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.

Tips:

道题给我们一个有序的正数数组nums，又给了我们一个正整数n，问我们最少需要给nums加几个数字，使其能组成[1,n]之间的所有数字，注意数组中的元素不能重复使用，否则的话只有要有1，就能组成所有的数字了

我们定义一个变量miss，用来表示[0,n]之间最小的不能表示的值，那么初始化为1，为啥不为0呢，因为n=0没啥意义，直接返回0了。那么此时我们能表示的范围是[0, miss)，表示此时我们能表示0到miss-1的数，如果此时的num <= miss，那么我们可以把我们能表示数的范围扩大到[0, miss+num)，如果num>miss，那么此时我们需要添加一个数，为了能最大限度的增加表示数范围，我们加上miss它本身，以此类推直至遍历完整个数组，我们可以得到结果。下面我们来举个例子说明：

给定nums = [1, 2, 4, 11, 30], n = 50，我们需要让[0, 50]之间所有的数字都能被nums中的数字之和表示出来。

首先使用1, 2, 4可能表示出0到7之间的所有数，表示范围为[0, 8)，但我们不能表示8，因为下一个数字11太大了，所以我们要在数组里加上一个8，此时能表示的范围是[0, 16)，那么我们需要插入16吗，答案是不需要，因为我们数组有1和4，可以组成5，而下一个数字11，加一起能组成16，所以有了数组中的11，我们此时能表示的范围扩大到[0, 27)，但我们没法表示27，因为30太大了，所以此时我们给数组中加入一个27，那么现在能表示的范围是[0, 54)，已经满足要求了，我们总共添加了两个数8和27，所以返回2即可。

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.

Code:

public class Solution {
    // note the array is sorted
    public int minPatches(int[] nums, int n) {
        int m = nums.length, miss = 1, patch = 0;
        for (int i = 0; miss > 0 && miss <= n;) { //still run loop @ i >= m
            if (i < m && nums[i] <= miss)
                miss += nums[i++];
            else {
                miss += miss; //path an element with value miss
                patch++;
            }
        }
        return patch;
    }
}