Given a 2D binary matrix of N rows and M columns. The task is to check whether the matrix is horizontal symmetric, vertical symmetric or both. The matrix is said to be horizontal symmetric if the first row is same as the last row, the second row is same as the second last row and so on. And the matrix is said to be vertical symmetric if the first column is same as the last column, the second column is same as the second last column and so on. Print “VERTICAL” if the matrix is vertical symmetric, “HORIZONTAL” if the matrix is vertical symmetric, “BOTH” if the matrix is vertical and horizontal symmetric and “NO” if not symmetric.
Input : N = 3 M = 3 0 1 0 0 0 0 0 1 0 Output : Both First and third row are same and also second row is in middle. So Horizontal Symmetric. Similarly, First and third column are same and also second column is in middle, so Vertical Symmetric. Input : 0 0 1 1 1 0 0 0 1. Output : Horizontal
The idea is to use to pointers indicating two rows (or columns) and compare each cell of both the pointed rows (or columns).
For Horizontal Symmetry, initialize one pointer i = 0 and another pointer j = N – 1.
Now, compare each element of i-th row and j-th row. Increase i by 1 and decrease j by 1 in each loop cycle. If at least one not identical element is found, mark matrix as not horizontal symmetric.
Similarly, for Vertical Symmetry, initialize one pointer i = 0 and another pointer j = M – 1.
Now, compare each element of i-th column and j-th column. Increase i by 1 and decrease j by 1 in each loop cycle. If at least one not identical element is found, mark matrix as not vertical symmetric.
Below is the implementation of above idea :
Time Complexity: O(N*M).
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