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N Queen in O(n) space

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Given n, of a n x n chessboard, find the proper placement of queens on chessboard.
Previous Approach : N Queen

Algorithm : 

Place(k, i)
// Returns true if a queen can be placed
// in kth row and ith column. Otherwise it
// returns false. X[] is a global array
// whose first (k-1) values have been set.
// Abs( ) returns absolute value of r
{
   for j := 1 to k-1 do

        // Two in the same column
        // or in the same diagonal
        if ((x[j] == i)  or
            (abs(x[j] – i) = Abs(j – k)))
           then return false;

   return true;
}

Algorithm nQueens(k, n) : 

// Using backtracking, this procedure prints all 
// possible placements of n queens on an n×n 
// chessboard so that they are nonattacking.
{
      for i:= 1 to n do
      {
         if Place(k, i) then
         {
             x[k] = i;
             if (k == n)
                write (x[1:n]);
             else 
               NQueens(k+1, n);
         }
      }
} 

Implementation:

C++

// CPP code to for n Queen placement
#include <bits/stdc++.h>
 
#define breakLine cout << "\n---------------------------------\n";
#define MAX 10
 
using namespace std;
 
int arr[MAX], no;
 
void nQueens(int k, int n);
bool canPlace(int k, int i);
void display(int n);
 
// Function to check queens placement
void nQueens(int k, int n){
 
    for (int i = 1;i <= n;i++){
        if (canPlace(k, i)){
            arr[k] = i;
            if (k == n)
                display(n);
            else
                nQueens(k + 1, n);
        }
    }
}
 
// Helper Function to check if queen can be placed
bool canPlace(int k, int i){
    for (int j = 1;j <= k - 1;j++){
        if (arr[j] == i ||
            (abs(arr[j] - i) == abs(j - k)))
           return false;
    }
    return true;
}
 
// Function to display placed queen
void display(int n){
    breakLine
    cout << "Arrangement No. " << ++no;
    breakLine
 
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= n; j++){
            if (arr[i] != j)
                cout << "\t_";
            else
                cout << "\tQ";
        }
        cout << endl;
    }
 
    breakLine
}
 
// Driver Code
int main(){
    int n = 4;   
    nQueens(1, n);   
    return 0;
}

                    

Java

// Java code to for n Queen placement
class GfG
{
 
    static void breakLine()
    {
        System.out.print("\n---------------------------------\n");
    }
    static int MAX = 10;
 
    static int arr[] = new int[MAX], no;
 
    // Function to check queens placement
    static void nQueens(int k, int n)
    {
 
        for (int i = 1; i <= n; i++)
        {
            if (canPlace(k, i))
            {
                arr[k] = i;
                if (k == n)
                {
                    display(n);
                }
                else
                {
                    nQueens(k + 1, n);
                }
            }
        }
    }
 
    // Helper Function to check if queen can be placed
    static boolean canPlace(int k, int i)
    {
        for (int j = 1; j <= k - 1; j++)
        {
            if (arr[j] == i ||
                (Math.abs(arr[j] - i) == Math.abs(j - k)))
            {
                return false;
            }
        }
        return true;
    }
 
    // Function to display placed queen
    static void display(int n)
    {
        breakLine();
        System.out.print("Arrangement No. " + ++no);
        breakLine();
 
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (arr[i] != j)
                {
                    System.out.print("\t_");
                }
                else
                {
                    System.out.print("\tQ");
                }
            }
            System.out.println("");
        }
 
        breakLine();
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 4;
        nQueens(1, n);
    }
}
 
// This code is contributed by 29AjayKumar

                    

Python3

# Python code to for n Queen placement
class GfG:
    def __init__(self):
        self.MAX = 10
        self.arr = [0] * self.MAX
        self.no = 0
 
    def breakLine(self):
        print("\n------------------------------------------------")
 
    def canPlace(self, k, i):
         
        # Helper Function to check
        # if queen can be placed
        for j in range(1, k):
            if (self.arr[j] == i or
               (abs(self.arr[j] - i) == abs(j - k))):
                return False
        return True
 
    def display(self, n):
         
        # Function to display placed queen
        self.breakLine()
        self.no += 1
        print("Arrangement No.", self.no, end = " ")
        self.breakLine()
 
        for i in range(1, n + 1):
            for j in range(1, n + 1):
                if self.arr[i] != j:
                    print("\t_", end = " ")
                else:
                    print("\tQ", end = " ")
            print()
 
        self.breakLine()
 
    def nQueens(self, k, n):
         
        # Function to check queens placement
        for i in range(1, n + 1):
            if self.canPlace(k, i):
                self.arr[k] = i
                if k == n:
                    self.display(n)
                else:
                    self.nQueens(k + 1, n)
 
# Driver Code
if __name__ == '__main__':
    n = 4
    obj = GfG()
    obj.nQueens(1, n)
 
# This code is contributed by vibhu4agarwal

                    

C#

// C# code to for n Queen placement
using System;
 
class GfG
{
 
    static void breakLine()
    {
        Console.Write("\n---------------------------------\n");
    }
    static int MAX = 10;
 
    static int []arr = new int[MAX];
    static int no;
 
    // Function to check queens placement
    static void nQueens(int k, int n)
    {
 
        for (int i = 1; i <= n; i++)
        {
            if (canPlace(k, i))
            {
                arr[k] = i;
                if (k == n)
                {
                    display(n);
                }
                else
                {
                    nQueens(k + 1, n);
                }
            }
        }
    }
 
    // Helper Function to check if queen can be placed
    static bool canPlace(int k, int i)
    {
        for (int j = 1; j <= k - 1; j++)
        {
            if (arr[j] == i ||
                (Math.Abs(arr[j] - i) == Math.Abs(j - k)))
            {
                return false;
            }
        }
        return true;
    }
 
    // Function to display placed queen
    static void display(int n)
    {
        breakLine();
        Console.Write("Arrangement No. " + ++no);
        breakLine();
 
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (arr[i] != j)
                {
                    Console.Write("\t_");
                }
                else
                {
                    Console.Write("\tQ");
                }
            }
            Console.WriteLine("");
        }
 
        breakLine();
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 4;
        nQueens(1, n);
    }
}
 
// This code contributed by Rajput-Ji

                    

Javascript

<script>
 
// JavaScript program to for n Queen placement
 
    function breakLine()
    {
        document.write("<br />");
        document.write("---------------------------------");
        document.write("<br />");
    }
    let MAX = 10;
   
    let arr = [];
    let no = 0;
   
    // Function to check queens placement
    function nQueens(k, n)
    {
   
        for (let i = 1; i <= n; i++)
        {
            if (canPlace(k, i))
            {
                arr[k] = i;
                if (k == n)
                {
                    display(n);
                }
                else
                {
                    nQueens(k + 1, n);
                }
            }
        }
    }
   
    // Helper Function to check if queen can be placed
    function canPlace(k, i)
    {
        for (let j = 1; j <= k - 1; j++)
        {
            if (arr[j] == i ||
                (Math.abs(arr[j] - i) == Math.abs(j - k)))
            {
                return false;
            }
        }
        return true;
    }
   
    // Function to display placed queen
    function display(n)
    {
        breakLine();
        document.write("Arrangement No. " + ++no);
        breakLine();
   
        for (let i = 1; i <= n; i++)
        {
            for (let j = 1; j <= n; j++)
            {
                if (arr[i] != j)
                {
                    document.write("\t_");
                }
                else
                {
                    document.write("\tQ");
                }
            }
            document.write("<br/>");
        }
   
        breakLine();
    }
  
// Driver code
 
        let n = 4;
        nQueens(1, n);
 
</script>

                    

Output: 
---------------------------------
Arrangement No. 1
---------------------------------
    _    Q    _    _
    _    _    _    Q
    Q    _    _    _
    _    _    Q    _

---------------------------------

---------------------------------
Arrangement No. 2
---------------------------------
    _    _    Q    _
    Q    _    _    _
    _    _    _    Q
    _    Q    _    _

---------------------------------

 

The time and space complexity of the given code for the N-Queens problem can be analyzed as follows:

Time Complexity:
The algorithm uses backtracking to generate all possible solutions for placing N queens on an N x N chessboard. The backtracking algorithm recursively explores all possible solutions by checking whether a queen can be placed in each column of the current row. The time complexity of the algorithm can be expressed as O(N!) because in the worst case scenario, every queen must be tried in every column of every row.

Space Complexity:
The space complexity of the algorithm depends on the size of the input problem, which is N. In the given code, an array ‘arr’ of size N is used to store the column index of the queen in each row. Additionally, a variable ‘no’ is used to count the number of valid solutions found. Therefore, the space complexity of the algorithm can be expressed as O(N).

In summary, the time complexity of the algorithm is O(N!), and the space complexity is O(N).



Last Updated : 30 Apr, 2023
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