Successful Pairs of Spells and Potions

lcpy gen -n 2300   # by problem number
lcpy gen -s successful_pairs_of_spells_and_potions   # by problem name

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.

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.

class Solution:
    # Time: O((n + m) log m) - sorting potions plus a binary search per spell
    # Space: O(m) - sorted copy of the potions
    def successful_pairs(self, spells: list[int], potions: list[int], success: int) -> list[int]:
        potions.sort()
        n = len(potions)
        pairs: list[int] = []
        for spell in spells:
            lo, hi = 0, n
            while lo < hi:
                mid = (lo + hi) // 2
                if spell * potions[mid] >= success:
                    hi = mid
                else:
                    lo = mid + 1
            pairs.append(n - lo)
        return pairs