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I'm trying to figure out the time complexity of this implementation of classic N-queens problem on geeksforgeeks. The goal is to find just one such non-attacking solution(as opposed to finding all of them). The relevant code is briefed below.

bool isSafe(int board[N][N], int row, int col)
{
    int i, j;

    /* Check this row on left side */
    for (i = 0; i < col; i++)
        if (board[row][i])
            return false;

    /* Check upper diagonal on left side */
    for (i=row, j=col; i>=0 && j>=0; i--, j--)
        if (board[i][j])
            return false;

    /* Check lower diagonal on left side */
    for (i=row, j=col; j>=0 && i<N; i++, j--)
        if (board[i][j])
            return false;

    return true;
}

/* A recursive utility function to solve N
   Queen problem */
bool solveNQUtil(int board[N][N], int col)
{
    /* base case: If all queens are placed
      then return true */
    if (col >= N)
        return true;

    /* Consider this column and try placing
       this queen in all rows one by one */
    for (int i = 0; i < N; i++)
    {
        /* Check if queen can be placed on
          board[i][col] */
        if ( isSafe(board, i, col) )
        {
            /* Place this queen in board[i][col] */
            board[i][col] = 1;

            /* recur to place rest of the queens */
            if ( solveNQUtil(board, col + 1) )
                return true;

            /* If placing queen in board[i][col]
               doesn't lead to a solution, then
               remove queen from board[i][col] */
            board[i][col] = 0; // BACKTRACK
        }
    }

     /* If queen can not be place in any row in
        this colum col  then return false */
    return false;
}

/* This function solves the N Queen problem using
   Backtracking. It mainly uses solveNQUtil() to
   solve the problem. It returns false if queens
   cannot be placed, otherwise return true and
   prints placement of queens in the form of 1s.
   Please note that there may be more than one
   solutions, this function prints one  of the
   feasible solutions.*/
bool solveNQ()
{
    int board[N][N] = { {0, 0, 0, 0},
        {0, 0, 0, 0},
        {0, 0, 0, 0},
        {0, 0, 0, 0}
    };

    if ( solveNQUtil(board, 0) == false )
    {
      printf("Solution does not exist");
      return false;
    }

    printSolution(board);
    return true;
}

From what I can tell, the recurrence is T(n) = n*T(n-1) + N*n, which leads to O((n+1)!). When one more recursive call is made, at least one more isSafe() should return false. So the number of recursive calls decreases by at least 1 each time. And for the overhead in isSafe(), since col rows has been filled already, there are at most min(col, N-col+1) iterations of for (i = 0; i < col; i++). But the for loop in solveNQUtil() runs a fixed number of N. That's why N*n exists.

But many people argue that N-queen runs O(n!) with recurrence T(n)= n*(T(n-1) + O(n)), such as here and comment on geeksforgeeks. So I start doubting myself.. surely O((n+1)!) != O(n!).

Could anyone here throw some light on this? Have been struggling with it for a while.

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    $\begingroup$ The difference between $n!$ and $(n+1)!$ is pretty small for such large time complexities. I would just ignore this small point. When people say $O(n!)$ they don't necessarily exclude $O((n+1)!)$. $\endgroup$ – Yuval Filmus Nov 20 '15 at 15:33
  • $\begingroup$ Can you please explain nT(n-1) term in the complexity relation, from the code it seems it should be NT(n-1)? Sorry I'm not able to add this question as a comment as I dont have enough reputation in stackoverflow to comment on answers(not yet), so that's why I'm posting this here. $\endgroup$ – Suthari Sandeep Jan 14 at 9:29
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Your analysis would be reasonable for rooks, but not for queens. For example, in the second row there are not n-1 possible positions, but n-2 in two cases and n-3 in most cases. Now big-O provides an upper bound, but it is an upper bound that vastly overestimates the effort needed.

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