## 1. Problem

2300. 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

## 2. Solution

I solved this problem like this.

Using binary search, we can solve this problem!

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