Successful pairs of spells and potions
class Solution:
def successfulPairs(self, spells: List[int], potions: List[int], success: int) -> List[int]:
def bisec(nums, target):
left = 0
right = len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid] < target:
left = mid + 1
else:
right = mid
return left
potions.sort()
# print(potions)
res = []
for i in spells:
minval = math.ceil(success / i)
index = bisec(potions, minval)
print(i, index, minval)
if potions[index] < minval:
res.append(0)
else:
res.append(len(potions) - index)
return res
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.lengthm == potions.length1 <= n, m <= 1051 <= spells[i], potions[i] <= 1051 <= success <= 1010