[] primes(int min, int max){
int counter=0;
int arcount=0;
for (int i=min;i<max;i++){
if (IsPrime(i)){
counter++;
}
}
int [] arr= new int[counter];
for (int i=min;i<max;i++){
if (IsPrime(i)){
arr[arcount]=i;
arcount++;
}
}
return arr;
} // End Primes
public static String tostring (int [] arr){
String ans="";
for (int i=0; i<arr.length;i++){
ans= ans+arr[i]+ " ";
}
return ans;
}
public static int closestPrime(long num){
long e = 0 , d = 0 , f = num;
for (int i = 2; i <= num + 1 ; i++){
if ((num + 1) % i == 0){
if ((num + 1) % i == 0 && (num + 1) == i){
d = num + 1;
break;
}
num++;
i = 1;
}
}
num = f;
for (int i = 2; i < num; i++){
if ((num - 1) % i == 0){
if ((num - 1) % i == 0 && (num - 1) == i){
e = num - 1;
break;
}
num--;
i = 1;
}
}
num = f;
if (d - num < num - e) System.out.println("Closest Prime: "+d);
else System.out.println("Closest Prime: "+e);
return (int) num;
} // End closestPrime
}//end class
The goal of my code is to be faster (and correct). I'm having difficulties achieving this. Suggestions?
**New code:
class Primes {
public static boolean IsPrime(int num) {
if (num==1){
return false;
}
for (int i=2; i<Math.sqrt(num);i++) {
if (num%i==0) {
return false;
}
}
return true;
}
// End boolen IsPrime
public static int[] primes(int min, int max){
int size=0;
int [] arrtemp= new int[max-min];
for (int i=min;i<max;i++){
if (IsPrime(i)){
arrtemp[size]=i;
size++;
}
}
int [] arr= new int[size];
for (int i=0;i<size;i++){
arr[i]=arrtemp[i];
}
return arr;
}
public static String tostring (int [] arr){
String ans="";
for (int i=0; i<arr.length;i++){
ans= ans+arr[i]+ " ";
}
return ans;
}
public static int closestPrime(int num) {
int count=1;
for (int i=num;;i++){
int plus=num+count, minus=num-count;
if (IsPrime(minus)){
return minus;
}
if (IsPrime(plus)) {
return plus;
}
count=count+1;
}
} // End closestPrime
}//end class
I did try to make it a bit better. what do you think, it can be improved more? (the speed test is still high...)

In your primes function you:
Check if the current number is divisible by two
Check to see if it's prime
Create an array to put your output in.
Check every number in the range again for primality before putting it in your array.
The problem is in the last step. By double-checking whether each number is prime, you're duplicating your most expensive operations.
You could use a dynamic data structure and add prime numbers to it as you find them. That way you only need to check once.
Alternatively, you could create a boolean array which is the size of your input range. Then as you find primes, set the corresponding array value to true.
UPDATE:
There are still a number of improvements you can make, but some will require more work than others to implement. Look at the specifics of your test and see what fits your needs.
Low-hanging fruit:
Use an ArrayList to collect primes as you find them in primes, as opposed to looping over the values twice.
In closestPrime, you're checking every single value on either side of num: half of these are even, thus not prime. You could adapt your code to check only odd numbers for primality.
Trickier to implement:
Try a more advanced algorithm for IsPrime: check out the Sieve of Eratosthenes
Above all, you should spend some time figuring out exactly where the bottlenecks are in your code. Oftentimes performance problems are caused by code we thought was perfectly fine. You might consider looking into the code-profiling options available in your development environment.

You make quite a few calls to isPrime(), each of which is very expensive. Your first step should be to minimize the number of times you do that, since the result doesn't change for any given number, there's no point calling more than once. You can do this with memoization by storing the values once they're computed:
ArrayList<Integer> memoList = new ArrayList<Integer>();
for(int i = 0; i < max; i++) {
if(isPrime(i)) {
memoList.add(i);
}
}
Now memoList holds all the primes you need, up to max, and you can loop over them rapidly without needing to recompute them every time.
Secondly, you can improve your isPrime() method. Your solution loops over every odd number from 3 to sqrt(n), but why not just loop over the primes, now that we know them?
public static boolean IsPrime(long num) {
for(int p : memoList) {
if(num % p == 0) {
return false;
}
}
return true;
}
These changes should dramatically improve how quickly your code runs, but there has been a lot of research into even more efficient ways of calculating primes. The Wikipedia page on Prime Numbers has some very good information on further tactics (prime sieves, in particular) you can experiment with.
Remember that as this is homework you should be sure to cite this page when you turn it in. You're welcome to use and expand upon this code, but not citing this question and any other resources you use is plagiarism.

I see a couple problems with your answer. First, 2 is a prime number. Your first conditional in IsPrime breaks this. Second, in your primes method, you are cycling through all number from min to max. You can safely ignore all negative numbers and all even numbers (as you do in IsPrime). It would make more sense to combine these two methods and save all the extra cycles.