Passed all but one test case:

// Complete the solve function below.
static String solve(int d, int k) {
        //get integer format of radius
        int r = (int)Math.sqrt(d); 

        int count = 0;

        //loop for all integer values on X axis
        for(int i = -r; i<r; i++){

                //calculate the float/double value of Y
                // Let's say, it could be 2.0 or 2.3
                double y = Math.sqrt(d - i*i);

            // convert to INT so it will become either 2.0 in above cases
                int intY = (int)y;

                //now compare the int version with original version.
                //if they are same, then Y is also integer and hence we got our point on
                //the circle. Count is incremented by 2 because, this new point is applicable both above and below x axis.
                if((float)intY == y ){
                        count+=2;
                }

        }  
        if(k >= count){
                return "possible";
        } else {
                return "impossible";

    }  
}

some tese case fail this code .....can you tell why that.....??

int solve(long int d,long int k) { int r=sqrt(d); int t,p,ans=4; //char *possible,*impossible;

if((r==1 && k>=4) || (r==2 && k>=4))
        return 1;


if(r>2)
{

        p=2,t=2;
        while((r-2)>2 && (r-t)>=2)
        {
                p=floor((r-t)/2);
                t=(t+p);
                ans+=4;
        }


}
 if(ans<=k)
            return 1;
        else
            return 0;        

}

//THis code clear all test cases

static String solve(int d, int k)

{

    long count=0;

    for(long i=0;i<Math.sqrt(d);i++)

    {

        long p=(long)Math.sqrt((long)d-i*i);

        if((p*p+i*i)==d)

        {

            count+=4;

        }

    }


    if(k>=count)

    {

        return "possible";

    }

    return "impossible";

}

can anyone help me out... i get TLE on testcase 2. Is my code efficient?

import math
t=int(input())
for _ in range(t):
    d,k=[int(x) for x in input().split()]
    ans=0
    r=math.ceil(d**.5)
    for i in range(0,r):
        x=(d-i**2)**.5
        #y=int(math.sqrt(d-i**2))
        if x==int(x):
            ans+=4
            #print(i,y)
    if k<ans:
        print("impossible")
    else:
        print("possible")

Could be solved easily with https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares and where = number of decompositions of in two squares. The French version of the above page gives this formula and the meaning of and .