c++, simple brute-force solution explained
Approach:\n\nInitialization:\n\nDefine variables for image dimensions (n, m) and threshold (threshold).\nCreate a 3D vector res to store intermediate results fo
# Approach:\n\n**Initialization:**\n\nDefine variables for image dimensions (n, m) and threshold (threshold).\nCreate a 3D vector res to store intermediate results for each pixel.\nCreate vectors indx and indy for neighboring pixel coordinates.\n\n**Iterate through 3x3 regions:**\n\n- Iterate through each 3x3 region of the image (excluding border pixels).\n- For each region:\n    1. Check for smoothness:\n        - Calculate the absolute differences between adjacent pixels within the region.\n        - If any difference exceeds the threshold, consider the region "not smooth" and move to the next region.\n    1. Calculate average:\n        - If the region is smooth, calculate the average of all pixel values within the region.\n        - Store this average in the res vector for each pixel within the region.\n    1. Combine results and handle overlapping regions:\n\n        - Create a 2D vector ans to store the final results.\n        - For each pixel:\n            1. If it belongs to multiple regions (multiple averages in res), calculate the overall average of those values.\n            1. If it doesn\'t belong to any region, keep its original value from the input image.\n    1. Return the final result:\n\n        - Return the ans vector containing the modified pixel values.\n\n# Intuition:\n\n1. The code aims to create a "smoothed" version of the input image by:\n    - Identifying 3x3 regions that are relatively uniform in pixel values (differences below a threshold).\n    - Replacing the pixels within those regions with their average value.\n    - Handling cases where pixels belong to multiple smooth regions.\n\n# Time Complexity:\n\n1. The nested loops iterate over all pixels and their neighbors, resulting in a time complexity of O(n * m) where n and m are the image dimensions.\n1. However, the potential inner loop for checking smoothness typically breaks early, effectively reducing the overall complexity to O(n * m).\n\n# Key Points:\n\n1. The threshold value plays a crucial role in determining the level of smoothing.\n1. The code effectively addresses overlapping regions by averaging multiple calculated averages for a pixel.\n1. The time complexity is reasonable for image processing tasks.\n\n# Code\n```\nclass Solution {\npublic:\n    vector<vector<int>> resultGrid(vector<vector<int>>& image, int threshold) {\n        int n = image.size();\n        int m = image[0].size();\n        \n        vector<vector<vector<int>>> res(n, vector<vector<int>>(m));\n        \n        vector<int> indx = {0, -1, +1, 0};\n        vector<int> indy = {-1, 0, 0, +1};\n        for(int i = 0; i < n - 2; i++){\n            for(int j = 0 ; j < m - 2; j++){\n                int sum = 0;\n                int c = 0;\n                for(int x = i; x < i + 3; x++){\n                    for(int y = j; y < j + 3; y++){\n                        for(int k = 0; k < 4; k++){\n                            if(x + indx[k] >= i && x + indx[k] < i + 3 && y + indy[k] >= j && y + indy[k] < j + 3){\n                                if(abs(image[x][y] - image[x + indx[k]][y + indy[k]]) > threshold){c = 1; break;}\n                            }\n                        } if(c == 1) break;\n                        else sum += image[x][y];\n                    } if(c == 1) break;\n                } if(c == 1) sum = 0;\n                else{\n                    for(int p = i; p < i + 3; p++){\n                    for(int q = j; q < j + 3; q++){\n                        res[p][q].push_back(sum/9);\n                    }\n                }  \n            }\n        }\n    }\n        \n        vector<vector<int>> ans(n, vector<int>(m));\n                    for(int x = 0; x < n; x++){\n                    for(int y = 0; y < m; y++){\n                        int z = res[x][y].size();\n                        int tt = 0;\n                        for(auto it : res[x][y]) tt += it;\n                        if(z == 0) ans[x][y] = image[x][y];\n                        else ans[x][y] = tt/z;\n                    }\n                } return ans;\n    }\n};\n```
['C++']