Given an integer N, the task is to print all the unique combinations of putting N pieces in an NxN board.
Note: Print (“*”) for pieces and (“-“) for an empty space.
Example:
Input: N = 2
Output:
* *
– –* –
* –* –
– *– *
* –– *
– *– –
* *
Explanation: The total number of empty spaces are 2*2=4 and the pieces to be set is 2 so there are 4C2 combinations ((4!/(2!*2!))=6) possible which is represented above.Input: N = 1
Output: *
Approach: This problem can be solved by using recursion to generate all possible solutions. Now, follow the steps below to solve this problem:
- Create a function named allCombinations, which will generate all possible solutions.
- It will take an integer piecesPlaced denoting the number of total pieces placed, integer N denoting the number of pieces needed to be placed, two integers row and col denoting the row and column where the current piece is going to be placed and a string ans for storing the matrix where pieces are placed, as arguments.
- Now, the initial call to allCombinations will pass 0 as piecesPlaced, N, 0 and 0 as row and col and an empty string as ans.
- In each call, check for the base case, that is:
- If row becomes N and all pieces are placed, i.e. piecesPlaced=N. Then print the ans and return. Else if piecesPlaced is not N, then just return from this call.
- Now make two calls:
- One to add a ‘*’ at the current position, and one to leave that position and add ‘-‘.
- After this, the recursive calls will print all the possible solutions.
Below is the implementation of the above approach.
C++
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print all// combinations of setting N// pieces in N x N boardvoid allCombinations(int piecesPlaced, int N, int row, int col, string ans){ // If the total 2d array's space // is exhausted then backtrack. if (row == N) { // If all the pieces are // placed then print the answer. if (piecesPlaced == N) { cout << ans; } return; } int nr = 0; int nc = 0; // Declare one string // that will set the piece. string x = ""; // Declare one string that // will leave the space blank. string y = ""; // If the current column // is out of bounds then // increase the row // and set col to 0. if (col == N - 1) { nr = row + 1; nc = 0; x = ans + "*\n"; y = ans + "-\n"; } // Else increase the col else { nr = row; nc = col + 1; x = ans + "*\t"; y = ans + "-\t"; } // Set the piece in the // box and move ahead allCombinations(piecesPlaced + 1, N, nr, nc, x); // Leave the space blank // and move forward allCombinations(piecesPlaced, N, nr, nc, y);}// Driver Codeint main(){ int N = 2; allCombinations(0, N, 0, 0, ""); return 0;} // This code is contributed by rakeshsahni. |
Java
// Java Program for the above approachimport java.io.*;import java.util.*;public class main { // Function to print all // combinations of setting N // pieces in N x N board public static void allCombinations( int piecesPlaced, int N, int row, int col, String ans) { // If the total 2d array's space // is exhausted then backtrack. if (row == N) { // If all the pieces are // placed then print the answer. if (piecesPlaced == N) { System.out.println(ans); } return; } int nr = 0; int nc = 0; // Declare one string // that will set the piece. String x = ""; // Declare one string that // will leave the space blank. String y = ""; // If the current column // is out of bounds then // increase the row // and set col to 0. if (col == N - 1) { nr = row + 1; nc = 0; x = ans + "*\n"; y = ans + "-\n"; } // Else increase the col else { nr = row; nc = col + 1; x = ans + "*\t"; y = ans + "-\t"; } // Set the piece in the // box and move ahead allCombinations( piecesPlaced + 1, N, nr, nc, x); // Leave the space blank // and move forward allCombinations(piecesPlaced, N, nr, nc, y); } // Driver Code public static void main(String[] args) throws Exception { int N = 2; allCombinations(0, N, 0, 0, ""); }} |
Python3
# Python Program for the above approach# Function to print all# combinations of setting N# pieces in N x N boarddef allCombinations(piecesPlaced, N, row, col, ans): # If the total 2d array's space # is exhausted then backtrack. if row == N: # If all the pieces are # placed then print the answer. if piecesPlaced == N: print(ans) return; nr = 0 nc = 0 # Declare one string # that will set the piece. x = "" # Declare one string that # will leave the space blank. y = "" # If the current column # is out of bounds then # increase the row # and set col to 0. if col == N - 1: nr = row + 1 nc = 0 x = ans + "*\n" y = ans + "-\n" # Else increase the col else: nr = row nc = col + 1 x = ans + "* " y = ans + "- " # Set the piece in the # box and move ahead allCombinations(piecesPlaced + 1, N, nr, nc, x); # Leave the space blank # and move forward allCombinations(piecesPlaced, N, nr, nc, y);# Driver CodeN = 2allCombinations(0, N, 0, 0, "")# This code is contributed by rdtank. |
C#
// C# Program for the above approachusing System;public class main { // Function to print all // combinations of setting N // pieces in N x N board public static void allCombinations(int piecesPlaced, int N, int row, int col, String ans) { // If the total 2d array's space // is exhausted then backtrack. if (row == N) { // If all the pieces are // placed then print the answer. if (piecesPlaced == N) { Console.WriteLine(ans); } return; } int nr = 0; int nc = 0; // Declare one string // that will set the piece. String x = ""; // Declare one string that // will leave the space blank. String y = ""; // If the current column // is out of bounds then // increase the row // and set col to 0. if (col == N - 1) { nr = row + 1; nc = 0; x = ans + "*\n"; y = ans + "-\n"; } // Else increase the col else { nr = row; nc = col + 1; x = ans + "*\t"; y = ans + "-\t"; } // Set the piece in the // box and move ahead allCombinations(piecesPlaced + 1, N, nr, nc, x); // Leave the space blank // and move forward allCombinations(piecesPlaced, N, nr, nc, y); } // Driver Code public static void Main(string[] args) { int N = 2; allCombinations(0, N, 0, 0, ""); }}// This code is contributed by ukasp. |
Javascript
// Javascript Program for the above approach// Function to print all// combinations of setting N// pieces in N x N boardfunction allCombinations(piecesPlaced, N, row, col, ans) { // If the total 2d array's space // is exhausted then backtrack. if (row == N) { // If all the pieces are // placed then print the answer. if (piecesPlaced == N) { document.write(ans); } return; } let nr = 0; let nc = 0; // Declare one string // that will set the piece. let x = ""; // Declare one string that // will leave the space blank. let y = ""; // If the current column // is out of bounds then // increase the row // and set col to 0. if (col == N - 1) { nr = row + 1; nc = 0; x = ans + "*<br>"; y = ans + "-<br>"; } // Else increase the col else { nr = row; nc = col + 1; x = ans + "* "; y = ans + "- "; } // Set the piece in the // box and move ahead allCombinations(piecesPlaced + 1, N, nr, nc, x); // Leave the space blank // and move forward allCombinations(piecesPlaced, N, nr, nc, y);}// Driver Codelet N = 2;allCombinations(0, N, 0, 0, "");// This code is contributed by Saurabh Jaiswal |
* * - - * - * - * - - * - * * - - * - * - - * *
Time Complexity: O(2^M), where M=N*N
Auxiliary Space:
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