C++ Program to Efficiently Compute Sums of Diagonals of a Matrix
Given a 2D square matrix, find the sum of elements in Principal and Secondary diagonals. For example, consider the following 4 X 4 input matrix.
A00 A01 A02 A03
A10 A11 A12 A13
A20 A21 A22 A23
A30 A31 A32 A33
The primary diagonal is formed by the elements A00, A11, A22, A33.
- Condition for Principal Diagonal: The row-column condition is row = column.
The secondary diagonal is formed by the elements A03, A12, A21, A30.
- Condition for Secondary Diagonal: The row-column condition is row = numberOfRows – column -1.
Examples :
Input:
4
1 2 3 4
4 3 2 1
7 8 9 6
6 5 4 3
Output:
Principal Diagonal: 16
Secondary Diagonal: 20
Input:
3
1 1 1
1 1 1
1 1 1
Output:
Principal Diagonal: 3
Secondary Diagonal: 3
Method 1 (Brute Force) :
In this method, we use two loops i.e. a loop for columns and a loop for rows and in the inner loop we check for the condition stated above:
C++
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
void printDiagonalSums( int mat[][MAX], int n)
{
int principal = 0, secondary = 0;
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
{
if (i == j)
principal += mat[i][j];
if ((i + j) == (n - 1))
secondary += mat[i][j];
}
}
cout << "Principal Diagonal:" <<
principal << endl;
cout << "Secondary Diagonal:" <<
secondary << endl;
}
int main()
{
int a[][MAX] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{1, 2, 3, 4},
{5, 6, 7, 8}};
printDiagonalSums(a, 4);
return 0;
}
|
Output:
Principal Diagonal:18
Secondary Diagonal:18
Time Complexity: O(N*N), as we are using nested loops to traverse N*N times.
Auxiliary Space: O(1), as we are not using any extra space.
Method 2 (Efficient Approach) :
In this method we use one loop i.e. a loop for calculating sum of both the principal and secondary diagonals:
C++
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
void printDiagonalSums( int mat[][MAX],
int n)
{
int principal = 0, secondary = 0;
for ( int i = 0; i < n; i++)
{
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
cout << "Principal Diagonal:" <<
principal << endl;
cout << "Secondary Diagonal:" <<
secondary << endl;
}
int main()
{
int a[][MAX] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{1, 2, 3, 4},
{5, 6, 7, 8}};
printDiagonalSums(a, 4);
return 0;
}
|
Output :
Principal Diagonal:18
Secondary Diagonal:18
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
Please refer complete article on Efficiently compute sums of diagonals of a matrix for more details!
Last Updated :
17 Jan, 2023
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