🔥 🔥 🔥 💯 Easy to understand || Greedy Approach || Detailed Explanation || Python,C++,Java🔥 🔥 🔥
Intuition\n Describe your first thoughts on how to solve this problem. \n The code works like providing change with limited coin denominations. Suppose you need
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n* The code works like providing change with limited coin denominations. Suppose you need to cover every amount up to \uD835\uDC5B cents. If you can\'t make exact change for a particular amount miss, it indicates you lack a coin of value less than or equal to miss. To fill this gap, you add a coin of that exact missing amount. This addition allows you to now cover new amounts up to 2 * miss. Repeat this process until you can provide change for every amount up to \uD835\uDC5B. This method ensures you add the minimum number of new coins needed to cover any shortages.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n* **Initialize Variables:** Start with miss set to 1, added to 0, and index i at 0 to track the smallest sum that cannot be formed and the number of patches added.\n\n* **Iterate with Condition:** Continue the loop as long as miss (the target sum we need to achieve) is less than or equal to n.\n\n* **Use Existing Numbers:** Check if the current number in the list (nums[i]) can be used to form miss. If yes, add it to miss to extend the range of formable sums and increment the index i.\n\n* **Add Necessary Numbers:** If nums[i] is greater than miss or all numbers are already used, add miss itself to cover the smallest unformable sum, and double the value of miss.\n\n* **Increment Added Count:** Whenever a number is added to cover a gap, increase the added counter.\n\n* **Return Total Patches:** Once miss exceeds n, return the total count of added numbers as the result, representing the minimum patches needed to form every number up to n.\n\n# Example:\n\n* Input: [1, 5, 10]\n\nStep\tCurrent nums\tMiss\tAction\tReason\tNew Coverage\n1. \t[1, 5, 10]\t1\tUse 1\t1 is available and matches miss\tSums up to 1\n2.\t[1, 5, 10]\t2\tAdd 2\t5 is too large to form sum 2\tSums up to 3\n3.\t[1, 5, 10]\t4\tAdd 4\t5 is too large to form sum 4\tSums up to 7\n4.\t[1, 5, 10]\t8\tUse 5\tCan use 5 with existing numbers to form 8\tSums up to 12\n5.\t[1, 5, 10]\t13\tUse 10\tCan use 10 with existing numbers to form 13\tSums up to 22\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n    O(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n    O(1)\n\n```python []\nclass Solution:\n    def minPatches(self, nums: List[int], n: int) -> int:\n        c=1\n        res=0\n        i=0\n        while c<=n:\n            if i<len(nums) and nums[i]<=c:\n                c+=nums[i]\n                i+=1\n            else:\n                c+=c\n                res+=1\n        return res\n```\n```C++ []\nclass Solution {\npublic:\n    int minPatches(vector<int>& nums, int n) {\n        long long miss = 1;\n        int result = 0;\n        size_t i = 0;\n\n        while (miss <= n) {\n            if (i < nums.size() && nums[i] <= miss) {\n                miss += nums[i];\n                i++;\n            } else {\n                miss += miss;\n                result++;\n            }\n        }\n\n        return result;\n    }\n};\n```\n```java []\nclass Solution {\n    public int minPatches(int[] nums, int n) {\n        long miss = 1;\n        int result = 0;\n        int i = 0;\n\n        while (miss <= n) {\n            if (i < nums.length && nums[i] <= miss) {\n                miss += nums[i];\n                i++;\n            } else {\n                miss += miss;\n                result++;\n            }\n        }\n\n        return result;\n    }\n}\n```\n\n
['C++', 'Java', 'Python3']