# Generate a matrix having sum of secondary diagonal equal to a perfect square

Given an integer N, the task is to generate a matrix of dimensions N x N using positive integers from the range [1, N] such that the sum of the secondary diagonal is a perfect square.

Examples:

Input: N = 3
Output:
1 2 3
2 3 1
3 2 1
Explanation:
The sum of secondary diagonal = 3 + 3 + 3 = 9(= 32).

Input: N = 7
Output:
1 2 3 4 5 6 7
2 3 4 5 6 7 1
3 4 5 6 7 1 2
4 5 6 7 1 2 3
5 6 7 1 2 3 4
6 7 1 2 3 4 5
7 1 2 3 4 5 6
Explanation:
The sum of secondary diagonal = 7 + 7 + 7 + 7 + 7 + 7 + 7 = 49(= 72).

Approach: Since the generated matrix needs to be of dimensions N x N, therefore, to make the sum of elements in the secondary diagonal a perfect square, the idea is to assign N at each index of the secondary diagonal. Therefore, the sum of all N elements in this diagonal is N2, which is a perfect square. Follow the steps below to solve the problem:

1. Initialize a matrix mat[][] of dimension N x N.
2. Initialize the first row of the matrix as {1 2 3 … N}.
3. Now for the remaining rows of the matrix, fill each row by circular left shift of the arrangement of the previous row of the matrix by 1.
4. Print the matrix after completing the above steps.

Below is the implementation of the above approach:

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to print the matrix whose sum` `// of element in secondary diagonal is a` `// perfect square` `void` `diagonalSumPerfectSquare(``int` `arr[], ``int` `N)` `{` `    `  `    ``// Iterate for next N - 1 rows` `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        `  `        ``// Print the current row after` `        ``// the left shift` `        ``for``(``int` `j = 0; j < N; j++) ` `        ``{` `            ``cout << (arr[(j + i) % 7]) << ``" "``;` `        ``}` `        ``cout << endl;` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    `  `    ``// Given N` `    ``int` `N = 7;`   `    ``int` `arr[N];`   `    ``// Fill the array with elements` `    ``// ranging from 1 to N` `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{` `        ``arr[i] = i + 1;` `    ``}`   `    ``// Function Call` `    ``diagonalSumPerfectSquare(arr, N);` `}`   `// This code is contributed by gauravrajput1`

 `// Java program for the above approach` `class` `GFG {`   `    ``// Function to print the matrix whose sum` `    ``// of element in secondary diagonal is a` `    ``// perfect square` `    ``static` `void` `diagonalSumPerfectSquare(``int``[] arr, ` `                                         ``int` `N)` `    ``{`   `        ``// Iterate for next N - 1 rows` `        ``for` `(``int` `i = ``0``; i < N; i++)` `        ``{`   `            ``// Print the current row after` `            ``// the left shift` `            ``for` `(``int` `j = ``0``; j < N; j++)` `            ``{` `                ``System.out.print(arr[(j + i) % ``7``] + ``" "``);` `            ``}` `            ``System.out.println();` `        ``}` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] srgs)` `    ``{`   `        ``// Given N` `        ``int` `N = ``7``;`   `        ``int``[] arr = ``new` `int``[N];`   `        ``// Fill the array with elements` `        ``// ranging from 1 to N` `        ``for` `(``int` `i = ``0``; i < N; i++)` `        ``{` `            ``arr[i] = i + ``1``;` `        ``}`   `        ``// Function Call` `        ``diagonalSumPerfectSquare(arr, N);` `    ``}` `}`   `// This code is contributed by Amit Katiyar`

 `# Python3 program for the above approach`   `# Function to print the matrix whose sum` `# of element in secondary diagonal is a` `# perfect square` `def` `diagonalSumPerfectSquare(arr, N):` `  `  `    ``# Print the current row` `    ``print``(``*``arr, sep ``=``" "``)` `    `  `    ``# Iterate for next N - 1 rows` `    ``for` `i ``in` `range``(N``-``1``):` `       `  `        ``# Perform left shift by 1` `        ``arr ``=` `arr[i::] ``+` `arr[:i:]` `        `  `        ``# Print the current row after` `        ``# the left shift` `        ``print``(``*``arr, sep ``=``" "``)`   `# Driver Code`   `# Given N` `N ``=` `7`   `arr ``=` `[]`   `# Fill the array with elements` `# ranging from 1 to N` `for` `i ``in` `range``(``1``, N ``+` `1``):` `    ``arr.append(i)`   `# Function Call` `diagonalSumPerfectSquare(arr, N)`

 `// C# program for the` `// above approach` `using` `System;` `class` `GFG {`   `    ``// Function to print the matrix whose sum` `    ``// of element in secondary diagonal is a` `    ``// perfect square` `    ``static` `void` `diagonalSumPerfectSquare(``int``[] arr, ` `                                         ``int` `N)` `    ``{` `        ``// Iterate for next N - 1 rows` `        ``for` `(``int` `i = 0; i < N; i++)` `        ``{` `            ``// Print the current row after` `            ``// the left shift` `            ``for` `(``int` `j = 0; j < N; j++) ` `            ``{` `                ``Console.Write(arr[(j + i) % 7] + ``" "``);` `            ``}` `            ``Console.WriteLine();` `        ``}` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(String[] srgs)` `    ``{` `        ``// Given N` `        ``int` `N = 7;`   `        ``int``[] arr = ``new` `int``[N];`   `        ``// Fill the array with elements` `        ``// ranging from 1 to N` `        ``for` `(``int` `i = 0; i < N; i++) {` `            ``arr[i] = i + 1;` `        ``}`   `        ``// Function Call` `        ``diagonalSumPerfectSquare(arr, N);` `    ``}` `}`   `// This code is contributed by 29AjayKumar`

Output
```1 2 3 4 5 6 7
2 3 4 5 6 7 1
3 4 5 6 7 1 2
4 5 6 7 1 2 3
5 6 7 1 2 3 4
6 7 1 2 3 4 5
7 1 2 3 4 5 6
```

Time Complexity: O(N2)
Auxiliary Space: O(N)

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