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:
nums = [1, 3], n = 6
Return 1.

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:
nums = [1, 5, 10], n = 20
Return 2.
The two patches can be [2, 4].

Example 3:
nums = [1, 2, 2], n = 5
Return 0.

Understand the problem:

show the algorithm with an example,

let nums=[1 2 5 6 20], n = 50.

Initial value: with 0 nums, we can only get 0 maximumly.

Then we need to get 1, since nums[0]=1, then we can get 1 using [1]. now the maximum number we can get is 1. (actually, we can get all number no greater than the maximum number)

number used [1], number added []
can achieve 1~1

Then we need to get 2 (maximum number +1). Since nums[1]=2, we can get 2. Now we can get all number between 1 ~ 3 (3=previous maximum value + the new number 2). and 3 is current maximum number we can get.

number used [1 2], number added []
can achieve 1~3

Then we need to get 4 (3+1). Since nums[2]=5>4; we need to add a new number to get 4. The optimal solution is to add 4 directly. In this case, we could achieve maximumly 7, using [1,2,4].

number used [1 2], number added [4]
can achieve 1~7

Then we need to get 8 (7+1). Since nums[2]=5<8, we can first try to use 5. Now the maximum number we can get is 7+5=12. Since 12>8, we successfully get 8.

number used [1 2 5], number added [4]
can achieve 1~12

Then we need to get 13 (12+1), Since nums[3]=6<13, we can first try to use 6. Now the maximum number we can get is 12+6=18. Since 18>13, we successfully get 13.

number used [1 2 5 6], number added [4]
can achieve 1~18

Then we need to get 19 (18+1), Since nums[4]=20>19, we need to add a new number to get 19. The optimal solution is to add 19 directly. In this case, we could achieve maximumly 37.

number used [1 2 5 6], number added [4 19]
can achieve 1~37

Then we need to get 38(37+1), Since nums[4]=20<38, we can first try to use 20. Now the maximum number we can get is 37+20=57. Since 57>38, we successfully get 38.

number used [1 2 5 6 20], number added [4 19]
can achieve 1~57

Since 57>n=50, we can all number no greater than 50.

The extra number we added are 4 and 19, so we return 2.

The code is given as follows

Code (Java):

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