Construct a matrix with sum equal to the sum of diagonal elements
Given an integer N, the task is to construct a matrix of size N2 using positive and negative integers and excluding 0, such that the sum of the matrix is equal to the sum of the diagonal of the matrix.
Examples:
Input: N = 2
Output:
1 -2
2 4
Explanation:
Diagonal sum = (1 + 4) = 5
Matrix sum = (1 – 2 + 2 + 4) = 5Input: N = 5
Output:
1 2 3 5 10
3 1 4 -9 1
-19 6 1 5 -8
4 -7 2 1 12
-17 1 1 1 1
Explanation:
Diagonal sum = (1 + 1 + 1 + 1 + 1) = 5
Matrix sum = 5
Approach:
The approach to solving the problem is to traverse all indices of the matrix and print a positive element(say y) at the N diagonal positions and equally distribute a single-valued positive and negative integer(say x and -x) in the remaining N2 – N positions.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to construct matrix with // diagonal sum equal to matrix sum void constructmatrix( int N) { bool check = true ; for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) { // If diagonal position if (i == j) { cout << 1 << " " ; } else if (check) { // Positive element cout << 2 << " " ; check = false ; } else { // Negative element cout << -2 << " " ; check = true ; } } cout << endl; } } // Driver Code int main() { int N = 5; constructmatrix(5); return 0; } |
Java
// Java program to implement // the above approach public class Main { // Function to construct matrix with // diagonal sum equal to matrix sum public static void constructmatrix( int N) { boolean check = true ; for ( int i = 0 ; i < N; i++) { for ( int j = 0 ; j < N; j++) { // If diagonal position if (i == j) { System.out.print( "1 " ); } else if (check) { // Positive element System.out.print( "2 " ); check = false ; } else { // Negative element System.out.print( "-2 " ); check = true ; } } System.out.println(); } } // Driver Code public static void main(String[] args) { int N = 5 ; constructmatrix( 5 ); } } |
Python3
# Python3 program to implement # the above approach # Function to construct matrix with # diagonal sum equal to matrix sum def constructmatrix(N): check = bool ( True ) for i in range (N): for j in range (N): # If diagonal position if (i = = j): print ( 1 , end = " " ) elif (check): # Positive element print ( 2 , end = " " ) check = bool ( False ) else : # Negative element print ( - 2 , end = " " ) check = bool ( True ) print () # Driver code N = 5 constructmatrix( 5 ) # This code is contributed by divyeshrabadiya07 |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to construct matrix with // diagonal sum equal to matrix sum public static void constructmatrix( int N) { bool check = true ; for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) { // If diagonal position if (i == j) { Console.Write( "1 " ); } else if (check) { // Positive element Console.Write( "2 " ); check = false ; } else { // Negative element Console.Write( "-2 " ); check = true ; } } Console.WriteLine(); } } // Driver Code static public void Main () { int N = 5; constructmatrix(N); } } // This code is contributed by piyush3010 |
Javascript
<script> // Javascript program to implement // the above approach // Function to construct matrix with // diagonal sum equal to matrix sum function constructmatrix(N) { let check = true ; for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { // If diagonal position if (i == j) { document.write( "1 " ); } else if (check) { // Positive element document.write( "2 " ); check = false ; } else { // Negative element document.write( "-2 " ); check = true ; } } document.write( "<br/>" ); } } // Driver Code let N = 5; constructmatrix(5); </script> |
1 2 -2 2 -2 2 1 -2 2 -2 2 -2 1 2 -2 2 -2 2 1 -2 2 -2 2 -2 1
Time Complexity: O(N2)
Auxiliary Space: O(1)
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