Python3 Program to Maximize sum of diagonal of a matrix by rotating all rows or all columns
Given a square matrix, mat[][] of dimensions N * N, the task is find the maximum sum of diagonal elements possible from the given matrix by rotating either all the rows or all the columns of the matrix by a positive integer.
Examples:
Input: mat[][] = { { 1, 1, 2 }, { 2, 1, 2 }, { 1, 2, 2 } }
Output: 6Â
Explanation:Â
Rotating all the columns of matrix by 1 modifies mat[][] to { {2, 1, 2}, {1, 2, 2}, {1, 1, 2} }.Â
Therefore, the sum of diagonal elements of the matrix = 2 + 2 + 2 = 6 which is the maximum possible.
Input: A[][] = { { -1, 2 }, { -1, 3 } }
Output: 2
Approach: The idea is to rotate all the rows and columns of the matrix in all possible ways and calculate the maximum sum obtained. Follow the steps to solve the problem:
- Initialize a variable, say maxDiagonalSum to store the maximum possible sum of diagonal elements the matrix by rotating all the rows or columns of the matrix.
- Rotate all the rows of the matrix by a positive integer in the range [0, N – 1] and update the value of maxDiagonalSum.
- Rotate all the columns of the matrix by a positive integer in the range [0, N – 1] and update the value of maxDiagonalSum.
- Finally, print the value of maxDiagonalSum.
Below is the implementation of the above approach:
Python3
import sys
N = 3
def findMaximumDiagonalSumOMatrixf(A):
maxDiagonalSum = - sys.maxsize - 1
for i in range (N):
curr = 0
for j in range (N):
curr + = A[j][(i + j) % N]
maxDiagonalSum = max (maxDiagonalSum,
curr)
for i in range (N):
curr = 0
for j in range (N):
curr + = A[(i + j) % N][j]
maxDiagonalSum = max (maxDiagonalSum,
curr)
return maxDiagonalSum
if __name__ = = "__main__" :
mat = [ [ 1 , 1 , 2 ],
[ 2 , 1 , 2 ],
[ 1 , 2 , 2 ] ]
print (findMaximumDiagonalSumOMatrixf(mat))
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Time Complexity: O(N2)Â
Auxiliary Space: O(1)
Please refer complete article on Maximize sum of diagonal of a matrix by rotating all rows or all columns for more details!
Last Updated :
27 Jan, 2022
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