A robust solution in python#
Intuition\nTo solve the problem of determining the minimum number of patches required to ensure that every number in the range \n[1,n] can be formed by the sum 
# Intuition\nTo solve the problem of determining the minimum number of patches required to ensure that every number in the range \n[1,n] can be formed by the sum of some elements in the sorted array nums, we can use a greedy algorithm.\n# Approach\n**Key Idea:**\nWe maintain a variable *miss* which represents the smallest number that cannot be formed with the current set of elements in nums. Initially, miss is set to 1.\n\nWe iterate through the array and for each number num:\n\n    1) If num is less than or equal to miss, it means we can extend the range of numbers we can form to at least miss + num - 1.\n    2) If num is greater than miss, it means we need to add a number equal to miss to cover the gap.\n\nWe repeat this process until miss exceeds n.\n\nSteps:\n\n1) Initialize miss to 1 and a counter patches to 0.\n2) Iterate through the array nums and for each number:\n    => If num is less than or equal to miss, update miss to miss + num.\n    =>If num is greater than miss, increment patches, add miss to cover the gap, and update miss to miss * 2.\n3) Continue this process until miss exceeds n.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:  \nO(logn) in the worst case, making it efficient for large values of n.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n1. Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n![0cc080cf-616a-4180-b9e6-f46c49f183a8_1718170157.6093411.jpeg](https://assets.leetcode.com/users/images/34e5d744-e0f0-4c55-8124-8bdce1bc03ca_1718507265.4834826.jpeg)\n\n# Code\n```\nfrom typing import List\n\nclass Solution:\n    def minPatches(self, nums: List[int], n: int) -> int:\n        patches = 0\n        miss = 1\n        i = 0\n        \n        while miss <= n:\n            if i < len(nums) and nums[i] <= miss:\n                miss += nums[i]\n                i += 1\n            else:\n                miss += miss\n                patches += 1\n        \n        return patches\n\n\n```
['Iterator', 'Python3']