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 (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.
Please refer complete article on Efficiently compute sums of diagonals of a matrix for more details!