[LeetCode] 2300. Successful Pairs of Spells and Potions

Posted 2023-04-03 CNoodle

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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

咒语和药水的成功对数。

给你两个正整数数组 spells 和 potions ，长度分别为 n 和 m ，其中 spells[i] 表示第 i 个咒语的能量强度，potions[j] 表示第 j 瓶药水的能量强度。

同时给你一个整数 success 。一个咒语和药水的能量强度相乘如果大于等于 success ，那么它们视为一对成功的组合。

请你返回一个长度为 n 的整数数组 pairs，其中 pairs[i] 是能跟第 i 个咒语成功组合的药水数目。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/successful-pairs-of-spells-and-potions
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是排序 + 二分。题意不难理解，注意看第一个例子，当发现 5 * 2 = 10 > 4 的时候，包括 2 在内后面所有的乘积都可以被判为 success，所以这里考虑对 potion 数组排序。

第一次做的时候用了两层 for 循环，发现最后有几个很长的 case 超时。优化的思路只能往二分上靠。这道题能用二分优化的前提是被二分的数据是有序的，因为我们对 potion 数组已经排过序，所以可以用二分。

时间O(mlogn)

空间O(n) - output array

Java实现

 class Solution 
     public int[] successfulPairs(int[] spells, int[] potions, long success) 
         Arrays.sort(potions);
         int m = spells.length;
         int n = potions.length;
         int[] res = new int[m];
 
         int i = 0;
         while (i < m) 
             long spell = (long) spells[i];
             int left = 0;
             int right = n;
             while (left < right) 
                 int mid = left + (right - left) / 2;
                 if (spell * potions[mid] < success) 
                     left = mid + 1;
                  else 
                     right = mid;
                 
             
             res[i] = n - left;
             i++;
         
         return res;