2300. Successful Pairs of Spells and Potions

Question

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<sup>th</sup> spell and potions[j] represents the strength of the j<sup>th</sup> 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<sup>th</sup> spell.

Solution 1

排序+二分搜索。同时记录success与spells[i]的比值来减小计算量。

排序

首先对potions[]进行排序，这样可以使用二分搜索查找分界值。
数组scales[]记录success与spells[i]的比值，以此为界大于等于scales[i]的位置都可以计入ret[]数组。

二分搜索

这里有一些tricky。

由于Arrays.binarySearch()方法无法返回重复的数字，因此在搜索时我们将查找值scale减去一个小值，保证在搜索时一定返回负值。（查找值的插入位置的负数-1）
将ret[i]记录为potions[]的总数减去正确的插入位置即可。

Code

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        int[] ret = new int[spells.length];
        
        Arrays.sort(potions);
        double[] scales = new double[potions.length];
        
        for(int i = 0; i < potions.length; i++){
            scales[i] = (double) potions[i];
        }
        
        for(int i = 0; i < spells.length; i++){
            double scale = (double) success / spells[i] - 0.000001;   //确保浮点数不在scale中出现，binarySearch方法返回的结果必定为上一个插入位置
            int index = Arrays.binarySearch(scales, scale);
            ret[i] = potions.length + (index + 1);
        }
        return ret;
    }
}

Solution 2

由于Arrays.binarySearch()无法正确的搜索有重复元素的数组，因此采用辅助方法binarySearch()来搜索最左侧的下标。

直接在binarySearch()方法中查找target，比较的对象为spell和potions[i]的乘积。

为了搜寻重复的第一个元素，当遇到target时不直接返回，而是继续修改right的位置，直到left等于right。

如果未搜索到，则返回数组的总长度。

Code

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        int[] ret = new int[spells.length];
        Arrays.sort(potions);
        for(int i = 0; i < spells.length; i++){
            int index = binarySearch(potions, spells[i], success);
            ret[i] = potions.length - index;
        }
        return ret;
    }
    
    private int binarySearch(int[] potions, long spell, long target){
        int left = 0, right = potions.length - 1;
        while(left < right){
            int mid = left + (right - left) / 2;
            if(potions[mid] * spell < target) left = mid + 1;
            else right = mid;
        }
        return potions[left] * spell < target ? potions.length : left;
    }
}