README_EN.md

2176. Count Equal and Divisible Pairs in an Array

中文文档

Description

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) where 0 <= i < j < n, such that nums[i] == nums[j] and (i * j) is divisible by k.

 

Example 1:

Input: nums = [3,1,2,2,2,1,3], k = 2
Output: 4
Explanation:
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.

Example 2:

Input: nums = [1,2,3,4], k = 1
Output: 0
Explanation: Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.

 

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i], k <= 100

Solutions

Python3

class Solution:
    def countPairs(self, nums: List[int], k: int) -> int:
        n = len(nums)
        return sum(
            nums[i] == nums[j] and (i * j) % k == 0
            for i in range(n)
            for j in range(i + 1, n)
        )

Java

class Solution {
    public int countPairs(int[] nums, int k) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (nums[i] == nums[j] && (i * j) % k == 0) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countPairs(vector<int>& nums, int k) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (nums[i] == nums[j] && (i * j) % k == 0) ++ans;
            }
        }
        return ans;
    }
};

Go

func countPairs(nums []int, k int) int {
	n := len(nums)
	ans := 0
	for i, v := range nums {
		for j := i + 1; j < n; j++ {
			if v == nums[j] && (i*j)%k == 0 {
				ans++
			}
		}
	}
	return ans
}

TypeScript

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