What is wrong in this code

#1 
public static void main(String[] args) {

	Scanner sc=new Scanner(System.in);
	ArrayList<Integer> primeNumbers = new ArrayList<>();
	primeNumbers = PrimeUpto10000(primeNumbers);

	
	int t=sc.nextInt();
	ArrayList<Integer> al = new ArrayList<>();
	
	for(int i=1 ; i<=t ; i++)
	{
		int n=sc.nextInt();
		int temp=n;
		
		// if we have already solve for (n) in the previous test cases , then we wont solve again
		if(n>al.size())
		{
			
			int start;

			// for i>1
			if(al.size()!=0) start=al.get(al.size()-1) + 1;
			// for i==1
			else start = 1;
			
			// how many more numbers to add
			int size=al.size();
			
			// (start) will start from next number of last checked number
			// temp-size will tell how many more number to add
			for(int j=start ; j < start + (temp-size)  ;  j++)
			{
				boolean condition = IsItInCondition(j , primeNumbers);	
				
				// if number i.e. (j) is in conditions , then add this number
				if(condition)
				{
					al.add(j);
				}
				// else increase limit of (j) so that we can add minimum reqd elements in arraylist
				else
				{
					temp++;
				}
			}
		}
		
		// sysout number
		System.out.println(al.get(n-1));
	}
	
	
	
}

// checking number is acceptable or not
static boolean IsItInCondition(int number , ArrayList<Integer> primeNumbers)
{
	boolean temp=true;
	for(int i:primeNumbers)
	{
		// if prime number > number , then break
		if(i>number) break;
		
		// if this number is divisible by prime number other than 2,3,5
		// then this number is not acceptable according to questions condition
		if(number%i==0)
			temp=false;
	}
	
	return temp;
}

// finding all prime numbers upto 10000
static ArrayList<Integer> PrimeUpto10000(ArrayList<Integer> primeNumbers)
{
	for(int i=1 ; i<=10000 ; i++)
	{
		boolean prime=true;
		for(int j=2 ; j<=Math.sqrt(i) ; j++)
		{
			if(i%j==0)
			{
				prime=false;
				break;
			}
		}
		
		if(prime)
			primeNumbers.add(i);
	}
	
	// remove 1 from list
	primeNumbers.remove(0);
	// remove 2 from list
	primeNumbers.remove(0);
	// remove 3 from list
	primeNumbers.remove(0);
	// remove 5 from list
	primeNumbers.remove(0);

	return primeNumbers;
}
#2

Hey @Himanshu-Jhawar-2273952536067590
Try for this
1
176
correct output : 10125
your code gives : 10007 // ugly nhi h
logic :
Ugly number(i) = minimum of (2 * ugly number(f2), 3* ugly number(f3), 5*ugly number(f5))

Where, f2, f3, f5 are counters for 2, 3, 5 respectively which stores the index for f2th ugly number, f3th ugly number, f5th ugly number.

Then we will store our answer.

Algorithm:

Initializing f2, f3, f5 to maintain f2th ugly number, f3th ugly number and f5th ugly number to 1.
Creating a DP matrix in which ith cell represent ith ugly number.
Start filling the dp matrix from i = 2.
Ugly number(i) = minimum of (2 * ugly number(f2), 3* ugly number(f3), 5*ugly number(f5)).
Finding the answer using above relation and storing it at ith index.
Now, the time complexity O(N).