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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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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)
          )
  
  

• 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
  }
  

• 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;
  }
  
  

• 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
      }
  }
  
  

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