Don’t be discouraged my friend, starts are always frustating. I went through your code, but logic seems very confusing, that’s why it must be failing on strong test cases.
In these scenarios you need to find out which test cases ur solution will fail.
since range of n is till 10^4 , we need to solve this in O( n log n )
first step u did correct by sorting the array.
second step is where optimization is need in searching. and most optimized search algorithm is ?
Binary Search
so iterate through sorted array
for element a , check if a - k exists in array using binary search of log( n )
so for n elements, it will take n * log n time in total
public static void main (String[]args) throws java.lang.Exception
{
// your code goes here
Scanner in = new Scanner (System.in);
int t = in.nextInt ();
for (int i = 0; i < t; i++)
{
int n=in.nextInt();
int k=in.nextInt();
int[] a=new int[n];
for (int j=0;j<n;j++)
{
a[j]=in.nextInt();
}
Arrays.sort(a);
int r=-1;
for (int j=0;j<n;j++)
{
if (a[j]>=k) r=j-1;
}
if (r==-1) r=n-1;
int req=k-a[r];
Boolean con=false;
int j=0;
while (!con && a[r]>=k/2)
{
if (a[j]==req)
{
con=true;
}
else if (j<r-1)
{
j++;
continue;
}
else if (r>1)
{
j=0;
r--;
req=k-a[r];
continue;
}
break;
}
if (con) System.out.println("Yes");
else System.out.println("No");
}
}
This is how I did it in Java. My logic is that I find the largest number smaller than k. Then that number subtracted from k will be the other number that I require, if that number exists in the array, then my condition is satisfied and if not then I check for the number smaller than the one I used before
