What is the optimal approach for solving Count Array Pairs Divisible by K?

The optimal approach involves using number theory and frequency counting of remainders modulo k to efficiently find pairs that satisfy the divisibility condition.

How can modular arithmetic simplify the solution for this problem?

Modular arithmetic helps identify pairs where the sum of their remainders modulo k equals zero, streamlining the solution and avoiding brute force checking.

What is the time complexity of the optimal solution for Count Array Pairs Divisible by K?

The time complexity of the optimal solution is O(n), where n is the length of the array, due to the use of frequency counts and modular arithmetic.

Are there any edge cases I should consider for this problem?

Edge cases include situations where no valid pairs exist, such as when the product of any pair in the array is never divisible by k.

What is the role of frequency counting in solving this problem?

Frequency counting helps track how many elements in the array have specific remainders when divided by k, which simplifies the process of finding divisible pairs.

#2183

Hard

auto_awesome数组·数学

LeetCode 题解工作台

统计可以被 K 整除的下标对数目

给你一个下标从 0 开始、长度为 n 的整数数组 nums 和一个整数 k ，返回满足下述条件的下标对 (i, j) 的数目： 0 且 nums[i] * nums[j] 能被 k 整除。示例 1：输入： nums = [1,2,3,4,5], k = 2 输出： 7 解释：共有 7 对下标的…

数组数学数论

题目描述

给你一个下标从 0 开始、长度为 n 的整数数组 nums 和一个整数 k ，返回满足下述条件的下标对 (i, j) 的数目：

0 <= i < j <= n - 1 且
nums[i] * nums[j] 能被 k 整除。

示例 1：

输入：nums = [1,2,3,4,5], k = 2
输出：7
解释：
共有 7 对下标的对应积可以被 2 整除：
(0, 1)、(0, 3)、(1, 2)、(1, 3)、(1, 4)、(2, 3) 和 (3, 4)
它们的积分别是 2、4、6、8、10、12 和 20 。
其他下标对，例如 (0, 2) 和 (2, 4) 的乘积分别是 3 和 15 ，都无法被 2 整除。

示例 2：

输入：nums = [1,2,3,4], k = 5
输出：0
解释：不存在对应积可以被 5 整除的下标对。

提示：

1 <= nums.length <= 10⁵
1 <= nums[i], k <= 10⁵

lightbulb

解题思路

方法一

1

2

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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Ability to optimize brute force approaches.
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Understanding of number theory and modular arithmetic.
question_mark
Familiarity with efficient frequency counting techniques.

warning

常见陷阱

外企场景

error
Using a brute force approach without optimizing divisibility checks.
error
Overlooking how modular arithmetic can simplify the solution.
error
Not considering edge cases where no valid pairs exist, like when the product is never divisible by k.

swap_horiz

进阶变体

外企场景

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What if k is a prime number?
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How would the solution change if we wanted to count pairs whose product is divisible by a prime factor of k?
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Can this solution be adapted for non-zero index-based array pairs?

help

常见问题

外企场景

继续练习

#2197 替换数组中的非互质数 #2280 表示一个折线图的最少线段数 #2344 使数组可以被整除的最少删除次数