Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.

public class Solution {
    int rec = 0;
    public int totalNQueens(int n) {
        
        if(n<=0)
          return rec;
        int[] num = new int[n];
        for(int i=0;i<n;i++)
           num[i] = i;
        DFS(0,n,num);
        return rec;
    }

    public void DFS(int start,int N,int[] num)
    {
        if(start == N)
        {
            if(isValid(num))
              rec += 1;
           // System.out.println(rec);
        }
        else
        {
            for(int i=start;i<N;i++)
            {
                 int temp1=num[start];
                 num[start]=num[i];
                 num[i]=temp1;
                 
            // if(isValid(num,i));
                 DFS(start+1,N,num);
                 
                 int temp2=num[start];
                 num[start]=num[i];
                 num[i]=temp2;
            }
        }
    }
   
    public boolean isValid(int[] num)
    {
        for(int i=0;i<num.length;i++)
         for(int j=i+1;j<num.length;j++)
            if(Math.abs(num[i]-num[j])==Math.abs(i-j))
               return false;
         return true;
    }
    
}