LeetCode | 1385. Find the Distance Value Between Two Arrays两个数组间的距离值【Python】

LeetCode 1385. Find the Distance Value Between Two Arrays两个数组间的距离值【Easy】【Python】【暴力】

Problem

LeetCode

Given two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays.

The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i]-arr2[j]| <= d.

Example 1:

        Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
Output: 2
Explanation: 
For arr1[0]=4 we have: 
|4-10|=6 > d=2 
|4-9|=5 > d=2 
|4-1|=3 > d=2 
|4-8|=4 > d=2 
For arr1[1]=5 we have: 
|5-10|=5 > d=2 
|5-9|=4 > d=2 
|5-1|=4 > d=2 
|5-8|=3 > d=2
For arr1[2]=8 we have:
|8-10|=2 <= d=2
|8-9|=1 <= d=2
|8-1|=7 > d=2
|8-8|=0 <= d=2
    

Example 2:

        Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
Output: 2
    

Example 3:

        Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
Output: 1
    

Constraints:

• 1 <= arr1.length, arr2.length <= 500
• -10^3 <= arr1[i], arr2[j] <= 10^3
• 0 <= d <= 100

问题

力扣

给你两个整数数组 arr1 ， arr2 和一个整数 d ，请你返回两个数组之间的 距离值 。

「距离值」 定义为符合此描述的元素数目：对于元素 arr1[i] ，不存在任何元素 arr2[j] 满足 |arr1[i]-arr2[j]| <= d 。

示例 1：

        输入：arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
输出：2
解释：
对于 arr1[0]=4 我们有：
|4-10|=6 > d=2 
|4-9|=5 > d=2 
|4-1|=3 > d=2 
|4-8|=4 > d=2 
对于 arr1[1]=5 我们有：
|5-10|=5 > d=2 
|5-9|=4 > d=2 
|5-1|=4 > d=2 
|5-8|=3 > d=2
对于 arr1[2]=8 我们有：
|8-10|=2 <= d=2
|8-9|=1 <= d=2
|8-1|=7 > d=2
|8-8|=0 <= d=2
    

示例 2：

        输入：arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
输出：2
    

示例 3：

        输入：arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
输出：1
    

提示：

• 1 <= arr1.length, arr2.length <= 500
• -10^3 <= arr1[i], arr2[j] <= 10^3
• 0 <= d <= 100

思路

暴力

时间复杂度: O(n^2)
空间复杂度: O(1)

Python3代码

        from typing import List

class Solution:
    def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
        # solution one: 暴力
        res = 0
        for x in arr1:
            cnt = 0
            for y in arr2:
                if abs(x-y) <= d:
                    break
                else:
                    cnt += 1
                if cnt == len(arr2):
                    res += 1
        return res

        # solution two: 一行代码
        return sum(all(abs(a1 - a2) > d for a2 in arr2) for a1 in arr1)
    

GitHub链接

Python

原文：https://www.cnblogs.com/wonz/p/12544180.html