[LeetCode 330] Patching array

补丁最小化算法

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本文介绍了一种算法，用于确定给定正整数数组和一个整数n的情况下，需要添加多少个元素（补丁），使得从1到n的所有整数都能通过数组中元素的组合求和得到。该算法通过迭代检查并维护当前可以通过数组元素组合获得的最大和边界值来实现。

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.

Solution:

in order to know if sum of nums[] cover range from 1..n, must iterate 1..n

can use one variable boundary to hold the current max sum number we can get.

for example,

[1,2,3, 10]

cur sum boundary

1 1 2

2 3 4

3 6 7

we don't have 7 in the array, so patch 7, then we can get 13 before check next element

10 13 14

23 24

Once path 7, we can get 23 as total max

public int minPatches(int[] nums, int n) {
        if(n<=0) return 0;
        nums = (nums == null)? new int[0] :nums;
        int curIndex=0, res = 0;
        long boundaryVal = 1, sum =0;
        while(boundaryVal <= n) {
            if(curIndex < nums.length && nums[curIndex] <=boundaryVal) {
                sum += nums[curIndex++];
                boundaryVal= sum+1;
            } else {
                res++;
                sum += boundaryVal;
                boundaryVal = sum+1;
            }
        }
        return res;
    }