# 330.patching-array

## Statement

``````输入: nums = [1,3], n = 6

``````输入: nums = [1,5,10], n = 20

``````

``````输入: nums = [1,2,2], n = 5

``````

• `1 <= nums.length <= 1000`
• `1 <= nums[i] <= 104`
• `nums` 按 升序排列
• `1 <= n <= 231 - 1`

• Difficulty: Hard
• Tag: `Greedy` `Array`

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`

## Solution

``````from collections import Counter
from typing import List

class Solution:
def minPatches(self, nums: List[int], n: int) -> int:
base = 0
res = 0

for i in nums:
while base + 1 < i and base < n:
base <<= 1
base += 1
res += 1
base += i

while base < n:
base <<= 1
base += 1
res += 1

return res
``````