C# Solution for Patching Array Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this solution is that we aim to cover the range of sums from 1 to 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this solution is that we aim to cover the range of sums from 1 to n using the elements in nums and additional patches if necessary. However, this solution uses a dynamic approach to iteratively extend the maximum sum that can be covered (maxSumCovered). If there is a gap (i.e., a number that cannot be formed with the current set of numbers), we add a patch to bridge that gap.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tInitialization:\n\t\u2022\tpatches is initialized to 0 to count the number of patches added.\n\t\u2022\tmaxSumCovered is initialized to 0, representing the largest sum that can be formed with the current set of numbers.\n\t\u2022\ti is initialized to 0 to iterate through the nums array.\n2.\tIteration:\n\t\u2022\tWhile maxSumCovered is less than n:\n\t\u2022\tIf the current element in nums (i.e., nums[i]) is less than or equal to maxSumCovered + 1, it means we can use nums[i] to extend our range of sums. Update maxSumCovered to maxSumCovered + nums[i] and increment i.\n\t\u2022\tIf nums[i] is greater than maxSumCovered + 1 or there are no more elements left in nums, we add maxSumCovered + 1 as a patch. This new element bridges the gap and extends our coverage to 2 * maxSumCovered + 1, and we increment the patches count.\n3.\tReturn:\n\t\u2022\tThe total number of patches added is returned.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this solution is O(m + log n), where:\n\n\t\u2022\tm is the length of the array nums.\n\t\u2022\tlog n comes from the fact that each patching operation doubles the value of maxSumCovered, and the number of such operations required to reach n is log n.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this solution is O(1) since we are using only a fixed amount of additional space (for variables like maxSumCovered, patches, and i).\n\n# Code\n```\npublic class Solution {\n    public int MinPatches(int[] nums, int n) {\n        int patches = 0;\n        long maxSumCovered = 0;\n        int i = 0;\n        \n        while (maxSumCovered < n) {\n            if (i < nums.Length && nums[i] <= maxSumCovered + 1) {\n                maxSumCovered += nums[i];\n                i++;\n            } else {\n                maxSumCovered += maxSumCovered + 1;\n                patches++;\n            }\n        }\n        \n        return patches;\n    }\n}\n```
['C#']