330. Patching Array
Description
Given a sorted integer array nums
and an integer n
, add/patch elements to the array such that any number in the range [1, n]
inclusive can be formed by the sum of some elements in the array.
Return the minimum number of patches required.
Example 1:
Input: nums = [1,3], n = 6 Output: 1 Explanation: Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4. Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3], [1,2,3]. Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6]. So we only need 1 patch.
Example 2:
Input: nums = [1,5,10], n = 20 Output: 2 Explanation: The two patches can be [2, 4].
Example 3:
Input: nums = [1,2,2], n = 5 Output: 0
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 104
nums
is sorted in ascending order.1 <= n <= 231 - 1
Solutions
Solution 1: Greedy
Let's assume that the number $x$ is the smallest positive integer that cannot be represented. Then all the numbers in $[1,..x-1]$ can be represented. In order to represent the number $x$, we need to add a number that is less than or equal to $x$:
- If the added number equals $x$, since all numbers in $[1,..x-1]$ can be represented, after adding $x$, all numbers in the range $[1,..2x-1]$ can be represented, and the smallest positive integer that cannot be represented becomes $2x$.
- If the added number is less than $x$, let's assume it's $x'$, since all numbers in $[1,..x-1]$ can be represented, after adding $x'$, all numbers in the range $[1,..x+x'-1]$ can be represented, and the smallest positive integer that cannot be represented becomes $x+x' \lt 2x$.
Therefore, we should greedily add the number $x$ to cover a larger range.
We use a variable $x$ to record the current smallest positive integer that cannot be represented, initialized to $1$. At this time, $[1,..x-1]$ is empty, indicating that no number can be covered; we use a variable $i$ to record the current index of the array being traversed.
We perform the following operations in a loop:
- If $i$ is within the range of the array and $nums[i] \le x$, it means that the current number can be covered, so we add the value of $nums[i]$ to $x$, and increment $i$ by $1$.
- Otherwise, it means that $x$ is not covered, so we need to supplement a number $x$ in the array, and then update $x$ to $2x$.
- Repeat the above operations until the value of $x$ is greater than $n$.
The final answer is the number of supplemented numbers.
The time complexity is $O(m + \log n)$, where $m$ is the length of the array $nums$. The space complexity is $O(1)$.
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