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:
- Initialize a matrix mat[][] of dimension N x N.
- Initialize the first row of the matrix as {1 2 3 … N}.
- 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.
- Print the matrix after completing the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;Â
// Function to print the matrix whose sum// of element in secondary diagonal is a// perfect squarevoid 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;    }}Â
// Driver Codeint main(){Â Â Â Â Â Â Â Â Â // Given NÂ Â Â Â int N = 7;Â
    int arr[N];Â
    // Fill the array with elements    // ranging from 1 to N    for(int i = 0; i < N; i++)     {        arr[i] = i + 1;    }Â
    // Function Call    diagonalSumPerfectSquare(arr, N);}Â
// This code is contributed by gauravrajput1 |
Java
// Java program for the above approachclass GFG {Â
    // Function to print the matrix whose sum    // of element in secondary diagonal is a    // perfect square    static void diagonalSumPerfectSquare(int[] arr,                                          int N)    {Â
        // Iterate for next N - 1 rows        for (int i = 0; i < N; i++)        {Â
            // 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();        }    }Â
    // Driver Code    public static void main(String[] srgs)    {Â
        // Given N        int N = 7;Â
        int[] arr = new int[N];Â
        // Fill the array with elements        // ranging from 1 to N        for (int i = 0; i < N; i++)        {            arr[i] = i + 1;        }Â
        // Function Call        diagonalSumPerfectSquare(arr, N);    }}Â
// This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approachÂ
# Function to print the matrix whose sum# of element in secondary diagonal is a# perfect squaredef 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 =" ")Â
# Driver CodeÂ
# Given NN = 7Â
arr = []Â
# Fill the array with elements# ranging from 1 to Nfor i in range(1, N + 1):Â Â Â Â arr.append(i)Â
# Function CalldiagonalSumPerfectSquare(arr, N) |
C#
// C# program for the// above approachusing System;class GFG {Â
    // 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();        }    }Â
    // Driver Code    public static void Main(String[] srgs)    {        // Given N        int N = 7;Â
        int[] arr = new int[N];Â
        // Fill the array with elements        // ranging from 1 to N        for (int i = 0; i < N; i++) {            arr[i] = i + 1;        }Â
        // Function Call        diagonalSumPerfectSquare(arr, N);    }}Â
// This code is contributed by 29AjayKumar |
Javascript
<script>Â
// Javascript program to implement// the above approach      // Function to print the matrix whose sum    // of element in secondary diagonal is a    // perfect square    function diagonalSumPerfectSquare( arr,N)    {          // Iterate for next N - 1 rows        for (let i = 0; i < N; i++)        {              // Print the current row after            // the left shift            for (let j = 0; j < N; j++)            {                document.write(arr[(j + i) % 7] + " ");            }            document.write("<br/>");        }    } Â
Â
// Driver CodeÂ
        // Given N        let N = 7;          let arr = new Array(N).fill(0);          // Fill the array with elements        // ranging from 1 to N        for (let i = 0; i < N; i++)        {            arr[i] = i + 1;        }          // Function Call        diagonalSumPerfectSquare(arr, N);   // This code is contributed by avijitmondal1998.</script> |
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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