< Back 2300. Successful Pairs of Spells and Potions The input provides is an integer list of spells, an integer list of potions, and an integer, success. A pair of spells and potions are considered successful if the product of their strings is at least success. We're asked to find the number of potions that will form a successful pair with each spell. The potion input is not sorted - without sorting our time complexity to solve this problem using brute force would be O(n^2). We can solve this problem in O(n log(n)) by sorting the input first. With potions sorted in ascending order, for each spell we can conduct binary search to find the minimum potion that will form a successful pair with the spell. Once we find the minimum successful pairing, we know that all potions after the minimum are also successful, counting them in the answer. The solution is as follows: class Solution: def successfulPairs(self, spells: List[int], potions: List[int], success: int) -> List[int]: potions.sort() ans = [] n = len(potions) for spell in spells: l, r = 0, n - 1 while l <= r: m = (r + l) // 2 if potions[m] * spell < success: l = m + 1 else: r = m - 1 ans.append(n - l) return ans _ Time Complexity: O(nlog(n) + mlog(m)) - We sort n potions and conduct a binary search for m spells. _ Space Complexity: O(n + m) - Sorting in Python uses O(n) space. We use O(m) space for the answer list.