Java Solution, Beats 100.00%
Intuition \n\nThe problem requires adding the minimum number of patches (elements) to an array such that any number from 1 to n can be formed by the sum of some
# Intuition \n\nThe problem requires adding the minimum number of patches (elements) to an array such that any number from 1 to `n` can be formed by the sum of some elements in the array. The key is to iteratively ensure that every integer up to `n` can be constructed using the existing elements and the patches added so far.\n\n# Approach\n\n1. **Initialization**:\n   - Initialize `patches` to count the number of patches needed.\n   - Initialize `miss` to track the smallest number that we cannot form yet. Start with `miss = 1`.\n   - Use an index `i` to traverse through the `nums` array.\n\n2. **Iterate Until `miss` Exceeds `n`**:\n   - While `miss <= n`, do the following:\n     - If the current element in `nums` (i.e., `nums[i]`) is less than or equal to `miss`, it means we can extend our range of constructible numbers. Add `nums[i]` to `miss` and move to the next element (`i++`).\n     - If `nums[i]` is greater than `miss`, we need to add a patch (i.e., `miss` itself) to the array. This patch will double the current `miss` value (since `miss` is being added to itself), and increment the patch count (`patches++`).\n\n3. **Return the Number of Patches**:\n   - Once `miss` exceeds `n`, return the total number of patches added.\n\n# Code\n```java\nclass Solution {\n    public int minPatches(int[] nums, int n) {\n        int patches = 0;\n        int i = 0;\n        long miss = 1;\n        while(miss <= n) {\n            if(i < nums.length && nums[i] <= miss)\n                miss += nums[i++];\n            else {\n                miss += miss;\n                patches++;\n            }\n        }\n        return patches;\n    }\n}\n```\n\n#### Detailed Explanation\n\n- **Variables**:\n  - `patches` counts the number of patches added.\n  - `i` is the index for traversing `nums`.\n  - `miss` is the smallest number that cannot be formed yet, initialized to 1.\n\n- **Loop**:\n  - The loop continues until `miss` exceeds `n`.\n  - **Condition**: If the current element in `nums` (`nums[i]`) can be used to form `miss` (i.e., `nums[i] <= miss`), add it to `miss` and move to the next element (`i++`).\n  - **Else**: If `nums[i]` cannot form `miss`, add `miss` as a patch, double `miss`, and increment the patch count.\n\n# Complexity\n\n### Time Complexity\n\n- The loop runs while `miss` is less than or equal to `n`.\n- In each iteration, either:\n  - We use an element from `nums` (which runs in `O(nums.length)`) or,\n  - We double `miss` and increment patches.\n- Since `miss` doubles each time it is patched, the number of iterations in the worst case is `O(log n)`.\n- Thus, the overall time complexity is `O(nums.length + log n)`.\n\n### Space Complexity\n\n- The space complexity is `O(1)` because we are using a constant amount of extra space (i.e., variables `patches`, `i`, and `miss`).\n
['Java']