# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `int` `matrix; ` ` `  `// 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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

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)

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : AnkitRai01, Yash_R