[LeetCode] 2300. Successful Pairs of Spells and Potions

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

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。这道题属于在答案上二分的题。

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。

复杂度

时间O(nlogm)
空间O(n) - output array

代码

Java实现

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        Arrays.sort(potions);
        int n = spells.length;
        int m = potions.length;
        int[] res = new int[n];

        for (int i = 0; i < n; i++) {
            long spell = (long) spells[i];
            int left = 0;
            int right = m;
            while (left < right) {
                int mid = left + (right - left) / 2;
                if (spell * potions[mid] < success) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            res[i] = m - left;
        }
        return res;
    }
}