Easy to Understand | Detailed Explaination
Intuition\nThe goal is to check if shifting the elements in the even and odd rows of the given matrix mat by k units will result in the rows being similar. Spec
# Intuition\nThe goal is to check if shifting the elements in the even and odd rows of the given matrix `mat` by `k` units will result in the rows being similar. Specifically, the even rows are shifted left, and the odd rows are shifted right. The problem can be thought of as comparing rows in a matrix after circular shifts. For this, we need to handle the wraparound effect when shifting elements past the matrix boundaries.\n\n# Approach\n1. **Handling the Shifts**: \n   - We need to perform circular shifts on each row, which means after shifting elements, if an index goes beyond the matrix boundary, it should wrap around to the start of the row. This is handled using modulo operations.\n   - The shift behaves differently for even and odd rows:\n     - **Even rows** (`i % 2 == 0`): Elements are shifted to the left.\n     - **Odd rows** (`i % 2 == 1`): Elements are shifted to the right.\n2. **Modulo Calculation**: \n   - For each element in the matrix, we calculate its corresponding position after the shift:\n     - For even rows, the new index is calculated as `(j - k + col) % col`.\n     - For odd rows, the new index is calculated as `(j + k) % col`.\n   - The value at the original position is compared with the value at the new position. If any mismatch occurs, the function returns `false`.\n3. **Edge Case**: \n   - If `k` is zero or a multiple of the number of columns, no shifts are needed, and the function directly returns `true` since all rows would be similar to themselves.\n\n# Complexity\n- **Time complexity**: \n  - The matrix has `row` rows and `col` columns. We iterate over all the elements of the matrix and perform constant-time operations for each element (mainly index calculation and comparison).\n  - Thus, the time complexity is $$O(row \\times col)$$, where `row` is the number of rows in the matrix and `col` is the number of columns.\n\n- **Space complexity**: \n  - The space complexity is $$O(1)$$ since we only use a few extra variables for indexing and do not require additional space proportional to the input size.\n\n# Code\n```java\nclass Solution {\n    public boolean areSimilar(int[][] mat, int k) {\n        // Normalize k to handle cases where k is larger than the number of columns\n        k = k % mat[0].length;\n        \n        // If k is 0, no shifts are needed and the matrix is already similar\n        if (k == 0) return true;\n        \n        int row = mat.length, col = mat[0].length;\n        \n        // Loop through each element in the matrix\n        for (int i = 0; i < row; i++) {\n            for (int j = 0; j < col; j++) {\n                \n                // For even rows, compare with the left-shifted element\n                if (i % 2 == 0 && mat[i][j] != mat[i][(j - k + col) % col]) {\n                    return false;\n                }\n                \n                // For odd rows, compare with the right-shifted element\n                if (i % 2 == 1 && mat[i][j] != mat[i][(j + k) % col]) {\n                    return false;\n                }\n            }\n        }\n        // If no mismatches found, the rows are similar\n        return true;\n    }\n}\n```\n\n
['Java']