Intuitive and simple implementation
Intuition\n Describe your first thoughts on how to solve this problem. \nIt is a common implementation problem.\n# Approach\n Describe your approach to solving 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt is a common implementation problem.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe will first find the the min-max limits of region and check if the grid contains 9 elements.Then for each each element in the region, we would check for the condition of adjacency ,i.e, if its adjacent within the region.Those satisfying would be updated with their average. \n\n# Complexity\n- Time complexity: $$O(n*m*8*4)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n*m)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    vector<vector<int>> resultGrid(vector<vector<int>>& image, int threshold) {\n\n        vector<int> dr = {-1, -1, -1, 0, 0, 1, 1, 1};\n        vector<int> dc = {-1, 0, 1, -1, 1, -1, 0, 1};\n        vector<int> dir = {-1, 0, 1, 0, -1};\n        int n = image.size();\n        int m = image[0].size();\n        vector<vector<int>> ans(n, vector<int>(m, 0));\n        vector<vector<int>> mpfr(n, vector<int>(m, 0));\n        for (int i = 0; i < n; i++) {\n            for (int j = 0; j < m; j++) {\n                int total = 0;\n                int cnt = 0;\n                int minr = 1e9, minc = 1e9, maxr = 0, maxc = 0;\n                for (int k = 0; k < 8; k++) {\n                    int nr = i + dr[k];\n                    int nc = j + dc[k];\n                    if (nr >= 0 && nc >= 0 && nr < n && nc < m) {\n                        cnt++;\n                        total += image[nr][nc];\n                        minr = min(nr, minr);\n                        minc = min(nc, minc);\n                        maxr = max(nr, maxr);\n                        maxc = max(nc, maxc);\n                    }\n                }\n                if (cnt == 8) {\n                    total += image[i][j];\n                    int f = 0;\n                    for (int k = 0; k < 8; k++) {\n                        int nr = i + dr[k];\n                        int nc = j + dc[k];\n                        for (int p = 0; p < 4; p++) {\n                            int newr = nr + dir[p];\n                            int newc = nc + dir[p + 1];\n                            if (newr >= minr && newc >= minc && newr <= maxr &&\n                                newc <= maxc) {\n                                int diff =\n                                    abs(image[nr][nc] - image[newr][newc]);\n                                if (diff > threshold) {\n                                    f = 1;\n                                    break;\n                                }\n                            }\n                        }\n                        if (f)\n                            break;\n                    }\n                    if (!f) {\n                        for (int k = 0; k < 8; k++) {\n                            int nr = i + dr[k];\n                            int nc = j + dc[k];\n\n                            ans[nr][nc] += (total / 9);\n                            mpfr[nr][nc]++;\n                        }\n                        ans[i][j] += (total / 9);\n                        mpfr[i][j]++;\n                    }\n                }\n            }\n        }\n        for (int i = 0; i < n; i++) {\n            for (int j = 0; j < m; j++) {\n                if (mpfr[i][j] == 0) {\n                    ans[i][j] = image[i][j];\n                } else {\n                    ans[i][j] /= mpfr[i][j];\n                }\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']