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Abhishek Chaudhary

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# Successful Pairs of Spells and Potions

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.

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.

Constraints:

• `n == spells.length`
• `m == potions.length`
• `1 <= n, m <= 105`
• `1 <= spells[i], potions[i] <= 105`
• `1 <= success <= 1010`

SOLUTION:

``````import bisect

class Solution:
def successfulPairs(self, spells: List[int], potions: List[int], success: int) -> List[int]:
n = len(potions)
potions.sort()
return [n - bisect.bisect_left(potions, success / sp) for sp in spells]
``````