Successful Pairs of Spells and Potions

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 <= n, m <= 105
• 1 <= spells[i], potions[i] <= 105
• 1 <= success <= 1010

Code

Time Complexity: $O(m\log m+n\log m)$, Space Complexity: $O(1)$

class Solution {
public:
    vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
        sort(potions.begin(), potions.end());
 
        vector<int> res;
        for(auto spell: spells) {
            double v = (double)success / spell;
            res.push_back(potions.end() - lower_bound(potions.begin(), potions.end(), v));
        }
        return res;
    }
};

觀察 double v = (double)success / spell; ，其實可以用 long need = (success + spell - 1) / spell; 取代，就是在做無條件進位，就是我們要的，使用這個技巧（Koko Eating Bananas 有用過）就可以避免浮點數的運算。

class Solution {
public:
    vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
        sort(potions.begin(), potions.end());
 
        vector<int> res;
        for(auto spell: spells) {
            long need = (success + spell - 1) / spell;
            res.push_back(potions.end() - lower_bound(potions.begin(), potions.end(), need));
        }
        return res;
    }
};

Source

• Successful Pairs of Spells and Potions - LeetCode