Given a number N, the task is to create a square matrix of size N*N with values in range [1, N*N], such that the sum of each diagonal of an even sub-square matrix is even.
Examples:
Input: N = 3
Output:
1 2 3
4 5 6
7 8 9
Explanation:For each even sub-square matrix the sum of each diagonal is a even number.
1 2
4 5
sum of each diagonal is 6 and 6 i.e even number.Input: N = 4
Output:
1 2 3 4
6 5 8 7
9 10 11 12
14 13 16 15
Explanation:
For each even sub-square matrix the sum of each diagonal is a even number.
1 2
6 5
sum of each diagonal is 6 and 8 i.e even number.
Approach: The idea is to arrange elements from 1 to N*N in the below-given ways:
- Initialize odd and even by 1 and 2 elements respectively.
- Iterate two nested loop in the range [0, N].
- If the sum of indices in the two nested loops is even the print the value of odd and increment odd by 2 and if the sum is odd then print the value of even, and increment even by 2.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>using namespace std;// Function to print N*N order matrix// with all sub-matrix of even order// is sum of its diagonal also evenvoid evenSubMatrix(int N){ // Even index int even = 1; // Odd index int odd = 2; // Iterate two nested loop for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { // For even index the element // should be consecutive odd if ((i + j) % 2 == 0) { cout << even << " "; even += 2; } // for odd index the element // should be consecutive even else { cout << odd << " "; odd += 2; } } cout << "\n"; }}// Driver Codeint main(){ // Given order of matrix int N = 4; // Function call evenSubMatrix(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to print N*N order matrix// with all sub-matrix of even order// is sum of its diagonal also evenstatic void evenSubMatrix(int N){ // Even index int even = 1; // Odd index int odd = 2; // Iterate two nested loop for(int i = 0; i < N; i++) { for(int j = 0; j < N; j++) { // For even index the element // should be consecutive odd if ((i + j) % 2 == 0) { System.out.print(even + " "); even += 2; } // For odd index the element // should be consecutive even else { System.out.print(odd + " "); odd += 2; } } System.out.println(); }}// Driver codepublic static void main(String[] args){ // Given order of matrix int N = 4; // Function call evenSubMatrix(N);}}// This code is contributed by offbeat |
Python3
# Python3 program for the above approach# Function to print N*N order matrix# with all sub-matrix of even order# is sum of its diagonal also evendef evenSubMatrix(N): # Even index even = 1 # Odd index odd = 2 # Iterate two nested loop for i in range(N): for j in range(N): # For even index the element # should be consecutive odd if ((i + j) % 2 == 0): print(even, end = " ") even += 2 # For odd index the element # should be consecutive even else: print(odd, end = " ") odd += 2 print() # Driver Code# Given order of matrixN = 4# Function callevenSubMatrix(N)# This code is contributed by sanjoy_62 |
C#
// C# program for the above approachusing System;class GFG{// Function to print N*N order matrix// with all sub-matrix of even order// is sum of its diagonal also evenstatic void evenSubMatrix(int N){ // Even index int even = 1; // Odd index int odd = 2; // Iterate two nested loop for(int i = 0; i < N; i++) { for(int j = 0; j < N; j++) { // For even index the element // should be consecutive odd if ((i + j) % 2 == 0) { Console.Write(even + " "); even += 2; } // For odd index the element // should be consecutive even else { Console.Write(odd + " "); odd += 2; } } Console.WriteLine(); }}// Driver codepublic static void Main(String[] args){ // Given order of matrix int N = 4; // Function call evenSubMatrix(N);}}// This code is contributed by amal kumar choubey |
Javascript
<script>// Java script program for the above approach// Function to print N*N order matrix// with all sub-matrix of even order// is sum of its diagonal also evenfunction evenSubMatrix( N){ // Even index let even = 1; // Odd index let odd = 2; // Iterate two nested loop for(let i = 0; i < N; i++) { for(let j = 0; j < N; j++) { // For even index the element // should be consecutive odd if ((i + j) % 2 == 0) { document.write(even + " "); even += 2; } // For odd index the element // should be consecutive even else { document.write(odd + " "); odd += 2; } } document.write("<br>"); }}// Driver code // Given order of matrix let N = 4; // Function call evenSubMatrix(N);// This code is contributed by manoj</script> |
Output:
1 2 3 4 6 5 8 7 9 10 11 12 14 13 16 15
Time Complexity: O(N*N)
Auxiliary Space: O(1)
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