# 2300. Successful Pairs of Spells and Potions
###### tags: `leetcode`
## 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.

- 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
$1 \leq n, m \leq 10^5$
$1 \leq spells[i], potions[i] \leq 10^5$
$1 \leq success \leq 10^{10}$

## Solution
- The problem is about the seeking, so sorting the `potions` can help fasten up the speed

sort(potions.begin(), potions.end());

- After that, the bounding for the problem is actually the upper bound of the `success / spells[i]`. Use that as the criterion to find the lower bound for the array (inclusive the equal elements).

for (int i = 0; i < spells.size(); i++)
{
    auto bound = ceil(double(success) / double(spells[i]));
    auto temp = lower_bound(potions.begin(), potions.end(), bound);
    spells[i] = potions.end() - temp;
}