https://leetcode.com/problems/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.

https://discuss.leetcode.com/topic/35709/share-my-thinking-process

Share my thinking process

The question asked for the “minimum number of patches required”. In other words, it asked for an optimal solution. Lots of problems involving optimal solution can be solved by dynamic programming and/or greedy algorithm. I started with greedy algorithm which is conceptually easy to design. Typically, a greedy algorithm needs selection of best moves for a subproblem. So what is our best move?

Think about this example: nums = [1, 2, 3, 9]. We naturally want to iterate through nums from left to right and see what we would discover. After we encountered 1, we know 1…1 is patched completely. After encountered 2, we know 1…3 (1+2) is patched completely. After we encountered 3, we know 1…6 (1+2+3) is patched completely. After we encountered 9, the smallest number we can get is 9. So we must patch a new number here so that we don’t miss 7, 8. To have 7, the numbers we can patch is 1, 2, 3 … 7. Any number greater than 7 won’t help here. Patching 8 will not help you get 7. So we have 7 numbers (1…7) to choose from. I hope you can see number 7 works best here because if we chose number 7, we can move all the way up to 1+2+3+7 = 13. (1…13 is patched completely) and it makes us reach n as quickly as possible. After we patched 7 and reach 13, we can consider last element 9 in nums. Having 9 makes us reach 13+9 = 22, which means 1…22 is completely patched. If we still did’t reach n, we can then patch 23, which makes 1…45 (22+23) completely patched. We continue until we reach n.

#### java

https://discuss.leetcode.com/topic/35517/share-my-greedy-solution-by-java-with-simple-explanation-time-1-ms

Share my greedy solution by Java with simple explanation (time: 1 ms)

The variable max records the maximal value that can be formed by the elements in nums and patched numbers. If max is less than nums[i] - 1 which means we need to patch a new number, we then patch max + 1.

#### cpp

https://leetcode.com/discuss/82822/solution-explanation

Solution + explanation

8ms, 33.52%, 149/149, April.26th, 2016

Solution

Explanation

Let miss be the smallest sum in [0,n] that we might be missing. Meaning we already know we can build all sums in [0,miss). Then if we have a number num <= miss in the given array, we can add it to those smaller sums to build all sums in [0,miss+num). If we don’t, then we must add such a number to the array, and it’s best to add miss itself, to maximize the reach.

Example: Let’s say the input is nums = [1, 2, 4, 13, 43] and n = 100. We need to ensure that all sums in the range [1,100] are possible.

Using the given numbers 1, 2 and 4, we can already build all sums from 0 to 7, i.e., the range [0,8). But we can’t build the sum 8, and the next given number (13) is too large. So we insert 8 into the array. Then we can build all sums in [0,16).

Do we need to insert 16 into the array? No! We can already build the sum 3, and adding the given 13 gives us sum 16. We can also add the 13 to the other sums, extending our range to [0,29).

And so on. The given 43 is too large to help with sum 29, so we must insert 29 into our array. This extends our range to [0,58). But then the 43 becomes useful and expands our range to [0,101). At which point we’re done.

Another implementation, though I prefer the above one.

https://discuss.leetcode.com/topic/45320/c-8ms-greedy-solution-with-explanation

C++, 8ms, greedy solution with explanation

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)

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.

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].

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.

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.

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.

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.

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

https://discuss.leetcode.com/topic/36238/my-simple-accepted-c-solution

My simple accepted C++ solution

Idea:

1. Check the content if the current one is within sum +1, which is the total sum of all previous existing numbers. If yes, we proceed and update sum. If not, we patch one number that is within sum + 1.

2. Keep updating the sum until it reaches n.

#### python

https://discuss.leetcode.com/topic/42901/simple-9-line-python-solution

Simple 9-line Python Solution

https://discuss.leetcode.com/topic/35508/greedy-solution-in-python

Greedy solution in Python

I used a greedy algorithm. When traversing through the given number list, consider each number as a goal and resource. When in the for loop for the ith number, try to add some numbers so that you can represent every number in the range [ 1, nums[i] ). Then, add the ith number to your source for further loops.

To reach the goal, suppose all the resource (the numbers smaller than the goal) sums to a number sum, then, sum+1 is what we don’t have. So we need to add a sum+1 to our resource. And now you can represent all the numbers not bigger than sum+sum+1.

Solution mine:

56ms, 33.02%, 149 / 149, April.26th, 2016