# 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

• class Solution {
public int minPatches(int[] nums, int n) {
long x = 1;
int ans = 0;
for (int i = 0; x <= n;) {
if (i < nums.length && nums[i] <= x) {
x += nums[i++];
} else {
++ans;
x <<= 1;
}
}
return ans;
}
}

• class Solution {
public:
int minPatches(vector<int>& nums, int n) {
long long x = 1;
int ans = 0;
for (int i = 0; x <= n;) {
if (i < nums.size() && nums[i] <= x) {
x += nums[i++];
} else {
++ans;
x <<= 1;
}
}
return ans;
}
};

• class Solution:
def minPatches(self, nums: List[int], n: int) -> int:
x = 1
ans = i = 0
while x <= n:
if i < len(nums) and nums[i] <= x:
x += nums[i]
i += 1
else:
ans += 1
x <<= 1
return ans


• func minPatches(nums []int, n int) (ans int) {
x := 1
for i := 0; x <= n; {
if i < len(nums) && nums[i] <= x {
x += nums[i]
i++
} else {
ans++
x <<= 1
}
}
return
}

• function minPatches(nums: number[], n: number): number {
let x = 1;
let ans = 0;
for (let i = 0; x <= n; ) {
if (i < nums.length && nums[i] <= x) {
x += nums[i++];
} else {
++ans;
x *= 2;
}
}
return ans;
}