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]
Top comments (0)