C++ Solution
Intuition\nFor every element, which is less than or equal to its predecessor, we will perform the required number of operations to make it strictly increasing.\
# Intuition\nFor every element, which is less than or equal to its predecessor, we will perform the required number of operations to make it strictly increasing.\n\n# Approach\nFirst we will declare a variable m, and will assign the nums[0] value to it. Now, traverse the array from left to right, for every element, if nums[i] <= m , then we increment the i-th element and add the required no.of steps to the result.\n\n# Complexity\n- Time complexity:\n  O(N)\n\n- Space complexity:\n  O(1)\n\n# Code\n```\nclass Solution {\npublic:\n    int minOperations(vector<int>& nums) {\n        if(nums.size()==1)\n            return 0;\n        int ans=0, m=nums[0];\n        for(int i=1; i<nums.size(); i++){\n            if(nums[i]<=m){\n                ans+=(m+1-nums[i]);\n                m+=1;\n            }\n            else{\n                m=nums[i];\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']