C++ Program to Find difference between sums of two diagonals
Last Updated :
31 May, 2022
Given a matrix of n X n. The task is to calculate the absolute difference between the sums of its diagonal.
Examples:
Input : mat[][] = 11 2 4
4 5 6
10 8 -12
Output : 15
Sum of primary diagonal = 11 + 5 + (-12) = 4.
Sum of primary diagonal = 4 + 5 + 10 = 19.
Difference = |19 - 4| = 15.
Input : mat[][] = 10 2
4 5
Output : 7
Calculate the sums across the two diagonals of a square matrix. Along the first diagonal of the matrix, row index = column index i.e mat[i][j] lies on the first diagonal if i = j. Along the other diagonal, row index = n – 1 – column index i.e mat[i][j] lies on the second diagonal if i = n-1-j. By using two loops we traverse the entire matrix and calculate the sum across the diagonals of the matrix.
Below is the implementation of this approach:
C++
#include <bits/stdc++.h>
#define MAX 100
using namespace std;
int difference( int arr[][MAX], int n)
{
int d1 = 0, d2 = 0;
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
{
if (i == j)
d1 += arr[i][j];
if (i == n - j - 1)
d2 += arr[i][j];
}
}
return abs (d1 - d2);
}
int main()
{
int n = 3;
int arr[][MAX] =
{
{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}
};
cout << difference(arr, n);
return 0;
}
|
Output:
15
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.
We can optimize above solution to work in O(n) using the patterns present in indexes of cells.
C++
#include <bits/stdc++.h>
#define MAX 100
using namespace std;
int difference( int arr[][MAX], int n)
{
int d1 = 0, d2 = 0;
for ( int i = 0; i < n; i++)
{
d1 += arr[i][i];
d2 += arr[i][n-i-1];
}
return abs (d1 - d2);
}
int main()
{
int n = 3;
int arr[][MAX] =
{
{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}
};
cout << difference(arr, n);
return 0;
}
|
Output:
15
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 Find difference between sums of two diagonals for more details!
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