0ms 4MB C++ Simple Solution
\nclass Solution {\npublic:\n    vector<int> numMovesStones(int a, int b, int c) {\n\t\tvector<int> pos = {a,b,c};\n        sort(pos.begin(),pos.end());\n      
```\nclass Solution {\npublic:\n    vector<int> numMovesStones(int a, int b, int c) {\n\t\tvector<int> pos = {a,b,c};\n        sort(pos.begin(),pos.end());\n        int x = pos[0], y = pos[1], z = pos[2];\n        int max = (z-y-1)+(y-x-1),min = 1;\n        if(y-x>2 && z-y>2){\n            min = 2;\n        }else if(y-x==1 && z-y==1){\n            min = 0;\n        }\n        vector<int> res = {min,max};\n        return res;\n    }\n};\n```\n\nIn case of maximum moves, in each step we will only take 1 step.\n\nIn case of minimum moves it will be either of 0,1,2\n* case 0 :  if abs(x-y) = 1 and abs(y-z) = 1\n* case 1 :  if abs(x-y) = 1 or abs(y-z) = 1\n* case 2 :  Otherwise\n\n**Time Complexity : O(1)\nSpace Complexity : O(1)**
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