Given an integer N representing the size of the matrix, the task is to construct a square matrix N * N which have an element from 1 to N2 such that the parity of the sum of its diagonals is equal to the parity of integer N.
Examples:
Input: N = 4
Output:
1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13
Explanation:
Sum of diagonal = 32 and 36 and integer N = 4, all the numbers are even that is same parity.Input: N = 3
Output:
1 2 3
6 5 4
7 8 9
Explanation:
Sum of diagonal = 15 and integer N = 3, all the numbers are odd that is same parity.
Approach: The idea is to observe that on filling the elements in the matrix in an alternative fashion the parity of N and the sum of diagonals is the same. Start the counter from 1 and then fill the first row from 0 to N – 1 in increasing order, then fill the second row from index N – 1 to 0, and so on. Keep filling each element from value 1 to N2 in this alternate fashion to get the required matrix.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include<bits/stdc++.h>using namespace std;// Function to construct a N * N// matrix based on the given conditionvoid createMatrix(int N){ // Matrix with given sizM int M[N][N]; // Counter to insert elements // from 1 to N * N int count = 1; for (int i = 0; i < N; i++) { // Check if it is first row // or odd row of the matrix if (i % 2 == 0) { for (int j = 0; j < N; j++) { M[i][j] = count; count++; } } else { // Insert elements from // right to left for (int j = N - 1;j >= 0; j--){ M[i][j] = count; count += 1; } } } // Print the matrix for (int i = 0; i < N; i++) { // Traverse column for (int j = 0; j < N; j++) { cout << M[i][j] << " "; } cout << endl; }}// Driver Codeint main(){ // Given size of matrix N int N = 3; // Function Call createMatrix(N); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to construct a N * N// matrix based on the given conditionstatic void createMatrix(int N){ // Matrix with given sizM int M[][] = new int[N][N]; // Counter to insert elements // from 1 to N * N int count = 1; for (int i = 0; i < N; i++) { // Check if it is first row // or odd row of the matrix if (i % 2 == 0) { for (int j = 0; j < N; j++) { M[i][j] = count; count++; } } else { // Insert elements from // right to left for(int j = N - 1; j >= 0; j--){ M[i][j] = count; count += 1 ; } } } // Print the matrix for (int i = 0; i < N; i++) { // Traverse column for (int j = 0; j < N; j++) { System.out.print(M[i][j] + " "); } System.out.println(); }} // Driver Codepublic static void main(String[] args){ // Given size of matrix N int N = 3; // Function Call createMatrix(N);}}// This code is contributed by Ritik Bansal |
Python3
# Python3 program for the above approach# Function to construct a N * N# matrix based on the given conditiondef createMatrix(N): # Matrix with given size M = [[0] * N for i in range(N)] # Counter to insert elements # from 1 to N * N count = 1 for i in range(N): # Check if it is first row # or odd row of the matrix if (i % 2 == 0): for j in range(N): # Insert elements from # left to right M[i][j] = count count += 1 # Condition if it is second # row or even row else: # Insert elements from # right to left for j in range(N - 1, -1, -1): M[i][j] = count count += 1 # Print the matrix for i in range(N): # Traverse column for j in range(N): print(M[i][j], end = " ") print()# Driver Codeif __name__ == '__main__': # Given size of matrix N N = 3 # Function call createMatrix(N)# This code is contributed by mohit kumar 29 |
C#
// C# program for // the above approachusing System;class GFG{ // Function to construct a N * N// matrix based on the given conditionstatic void createMatrix(int N){ // Matrix with given sizM int[,] M = new int[N, N]; // Counter to insert elements // from 1 to N * N int count = 1; for (int i = 0; i < N; i++) { // Check if it is first row // or odd row of the matrix if (i % 2 == 0) { for (int j = 0; j < N; j++) { M[i, j] = count; count++; } } else { // Insert elements from // right to left for(int j = N - 1; j >= 0; j--) { M[i, j] = count; count += 1; } } } // Print the matrix for (int i = 0; i < N; i++) { // Traverse column for (int j = 0; j < N; j++) { Console.Write(M[i, j] + " "); } Console.WriteLine(); }} // Driver Codepublic static void Main(){ // Given size of matrix N int N = 3; // Function Call createMatrix(N);}}// This code is contributed by Chitranayal |
Javascript
<script>// javascript program for the above approach // Function to construct a N * N// matrix based on the given conditionfunction createMatrix(N){ // Matrix with given sizM var M = Array(N).fill(0).map(x => Array(N).fill(0)); // Counter to insert elements // from 1 to N * N var count = 1; for (i = 0; i < N; i++) { // Check if it is first row // or odd row of the matrix if (i % 2 == 0) { for (j = 0; j < N; j++) { M[i][j] = count; count++; } } else { // Insert elements from // right to left for(j = N - 1; j >= 0; j--){ M[i][j] = count; count += 1 ; } } } // Print the matrix for (i = 0; i < N; i++) { // Traverse column for (j = 0; j < N; j++) { document.write(M[i][j] + " "); } document.write("<br>"); }} // Driver Codevar N = 3;// Function CallcreateMatrix(N);// This code is contributed by 29AjayKumar </script> |
Output:
1 2 3 6 5 4 7 8 9
Time Complexity: O(N*N)
Auxiliary Space: O(1)
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