330.patching-array

Statement

简体中文English

Metadata

Link: 按要求补齐数组
Difficulty: Hard
Tag: 贪心 数组

给定一个已排序的正整数数组 nums ，和一个正整数 n 。从 [1, n] 区间内选取任意个数字补充到 nums 中，使得 [1, n] 区间内的任何数字都可以用 nums 中某几个数字的和来表示。

请返回 满足上述要求的最少需要补充的数字个数 。

示例 1:

输入: nums = [1,3], n = 6
输出: 1 
解释:
根据 nums 里现有的组合 [1], [3], [1,3]，可以得出 1, 3, 4。
现在如果我们将 2 添加到 nums 中， 组合变为: [1], [2], [3], [1,3], [2,3], [1,2,3]。
其和可以表示数字 1, 2, 3, 4, 5, 6，能够覆盖 [1, 6] 区间里所有的数。
所以我们最少需要添加一个数字。

示例 2:

输入: nums = [1,5,10], n = 20
输出: 2
解释: 我们需要添加 [2,4]。

示例 3:

输入: nums = [1,2,2], n = 5
输出: 0

提示：

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴
nums 按 升序排列
1 <= n <= 2³¹ - 1

Metadata

Link: Patching Array
Difficulty: Hard
Tag: Greedy Array

Given a sorted integer array nums and an integer n, add/patch elements to the array such that any number in the 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

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴
nums is sorted in ascending order.
1 <= n <= 2³¹ - 1

Solution

Python 3

from collections import Counter
from typing import List


class Solution:
    def minPatches(self, nums: List[int], n: int) -> int:
        base = 0
        res = 0

        for i in nums:
            while base + 1 < i and base < n:
                base <<= 1
                base += 1
                res += 1
            base += i

        while base < n:
            base <<= 1
            base += 1
            res += 1

        return res

最后更新: October 11, 2023