# LeetCode – Patching Array (Java)

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:

nums = [1, 3], n = 6

Return 1.

**Analysis**

Let miss be the smallest number that can not be formed by the sum of elements in the array. All elements in [0, miss) can be formed. The miss value starts with 1. If the next number nums[i]<=miss, then the boundary is increased to be [0, miss+nums[i]), because all numbers between the boundaries can be formed; if next number nums[i]>miss, that means there is a gap and we need to insert a number, inserting miss itself is a the choice because its boundary doubles and cover every number between the boundaries [0, miss+miss).

Here is an example.

Given nums=[1, 4, 10] and n=50.

miss=1; i=0, nums[i]<=miss, then miss=1+1=2 i=1, nums[i]>2, then miss = miss+miss = 4 i=1, nums[i]<=miss, then miss = miss+num[i] = 8 i=2, nums[i]>miss, then miss = miss+miss = 16 i=2, nums[i]>miss, then miss = miss+miss = 32 i=2, nums[i]>miss, then miss = miss+miss = 64 64 > 50. Done! 4 elements are added!

**Java Solution**

Writing the code is trivial, once we understand the algorithm.

public int minPatches(int[] nums, int n) { long miss = 1; int count = 0; int i = 0; while(miss <= n){ if(i<nums.length && nums[i] <= miss){ miss = miss + nums[i]; i++; }else{ miss += miss; count++; } } return count; } |

<pre><code> String foo = "bar"; </code></pre>

## Leave a comment

can you prove that this greedy algorithms can get the optimal solution?

can you prove this greedy algo get THE SOLUTION

nums = [99,100] , n =50

there is no guarantee that n is bigger than the elements in array

for any p,q nums[p]≤ n ≤ nums[q], elements sum from 1 to p should be <=n

so for above case , 10*11/2 = 55 and we simply eliminate 10 -1 = 9 unless missing is part is smaller than 3