Description
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.
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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.
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Constraints:
- n == spells.length
- m == potions.length
- 1 <= n, m <= 105
- 1 <= spells[i], potions[i] <= 105
- 1 <= success <= 1010
Solution
Python3
class Solution:
    def successfulPairs(self, spells: List[int], potions: List[int], success: int) -> List[int]:
        M = len(potions)
        res = []
        potions.sort()
 
        for spell in spells:
            index = bisect_left(potions, (success + spell - 1) // spell) 
            res.append(M - index)
 
        return res