330. Patching Array

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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:

Input: nums = [1, 3], n = 6
Output: 1 
Explanation:
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:

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

Example 3:

Input: nums = [1, 2, 2], n = 5.
Output: 0

Approach #1: C++. [easy understand]

class Solution {
public:
    int minPatches(vector<int>& nums, int n) {
        int size = nums.size();
        long pre_sum = 0;
        int ans = 0;
        for (int i = 0; i < size && pre_sum < n; ++i) {
            int v = nums[i];
            while (v > pre_sum + 1 && pre_sum < n) {
                pre_sum += (pre_sum + 1);
                ans += 1;
            }
            pre_sum += v;
        }
        if (pre_sum < n) {
            while (pre_sum < n) {
                pre_sum += (pre_sum + 1);
                ans += 1;
            }
        }
        
        return ans;
    }
};

Analysis:

We can use pre_sum as base to extend the sum. If current element nums[i] > pre_sum, we should extend the pre_sum by add (pre_sum + 1).

Approach #2: C++.  [This is more clear.]

int minPatches(vector<int>& nums, int n) {
    long miss = 1, added = 0, i = 0;
    while (miss <= n) {
        if (i < nums.size() && nums[i] <= miss) {
            miss += nums[i++];
        } else {
            miss += miss;
            added++;
        }
    }
    return added;
}

Analysis:

@Stefan Pochmann

330. Patching Array

标签：use   highlight   osi   cto   c++   end   analysis   miss   base   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/10204364.html