Python 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. 
 

1. Condition for Principal Diagonal: The row-column condition is row = column. 
   The secondary diagonal is formed by the elements A03, A12, A21, A30.
2. 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 (O(n ^ 2) :

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:
 

# A simple Python program to 
# find sum of diagonals
MAX = 100

def printDiagonalSums(mat, n):

    principal = 0
    secondary = 0;
    for i in range(0, n): 
        for j in range(0, n): 

            # Condition for principal diagonal
            if (i == j):
                principal += mat[i][j]

            # Condition for secondary diagonal
            if ((i + j) == (n - 1)):
                secondary += mat[i][j]
        
    print("Principal Diagonal:", principal)
    print("Secondary Diagonal:", secondary)

# Driver code
a = [[ 1, 2, 3, 4 ],
     [ 5, 6, 7, 8 ], 
     [ 1, 2, 3, 4 ],
      [ 5, 6, 7, 8 ]]
printDiagonalSums(a, 4)

# This code is contributed 
# by ihritik

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 (O(n) :

In this method we use one loop i.e. a loop for calculating sum of both the principal and secondary diagonals: 
 

# A simple Python3 program to find
# sum of diagonals
MAX = 100

def printDiagonalSums(mat, n):

    principal = 0
    secondary = 0
    for i in range(0, n): 
        principal += mat[i][i]
        secondary += mat[i][n - i - 1]
        
    print("Principal Diagonal:", principal)
    print("Secondary Diagonal:", secondary)

# Driver code
a = [[ 1, 2, 3, 4 ],
     [ 5, 6, 7, 8 ], 
     [ 1, 2, 3, 4 ],
     [ 5, 6, 7, 8 ]]
printDiagonalSums(a, 4)

# This code is contributed
# by ihritik

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.
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