2176. Count Equal and Divisible Pairs in an Array

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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

Solution 1: Enumeration

We first enumerate the index $j$ in the range $[0, n)$, and then enumerate the index $i$ in the range $[0, j)$. We count the number of pairs that satisfy $\textit{nums}[i] = \textit{nums}[j]$ and $(i \times j) \bmod k = 0$.

The time complexity is $O(n^2)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

Python3

class Solution:
    def countPairs(self, nums: List[int], k: int) -> int:
        ans = 0
        for j, y in enumerate(nums):
            for i, x in enumerate(nums[:j]):
                ans += int(x == y and i * j % k == 0)
        return ans

Java

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

C++

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

Go

func countPairs(nums []int, k int) (ans int) {
	for j, y := range nums {
		for i, x := range nums[:j] {
			if x == y && (i*j%k) == 0 {
				ans++
			}
		}
	}
	return
}

TypeScript

function countPairs(nums: number[], k: number): number {
    let ans = 0;
    for (let j = 1; j < nums.length; ++j) {
        for (let i = 0; i < j; ++i) {
            if (nums[i] === nums[j] && (i * j) % k === 0) {
                ++ans;
            }
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn count_pairs(nums: Vec<i32>, k: i32) -> i32 {
        let mut ans = 0;
        for j in 1..nums.len() {
            for (i, &x) in nums[..j].iter().enumerate() {
                if x == nums[j] && (i * j) as i32 % k == 0 {
                    ans += 1;
                }
            }
        }
        ans
    }
}

C

int countPairs(int* nums, int numsSize, int k) {
    int ans = 0;
    for (int j = 1; j < numsSize; ++j) {
        for (int i = 0; i < j; ++i) {
            ans += (nums[i] == nums[j] && (i * j % k) == 0);
        }
    }
    return ans;
}