2176. Count Equal and Divisible Pairs in an Array

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

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

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

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