Construct a matrix such that union of ith row and ith column contains every element from 1 to 2N-1

Given a number N, the task is to construct a square matrix of N * N where union of elements in some ith row with the ith column contains every element in the range [1, 2*N-1]. If no such matrix exists, print -1.

Note: There can be multiple possible solutions for a particular N

Examples:

Input: N = 6
Output:
6 4 2 5 3 1
10 6 5 3 1 2
8 11 6 1 4 3
11 9 7 6 2 4
9 7 10 8 6 5
7 8 9 10 11 6
Explanation:
The above matrix is of 6 * 6 in which every ith row and column contains elements from 1 to 11, like:
1st row and 1st column = {6, 4, 2, 5, 3, 1} & {6, 10, 8, 11, 9, 7} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
2nd row and 2nd column = {10, 6, 5, 3, 1, 2} & {4, 6, 11, 9, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
3rd row and 3rd column = {8, 11, 6, 1, 4, 3} & {2, 5, 6, 7, 10, 9} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
4th row and 4th column = {11, 9, 7, 6, 2, 4} & {5, 3, 1, 6, 8, 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
5th row and 5th column = {9, 7, 10, 8, 6, 5} & {3, 1, 4, 2, 6, 11} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
6th row and 6th column = {7, 8, 9, 10, 11, 6} & {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

Input: N = 6
Output: -1
Explanation:
There is no such matrix possible in which every i’th row and i’th column contains every elements from 1 to 9. Hence, the answer is -1.



Approach:
If we observe carefully, we can see that:

  • For any odd number except 1, the square matrix is not possible to generate
  • To generate the square matrix for even order, the idea is to fill up the upper half of the diagonal elements of the matrix in the range of 1 to N-1 and fill all the diagonal elements with the N and the lower half of the diagonal elements can be filled in number from range N + 1 to 2N – 1.

Below is the algorithm fr this approach:

  1. Matrix of odd order can’t be filled as observed except for N = 1
  2. For Matrix of even order,
    • Firstly, fill all diagonal elements equal to N.
    • Consider the two halves of the matrix diagonally bisected, each half can be filled with N-1 elements.
    • Fill upper half with elements from [1, N-1] and the lower half with elements from [N+1, 2N-1].
    • As it can be easily observed that there is a pattern that second row’s last element can be always 2.
    • Now, consecutive elements in the last column are at a difference of 2. Hence generalised form can be given as A[i]=[(N-2)+2i]%(N-1)+1, for all i from 1 to N-1
    • Simply add N to all the elements of the lower half.

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
int matrix[100][100];
  
// Function to find the square matrix
void printRequiredMatrix(int n)
{
    // For Matrix of order 1,
    // it will contain only 1
    if (n == 1) {
        cout << "1"
             << "\n";
    }
  
    // For Matrix of odd order,
    // it is not possible
    else if (n % 2 != 0) {
        cout << "-1"
             << "\n";
    }
  
    // For Matrix of even order
    else {
        // All diagonal elements of the
        // matrix can be N itself.
        for (int i = 0; i < n; i++) {
            matrix[i][i] = n;
        }
        int u = n - 1;
  
        // Assign values at desired
        // place in the matrix
        for (int i = 0; i < n - 1; i++) {
  
            matrix[i][u] = i + 1;
  
            for (int j = 1; j < n / 2; j++) {
  
                int a = (i + j) % (n - 1);
                int b = (i - j + n - 1) % (n - 1);
                if (a < b)
                    swap(a, b);
                matrix[b][a] = i + 1;
            }
        }
  
        // Loop to add N in the lower half
        // of the matrix such that it contains
        // elements from 1 to 2*N - 1
        for (int i = 0; i < n; i++)
            for (int j = 0; j < i; j++)
                matrix[i][j] = matrix[j][i] + n;
  
        // Loop to print the matrix
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
                cout << matrix[i][j] << " ";
            cout << "\n";
        }
    }
    cout << "\n";
}
  
// Driver Code
int main()
{
    int n = 1;
    printRequiredMatrix(n);
  
    n = 3;
    printRequiredMatrix(n);
  
    n = 6;
    printRequiredMatrix(n);
  
    return 0;
}

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Java

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// Java implementation of the above approach 
class GFG {
      
    static int matrix[][] = new int[100][100]; 
      
    // Function to find the square matrix 
    static void printRequiredMatrix(int n) 
    
        // For Matrix of order 1, 
        // it will contain only 1 
        if (n == 1) { 
            System.out.println("1");
        
      
        // For Matrix of odd order, 
        // it is not possible 
        else if (n % 2 != 0) { 
            System.out.println("-1");
        
      
        // For Matrix of even order 
        else
            // All diagonal elements of the 
            // matrix can be N itself. 
            for (int i = 0; i < n; i++) { 
                matrix[i][i] = n; 
            
            int u = n - 1
      
            // Assign values at desired 
            // place in the matrix 
            for (int i = 0; i < n - 1; i++) { 
      
                matrix[i][u] = i + 1
      
                for (int j = 1; j < n / 2; j++) { 
      
                    int a = (i + j) % (n - 1); 
                    int b = (i - j + n - 1) % (n - 1); 
                    if (a < b) {
                        int temp = a;
                        a = b; 
                        b = temp;
                    }
                    matrix[b][a] = i + 1
                
            
      
            // Loop to add N in the lower half 
            // of the matrix such that it contains 
            // elements from 1 to 2*N - 1 
            for (int i = 0; i < n; i++) 
                for (int j = 0; j < i; j++) 
                    matrix[i][j] = matrix[j][i] + n; 
      
            // Loop to print the matrix 
            for (int i = 0; i < n; i++) { 
                for (int j = 0; j < n; j++) 
                    System.out.print(matrix[i][j] + " "); 
                System.out.println() ;
            
        
    System.out.println();
    
      
    // Driver Code 
    public static void main (String[] args) 
    
        int n = 1
        printRequiredMatrix(n); 
      
        n = 3
        printRequiredMatrix(n); 
      
        n = 6
        printRequiredMatrix(n); 
      
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the above approach 
import numpy as np;
  
matrix = np.zeros((100,100)); 
  
# Function to find the square matrix 
def printRequiredMatrix(n) : 
  
    # For Matrix of order 1, 
    # it will contain only 1 
    if (n == 1) :
        print("1"); 
  
    # For Matrix of odd order, 
    # it is not possible 
    elif (n % 2 != 0) : 
        print("-1");
  
    # For Matrix of even order 
    else :
        # All diagonal elements of the 
        # matrix can be N itself. 
        for i in range(n) :
            matrix[i][i] = n; 
      
        u = n - 1
  
        # Assign values at desired 
        # place in the matrix 
        for i in range(n - 1) :
  
            matrix[i][u] = i + 1
  
            for j in range(1, n//2) :
  
                a = (i + j) % (n - 1); 
                b = (i - j + n - 1) % (n - 1); 
                if (a < b) :
                    a,b = b,a
                      
                matrix[b][a] = i + 1
  
        # Loop to add N in the lower half 
        # of the matrix such that it contains 
        # elements from 1 to 2*N - 1 
        for i in range(n) :
            for j in range(i) :
                matrix[i][j] = matrix[j][i] + n; 
  
        # Loop to print the matrix 
        for i in range(n) :
            for j in range(n) :
                print(matrix[i][j] ,end=" "); 
            print();
  
    print()
  
# Driver Code 
if __name__ == "__main__"
  
    n = 1
    printRequiredMatrix(n); 
  
    n = 3
    printRequiredMatrix(n); 
  
    n = 6
    printRequiredMatrix(n); 
  
    # This code is contributed by AnkitRai01

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C#

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// C# implementation of the above approach 
using System;
  
class GFG {
      
    static int [,]matrix = new int[100, 100]; 
      
    // Function to find the square matrix 
    static void printRequiredMatrix(int n) 
    
        // For Matrix of order 1, 
        // it will contain only 1 
        if (n == 1) { 
            Console.WriteLine("1");
        
      
        // For Matrix of odd order, 
        // it is not possible 
        else if (n % 2 != 0) { 
            Console.WriteLine("-1");
        
      
        // For Matrix of even order 
        else 
        
            // All diagonal elements of the 
            // matrix can be N itself. 
            for (int i = 0; i < n; i++) { 
                matrix[i, i] = n; 
            
            int u = n - 1; 
      
            // Assign values at desired 
            // place in the matrix 
            for (int i = 0; i < n - 1; i++) { 
      
                matrix[i, u] = i + 1; 
      
                for (int j = 1; j < n / 2; j++) { 
      
                    int a = (i + j) % (n - 1); 
                    int b = (i - j + n - 1) % (n - 1); 
                    if (a < b) {
                        int temp = a;
                        a = b; 
                        b = temp;
                    }
                    matrix[b, a] = i + 1; 
                
            
      
            // Loop to add N in the lower half 
            // of the matrix such that it contains 
            // elements from 1 to 2*N - 1 
            for (int i = 0; i < n; i++) 
                for (int j = 0; j < i; j++) 
                    matrix[i, j] = matrix[j, i] + n; 
      
            // Loop to print the matrix 
            for (int i = 0; i < n; i++) { 
                for (int j = 0; j < n; j++) 
                    Console.Write(matrix[i, j] + " "); 
                Console.WriteLine() ;
            
        
    Console.WriteLine();
    
      
    // Driver Code 
    public static void Main (String[] args) 
    
        int n = 1; 
        printRequiredMatrix(n); 
      
        n = 3; 
        printRequiredMatrix(n); 
      
        n = 6; 
        printRequiredMatrix(n); 
    
}
  
// This code is contributed by Yash_R

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Output:

1

-1

6 4 2 5 3 1 
10 6 5 3 1 2 
8 11 6 1 4 3 
11 9 7 6 2 4 
9 7 10 8 6 5 
7 8 9 10 11 6

Performance Analysis:

  • Time Complexity: As in the above approach, there is two loops to iterate over the whole N*N matrix which takes O(N2) time in worst case, Hence the Time Complexity will be O(N2).
  • Auxiliary Space Complexity: As in the above approach, there is no extra space used, Hence the auxiliary space complexity will be O(1)

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Improved By : AnkitRai01, Yash_R