100% EASY TO UNDERSTAND AND BGN FRIENDLY
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nThis solution checks if
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThis solution checks if a matrix mat is similar to its original form after performing a series of row shifts (leftShift or rightShift) for a given number of times k. The matrix\'s even-indexed rows are right-shifted, while the odd-indexed rows are left-shifted. After performing the shifts, the modified matrix is compared to the original one to check if they are identical.\n\nHere\u2019s a detailed breakdown of how the code works:\n\n1. Creating a Deep Copy of the Original Matrix\n\nint[][] original = new int[mat.length][mat[0].length];\n- for (int i = 0; i < mat.length; i++) {\n\n---\n\n    for (int j = 0; j < mat[i].length; j++) {\n        original[i][j] = mat[i][j];\n    }\n}\n\n---\n\n- **Purpose**:   The code creates a deep copy of the mat matrix. A deep copy is necessary because if we just assign original = mat, both variables will point to the same matrix, and changes in mat will affect original.\nThe original matrix will hold the unmodified version of mat, allowing us to compare the modified matrix with its original form after performing the shifts.\n2. Shifting the Rows\n\n- for (int i = 0; i < mat.length; i++) {\n\n---\n\n    for (int j = 0; j < k; j++) {\n        if (i == 0 || i % 2 == 0) {\n            rightShift(mat[i]);\n        } else {\n            leftShift(mat[i]);\n        }\n    }\n}\n\n\n\nPurpose: This loop performs the shifting of rows.\n\n\n\nRow selection: Rows with even indices (i == 0 or i % 2 == 0) are right-shifted, while rows with odd indices are left-shifted.\n\n\n\n- Nested loop: The inner loop runs k times, shifting each row multiple times as specified by the input k.\n\n3. Left and Right Shift Operations\n\n\nLeft Shift:\n\n```\npublic void leftShift(int[] nums) {\n    int first = nums[0];\n    for (int i = 0; i < nums.length - 1; i++) {\n        nums[i] = nums[i + 1];\n    }\n    nums[nums.length - 1] = first;\n}\n```\n---\nPurpose: This method shifts the elements of the array nums one position to the left.\n\n\nMechanism:\n- \nThe first element (nums[0]) is saved in a temporary variable first.\nThen, all other elements are shifted one position to the left.\nFinally, the first element is placed at the last position (nums[nums.length - 1]).\n\n\nRight Shift:\n\n```\npublic void rightShift(int[] nums) {\n    int last = nums[nums.length - 1];\n    for (int i = nums.length - 1; i > 0; i--) {\n        nums[i] = nums[i - 1];\n    }\n    nums[0] = last;\n}\n```\n\nPurpose: This method shifts the elements of the array nums one position to the right.\n\nMechanism:\n\n- The last element (nums[nums.length - 1]) is saved in a temporary variable last.\nAll other elements are shifted one position to the right.\nThe last element is placed at the first position (nums[0]).\n**4. Comparing the Modified Matrix with the Original**\n\n---\n\n\n     return areMatricesEqual(original, mat);\n```javascript []\nconsole.log(\'Hello world!\')\n```\n```python []\nprint(\'Hello world!\')\n```\n```ruby []\nputs \'Hello world!\'\n```\nPurpose: \n\n\nAfter the shifts, this method compares the modified matrix (mat) with the original matrix (original) to see if they are identical.\nMatrix Comparison Method:\n\n```\npublic boolean areMatricesEqual(int[][] mat1, int[][] mat2) {\n    for (int i = 0; i < mat1.length; i++) {\n        for (int j = 0; j < mat1[i].length; j++) {\n            if (mat1[i][j] != mat2[i][j]) {\n                return false;\n            }\n        }\n    }\n    return true;\n}\n```\nPurpose: This method checks if two 2D matrices mat1 and mat2 are the same.\n\n\nMechanism:\n\nIt iterates through every element of both matrices.\n\nIf any element is different between the two matrices, it returns false.\n\nIf all elements match, it returns true.\n\nOverall Flow:\nA deep copy of the matrix is created to preserve the original state.\nThe matrix rows are shifted k times \u2014 even rows are right-shifted, and odd rows are left-shifted.\nAfter the shifts, the modified matrix is compared with the original.\n- If the modified matrix matches the original, the method returns true; otherwise, it returns false.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n> Quote\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n    public boolean areSimilar(int[][] mat, int k) {\n        \n        int[][] original = new int[mat.length][mat[0].length];\n        for (int i = 0; i < mat.length; i++) {\n            for (int j = 0; j < mat[i].length; j++) {\n                original[i][j] = mat[i][j];\n            }\n        }\n\n       \n        for (int i = 0; i < mat.length; i++) {\n            for (int j = 0; j < k; j++) {\n                if (i == 0 || i % 2 == 0) {\n                    rightShift(mat[i]);\n                } else {\n                    leftShift(mat[i]);\n                }\n            }\n        }\n\n       \n        return areMatricesEqual(original, mat);\n    }\n\n    \n    public void leftShift(int[] nums) {\n        int first = nums[0];\n        for (int i = 0; i < nums.length - 1; i++) {\n            nums[i] = nums[i + 1];\n        }\n        nums[nums.length - 1] = first;\n    }\n\n   \n    public void rightShift(int[] nums) {\n        int last = nums[nums.length - 1];\n        for (int i = nums.length - 1; i > 0; i--) {\n            nums[i] = nums[i - 1];\n        }\n        nums[0] = last;\n    }\n\n    \n    public boolean areMatricesEqual(int[][] mat1, int[][] mat2) {\n        for (int i = 0; i < mat1.length; i++) {\n            for (int j = 0; j < mat1[i].length; j++) {\n                if (mat1[i][j] != mat2[i][j]) {\n                    return false;\n                }\n            }\n        }\n        return true;\n    }\n}\n\n```
['Java']