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Intuition\n\n\n\nC++ []\nclass Solution {\npublic:\n    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& image) {\n        for (auto& row : image)\n 
# Intuition\n![0ehh83fsnh811.jpg](https://assets.leetcode.com/users/images/f40dbdd1-4eda-4eda-9739-e4fb39ecd6ce_1716777765.1924446.jpeg)\n![Screenshot 2024-05-27 080218.png](https://assets.leetcode.com/users/images/e070c1b5-e108-4da6-912a-e9f174760eec_1716777774.2099407.png)\n![Screenshot 2024-05-27 080609.png](https://assets.leetcode.com/users/images/a52dc047-2394-42e1-9daf-03e683698379_1716777780.1087859.png)\n```C++ []\nclass Solution {\npublic:\n    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& image) {\n        for (auto& row : image)\n            reverse(row.begin(), row.end());\n        for (auto& x : image) {\n            for (auto& y : x)\n                if (y == 0)\n                    y = 1;\n                else\n                    y = 0;\n        }\n        return image;\n    }\n};\n```\n```python3 []\nclass Solution:\n    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:\n        n = len(image)\n        for i in range(n):\n            for j in range((n + 1) // 2):\n                temp = image[i][j]\n                image[i][j] = image[i][n - j - 1] ^ 1\n                image[i][n - j - 1] = temp ^ 1\n                \n        return image\n```\n```Java []\nclass Solution {\n    public int[][] flipAndInvertImage(int[][] A) {\n        int n = A.length;\n\n        for (int i = 0; i < n; ++i)\n            for (int j = 0; j < (n + 1) / 2; ++j) {\n                int temp = A[i][j];\n                A[i][j] = A[i][n - j - 1] ^ 1;\n                A[i][n - j - 1] = temp ^ 1;\n            }\n\n        return A;\n    }\n}\n```\n```C []\n/**\n * Return an array of arrays of size *returnSize.\n * The sizes of the arrays are returned as *returnColumnSizes array.\n * Note: Both returned array and *columnSizes array must be malloced, assume\n * caller calls free().\n */\nint** flipAndInvertImage(int** A, int ASize, int* AColSize, int* returnSize, int** returnColumnSizes) {\n    int i = 0, j = 0;\n    int n = ASize;\n    int m = AColSize[0];\n    int temp = 0;\n\n    *returnColumnSizes = malloc(sizeof(int) * ASize);\n    *returnSize = ASize;\n    for (i = 0; i < ASize; i++) {\n        (*returnColumnSizes)[i] = m;\n    }\n\n    for (i = 0; i < n; i++) {\n        for (j = 0; j < m / 2; j++) {\n            temp = A[i][ASize - 1 - j];\n            A[i][ASize - 1 - j] = A[i][j];\n            A[i][j] = temp;\n        }\n    }\n\n    for (i = 0; i < n; i++) {\n        for (j = 0; j < m; j++) {\n            if (A[i][j] == 0) {\n                A[i][j] = 1;\n            } else {\n                A[i][j] = 0;\n            }\n        }\n    }\n    return A;\n}\n```\n```python []\nclass Solution(object):\n    def flipAndInvertImage(self, image):\n        """\n        :type image: List[List[int]]\n        :rtype: List[List[int]]\n        """\n        n = len(image)\n        for i in range(n):\n            for j in range((n + 1) // 2):\n                temp = image[i][j]\n                image[i][j] = image[i][n - j - 1] ^ 1\n                image[i][n - j - 1] = temp ^ 1\n                \n        return image\n```\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n       O(n*m)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n       O(1)\n# Code\n```\n/**\n * Return an array of arrays of size *returnSize.\n * The sizes of the arrays are returned as *returnColumnSizes array.\n * Note: Both returned array and *columnSizes array must be malloced, assume\n * caller calls free().\n */\nint** flipAndInvertImage(int** A, int ASize, int* AColSize, int* returnSize, int** returnColumnSizes) {\n    int i = 0, j = 0;\n    int n = ASize;\n    int m = AColSize[0];\n    int temp = 0;\n\n    *returnColumnSizes = malloc(sizeof(int) * ASize);\n    *returnSize = ASize;\n    for (i = 0; i < ASize; i++) {\n        (*returnColumnSizes)[i] = m;\n    }\n\n    for (i = 0; i < n; i++) {\n        for (j = 0; j < m / 2; j++) {\n            temp = A[i][ASize - 1 - j];\n            A[i][ASize - 1 - j] = A[i][j];\n            A[i][j] = temp;\n        }\n    }\n\n    for (i = 0; i < n; i++) {\n        for (j = 0; j < m; j++) {\n            if (A[i][j] == 0) {\n                A[i][j] = 1;\n            } else {\n                A[i][j] = 0;\n            }\n        }\n    }\n    return A;\n}\n```\n![th.jpeg](https://assets.leetcode.com/users/images/0a4b32d3-8a2f-4098-b266-295df0b8a289_1716777889.0881238.jpeg)\n
['Array', 'Two Pointers', 'C', 'Matrix', 'Simulation', 'Python', 'C++', 'Java', 'Python3']