using adjacent difference.
Intuition\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 
# Intuition\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\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int minOperations(vector<int>& v) \n    {\n       int n=v.size(); \n       int number=0;\n       for(int i=1;i<n;i++)\n       {\n           if(v[i] <= v[i-1]){\n           int p=v[i-1]-v[i];\n           number=number+p+1;\n           v[i]=v[i]+p+1;\n           }\n\n       }\n       return number;\n    }\n};\n```
['C++']