Python Easy and Efficient Solution .
Intuition\n Describe your first thoughts on how to solve this problem. \nWe have to Return the minimum number of operations needed to make nums strictly increas
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe have to Return the minimum number of operations needed to make nums strictly increasing.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTaking the previous value and comparing with the current value and increasing the count with the previous value to meet the rule of \n*       nums[i] < nums[i+1] for all 0 <= i < nums.length - 1\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe code uses a single loop that iterates through the nums list, and for each element, it performs some constant time operations. Within the loop, there are a few simple conditional checks and arithmetic operations. The time complexity of these operations is linear in the length of the nums list, so the overall time complexity of this code is O(n), where \'n\' is the length of the nums list.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this code is primarily determined by the variables used. There are a few integer variables (count and prev) that are used to store intermediate results, and their memory usage does not depend on the size of the nums list. Therefore, the space complexity is O(1), which means it is constant and does not depend on the input size.\n\n\n# Code\n```\nclass Solution:\n    def minOperations(self, nums: List[int]) -> int:\n        if len(nums)==1:\n            return 0\n        count = prev= 0\n        for i in nums:\n            if i <= prev :\n                prev += 1\n                count += prev-i\n            else:\n                prev = i\n        return count\n```
['Python3']