Problem Explanation

Given an array of positive integers in increasing order and a target integer N, our goal is to be able to form all numbers from 1 to N (inclusive) using the sum of any combination of elements from the array. If the array as it is, cannot form all numbers from 1 to N, then we need to add elements to the array so that this becomes possible. The problem asks for the minimum number of such additional elements we need to add.

For example, if we have the array [1, 3] and N=6. We cannot form the number 2 and 6 using the current elements, so we need to add 2 to the array. The minimum number of additional elements (patches) needed here is 1 which is 2.

Solution Approach

To solve this problem, we follow a Greedy strategy. Start with a variable miss = 1, this variable represents the minimum sum in range [1,n] that we might miss. Now, iterate through the array, where at each step, we:

• Check if the current array element is smaller than or equal to miss
  • If yes, include this array element in the subset. So, we add the value of the current array element to miss. If we have traverse the array till the end, still we are unable to reach to the n then we can add a number into the array which is equal to miss. Then miss would be 2*miss
  • If no, the number miss is the number we need to patch/add. So, add miss to miss itself and increment counter of patches.

Continue this process until miss is smaller than or equal to n. Our answer will be the count of these patches.

The logic behind this approach relies on the fact that any n positive numbers can at most cover the range from 1 to n * (n + 1) / 2.

Example Walkthrough

Let's consider an example where we have the array [1, 3] and n=6.

At start, miss = 1 and we have 0 patches.

• At index 0, nums[0] = 1 <= miss = 1, so we add nums[0] to miss and miss becomes 2.
• Now, there is no 2 in the array and 2 <= miss = 2, so we add 2 to miss and increment patches by 1. Now, miss becomes 4.
• Next, at index 1, nums[1] = 3 <= miss = 4, so we add nums[1] to miss and miss becomes 7.
• Now miss = 7 is more than n = 6, so we stop here. Our answer will be patches = 1.

Python Solution

1
2python
3class Solution:
4    def minPatches(self, nums: List[int], n: int) -> int:
5        miss, patches, i = 1, 0, 0
6        size = len(nums)
7        while miss <= n:
8            if i < size and nums[i] <= miss:
9                miss += nums[i]
10                i += 1
11            else:
12                miss += miss
13                patches += 1
14        return patches

Java Solution

1
2java
3class Solution {
4    public int minPatches(int[] nums, int n) {
5        long miss = 1;
6        int patches = 0, i = 0;
7        
8        while (miss <= n) {
9            if (i < nums.length && nums[i] <= miss) {
10                miss += nums[i];
11                i++;
12            } else {
13                miss += miss;
14                patches++;
15            }
16        }
17        
18        return patches;
19    }
20}

JavaScript Solution

1
2javascript
3var minPatches = function(nums, n) {
4    let miss = 1, patches = 0, i = 0;
5    while (miss <= n) {
6        if (i < nums.length && nums[i] <= miss) {
7            miss += nums[i++];
8        } else {
9            miss += miss;
10            patches++;
11        }
12    }
13    return patches;
14};

C# Solution

1
2csharp
3public class Solution {
4    public int MinPatches(int[] nums, int n) {
5        int patches = 0, i = 0;
6        long miss = 1;
7        while (miss <= n) {
8            if (i<nums.Length && nums[i] <= miss) {
9                miss += nums[i++];
10            } else {
11                miss += miss;
12                patches++;
13            }
14        }
15        return patches;
16    }
17}

C++ Solution

1
2cpp
3class Solution {
4public:
5    int minPatches(vector<int>& nums, int n) {
6        long miss = 1, patches = 0, i = 0;
7        
8        while (miss <= n) {
9            if (i < nums.size() && nums[i] <= miss) {
10                miss += nums[i++];
11            } else {
12                miss += miss;
13                patches++;
14            }
15        }
16        
17        return patches;
18    }
19};

Time Complexity

The main operation of this algorithm is iterating over the array and comparing each element with the 'miss' variable. Hence, the overall time complexity would be O(n) for all solutions.

Space Complexity

Since this algorithm does not require a significant amount of extra space, the space complexity for Python, Java, JavaScript, C#, and C++ solutions is O(1).

Some Points to Note

In the Java solution, use long instead of int to prevent overflow.

In Python, you don't need to define size separately; this is because dynamically-typed languages such as Python fetch the length of the list on the fly.

In JavaScript, ensure that you increment your i within the conditional statement to prevent out of bounds exceptions.

In the C# and C++ solutions, make sure to use long instead of int to prevent overflow as well.

You can factor in exceptions; for instance, what should happen when n is 0 or nums is empty? To make your function safer, include checks for these cases at the beginning of your function.

Overall, the concept of this problem can be hard to grasp at first, particularly the greedy strategy and its implementation. It might help to visualize or walk through the process with several examples until the process makes sense to you. The payoff is an elegant, efficient approach to a complex problem.