2300.Successful Pairs of Spells and Potions === ###### tags: `Medium`,`Array`,`Two Pointers`,`Binary Search`,`Sorting` [2300. Successful Pairs of Spells and Potions](https://leetcode.com/problems/successful-pairs-of-spells-and-potions/) ### 題目描述 You are given two positive integer arrays `spells` and `potions`, of length `n` and `m` respectively, where `spells[i]` represents the strength of the i^th^ spell and `potions[j]` represents the strength of the j^th^ 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 i^th^ spell.* ### 範例 **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` <= 10^5^ * 1 <= `spells[i]`, `potions[i]` <= 10^5^ * 1 <= `success` <= 10^10^ ### 解答 #### C++ ##### 思路: 1. 題目: spell強度*potion強度 >= success算成功的pair, 計算每個spell對於每個potion成功的pair有幾個 2. 每個spell其對應的potion不變, 故以potion為主要思考點, 每個potion需要多少spell強度才算成功 3. 排序potion, 因 spell*potion >= success才算成功, 移項得potion >= success/spell, 所以可以binary search potion找小於success/spell的值都是不合格的 ```cpp= vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) { int size = potions.size(); vector<int> res; sort(begin(potions), end(potions)); for(long long sp : spells) { int invalid_size = lower_bound(begin(potions), end(potions), ceil((double)success/(double)sp))-potions.begin(); res.push_back(size-invalid_size); } return res; } ``` Time: $O(nlog(n))$ Extra Space: $O(n)$ > [name=XD] [time= Apr 4, 2023] #### Javascript ```javascript= function successfulPairs(spells, potions, success) { const sortedSpells = []; for (let i = 0; i < spells.length; i++) { sortedSpells.push([spells[i], i]); } const result = new Array(spells.length).fill(0); sortedSpells.sort((a, b) => a[0] - b[0]); potions.sort((a, b) => a - b); let right = potions.length - 1; for (const [spell, index] of sortedSpells) { while (right >= 0 && spell * potions[right] >= success) { right--; } result[index] = potions.length - right - 1; } return result; } ``` > Sorting + Two Pointers > 錯蠻多次的...只想到sorting後去做但還是一直TLE,樓上太神啦 > [name=Marsgoat] [time= Apr 21, 2023] ### Reference [回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)