Simple, Concise Detailed Descriptive Solution !!!
Intuition\nThe problem requires checking if cyclically shifting the rows of a matrix k times results in the matrix being identical to the original one. We can l
# Intuition\nThe problem requires checking if cyclically shifting the rows of a matrix `k` times results in the matrix being identical to the original one. We can leverage the properties of cyclic shifts and directly compare the original matrix with its cyclically shifted version.\n\nBy reducing `k` using modulo with the row length, we ensure we are only concerned with shifts that actually affect the row\'s arrangement. We don\'t need to differentiate between left or right shifts (for even and odd-indexed rows, respectively) because the goal is to see if the matrix returns to its original state after `k` shifts. If the values match after a cyclic shift, that\'s sufficient to return `True`.\n\n# Approach\n1. **Modulo Operation**: First, we reduce `k` by applying `k %= n`, where `n` is the number of columns in the matrix. This simplifies the problem by eliminating unnecessary full rotations.\n  \n2. **Row Comparison**: \n   - Iterate over each row in the matrix.\n   - For each row, compare each element with its counterpart after the `k`-shifted position.\n   - If the row values do not match their shifted counterparts, return `False`.\n\n3. **Final Return**: If all rows pass the comparison, the matrix is identical after `k` shifts, so return `True`.\n\n# Complexity\n- **Time complexity**: \n  The time complexity is $$O(m \\times n)$$, where `m` is the number of rows and `n` is the number of columns. This is because we need to iterate over each element of the matrix to check for cyclic similarity.\n\n- **Space complexity**: \n  The space complexity is $$O(1)$$ as no extra space is used except for a few variables.\n\n# Why This Works\nThe solution works because cyclically shifting a row `k` times either left or right will eventually yield the same result modulo the length of the row. Therefore, as long as we compare the original row and its cyclically shifted version without caring whether it was a left or right shift, the values should match if the matrix returns to its original state after `k` shifts.\n\nSince we are comparing each row element to its shifted counterpart, we don\'t need to explicitly distinguish between even (left shift) or odd (right shift) row indices\u2014just checking if the values match after the shift is enough.\n\n\n# Code\n\n```python3 []\nclass Solution:\n    def areSimilar(self, mat: List[List[int]], k: int) -> bool:\n        # hey so we can just compare original and the cyclic shift dont need to compare each shift seperatly \n        k %= len(mat[0])\n\n        for row in mat:\n            for j in range(len(mat[0])):\n                if row[j] != row[(j + k) % len(mat[0])]:\n                    return False\n\n        return True  \n```
['Python3']