# Leetcode 330: Patching Array

Given a sorted positive integer array *nums* and an integer *n*, add/patch elements to the array such that any number in 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

**Problem Analysis:**

Suppose k is the smallest number missed in nums, [1, …, k) is covered, so add the number k we can cover [1, …, 2*k), if existing nums[i] that k≤nums[i]<2*k, then we can update covering range to [1, …, 2*k+nums[i]).

Iterating the procedure and update until reaching n, we can find the minimum number of patches.

**Solution:**

`class Solution {`

public:

int minPatches(vector<int>& nums, int n) {

long long miss = 1;

int count = 0;

int i = 0;

while (miss <= n) {

if (i < nums.size() && nums[i] <= miss) {

miss += nums[i++];

} else {

miss += miss;

++count;

}

}

return count;

}

};

Time Complexity: O(n)

Space Complexity: O(1)