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Construct a matrix such that union of ith row and ith column contains every element from 1 to 2N-1
• Difficulty Level : Hard
• Last Updated : 28 May, 2021

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++

 `// 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`

## Javascript

 ``
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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