leetcode 2300. Successful Pairs of Spells and Potions(python)

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描述

You are given two positive integer arrays spells and potions, of length n and m respectively, where spells[i] represents the strength of the ith spell and potions[j] represents the strength of the jth potion. You are also given an integer success. A spell and potion pair is considered successful if the product of their strengths is at least success. Return an integer array pairs of length n where pairs[i] is the number of potions that will form a successful pair with the ith spell.

Example 1:

Input: spells = [5,1,3], potions = [1,2,3,4,5], success = 7
Output: [4,0,3]
Explanation:
- 0th spell: 5 * [1,2,3,4,5] = [5,10,15,20,25]. 4 pairs are successful.
- 1st spell: 1 * [1,2,3,4,5] = [1,2,3,4,5]. 0 pairs are successful.
- 2nd spell: 3 * [1,2,3,4,5] = [3,6,9,12,15]. 3 pairs are successful.
Thus, [4,0,3] is returned.
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Example 2:

Input: spells = [3,1,2], potions = [8,5,8], success = 16
Output: [2,0,2]
Explanation:
- 0th spell: 3 * [8,5,8] = [24,15,24]. 2 pairs are successful.
- 1st spell: 1 * [8,5,8] = [8,5,8]. 0 pairs are successful. 
- 2nd spell: 2 * [8,5,8] = [16,10,16]. 2 pairs are successful. 
Thus, [2,0,2] is returned.
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Note:

n == spells.length
m == potions.length
1 <= n, m <= 10^5
1 <= spells[i], potions[i] <= 10^5
1 <= success <= 10^10
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解析

根据题意,给定两个正整数数组 spell 和 potion ,长度分别为 n 和 m ,其中 spells[i] 代表第 i 个法术的强度 ,potions[j] 代表第 j 个药水的强度 。题目还给定一个整数 success 。 如果 spells[i] 与 potions[j] 的乘大于等于 success ,那么咒语和药水配对被认为是成功的。 返回一个长度为 n 的整数数组 pairs ,其中 pairs[i] 是与第 i 个法术成功配对的药水数量。

这道题其实就是考查二分法的常规用法,我们已知了 spells ,要求能与 spells[i] 成功配对的药水数量,所以我们将 potions 先进行升序的排序,然后我们遍历 spells[i] ,求出符合要求的药水强度的最小值为 math.ceil(success/spells[i]) ,所以我们只需要在 potions 中通过二分法查找大于等于该药水强度的数量,并将其加入到 result 即可,不断重复上面的操作,最后得到的 result 即为结果。

时间复杂度为 O(NlogN) ,空间复杂度为 O(N) 。

解答

class Solution:
    def successfulPairs(self, spells: List[int], potions: List[int], success: int) -> List[int]:
        N = len(potions)
        potions.sort()
        result = []
        for s in spells:
            target = math.ceil(success/s)
            idx = bisect.bisect_left(potions, target)
            result.append(N-idx)
        return result		
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运行结果

56 / 56 test cases passed.
Status: Accepted
Runtime: 1477 ms
Memory Usage: 37.1 MB
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原题链接

leetcode.com/contest/biw…

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