Asked in: INFOTECH
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All Testcases Passed ✔
Solution
def solution(matrix):
n = len(matrix)
first = []
second = []
// ... rest of solution available after purchaseExplanation
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To solve the problem of swapping the longest diagonals of a square matrix, the first step is to clearly understand what the two longest diagonals represent. In a square matrix of size n x n, the first longest diagonal (often called the main diagonal) consists of elements where the row index and column index are the same — that is, positions (0,0), (1,1), ..., (n-1,n-1). The second longest diagonal (often called the secondary diagonal) consists of elements where the row and column indices sum to n-1 — that is, positions (0,n-1), (1,n-2), ..., (n-1,0).
Your goal is to swap the elements along these two diagonals. This means that for every position i (ranging from 0 to n-1), the element at position (i, i) should be swapped with the element at position (i, n-1-i).
Here is a step-by-step approach you should consider:
1. **Understand the Matrix and its Dimensions**
The input is a square matrix, so the number of rows equals the number of columns. Knowing the dimension n allows you to precisely identify indices for diagonal elements.
2. **Identify the Diagonal Positions**
- The first diagonal elements are at (0,0), (1,1), ..., (n-1,n-1).
- The second diagonal elements are at (0,n-1), (1,n-2), ..., (n-1,0).
These two sets of indices are well-defined and symmetric.
3. **Iterate Over the Indices of Diagonals**
Use a single loop from i = 0 to n-1. For each i:
- Access the element at (i, i) — this is from the first diagonal.
- Access the element at (i, n-1-i) — this is from the second diagonal.
The elements at these positions are the ones to be swapped.
4. **Swap the Elements at Corresponding Diagonal Positions**
For each i, exchange the values at (i, i) and (i, n-1-i).
Swapping means temporarily storing one value, assigning the other, and then restoring the first.
5. **Take Care of Middle Element When n is Odd**
When n is odd, the middle element (at index (n//2, n//2)) belongs to both diagonals. This element would get swapped with itself, which is harmless and effectively a no-operation. Just ensure your swapping logic handles this naturally, typically by swapping the element with itself (no change).
6. **In-place Modification vs Creating a New Matrix**
Decide whether you want to swap the elements in-place in the given matrix or create a new matrix and store swapped elements. In-place is generally more efficient for memory. Since swapping involves only elements along the diagonals, you can modify the original matrix directly.
7. **Validation and Testing**
After performing the swaps, validate your implementation by testing with sample matrices:
- A small matrix (e.g., 2x2 or 3x3) to manually verify correctness.
- Larger matrices to check for indexing issues or off-by-one errors.
- Edge cases such as 1x1 matrices where diagonals coincide completely.
8. **Algorithmic Complexity**
This approach runs in O(n) time since you only iterate over the diagonals once. Space complexity is O(1) if done in-place.
In summary, the key idea is to traverse the matrix once, swap corresponding elements from the main and secondary diagonals, and handle special cases carefully. The problem mainly tests your understanding of matrix indexing and simple swapping operations.
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