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;// 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; }}// Driver Codeint 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
// Java program for the above approachclass 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
# Python3 program for the above approach# 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 =" ")# Driver Code# Given NN = 7arr = []# Fill the array with elements# ranging from 1 to Nfor i in range(1, N + 1): arr.append(i)# Function CalldiagonalSumPerfectSquare(arr, N) |
C#
// C# program for the// above approachusing 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 |
Javascript
<script>// 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/>"); } } // Driver Code // 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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