Finding Nth root of M using binary search in Java

Question

I have written a code in java to find nth root of m, where n and m both are integers. 1 <= n <= 30 , 1 <= m <= 10^9

If the root is not integer then I have to return -1.

My code works for the smaller values of m and n. But it fails for n=6, m=4096 . My code returns -1, whereas the correct answer is 4. I think it happens because of overflow in the isProdGreater() method. How can I prevent this overflow?

  public int NthRoot(int n, int m)
    {
        // code here
        int l = 1, h = m;
        
        int mid;
        while(l<h){
            mid = l + (h-l)/2;
            //long prod = pow(mid,n);
            if(isProdGreater(mid, n, m)){
                h = mid;
            }else{
                l = mid+1;
            }
        }
        
        if(l*l == m) return l;
        return -1;
    }
    
    boolean isProdGreater(int x, int n, int target){
        long a = 1;
        for(int i=0; i<n; i++){
            
            if(a*x >= target) return true;
            a = a*x;
        }
        return false;
    }

Answer 1

The problem is your last checking

if(l*l == m) return l;

This is checking the square, not the nth power. You probably want

if(Math.pow(l,n) == m) return l;