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: Hash Map

• Time complexity: O(n²)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
const countPairs = function (nums, k) {
  const n = nums.length;
  const numMap = new Map();
  let result = 0;

  const getDivisibleCount = (indices, pair) => {
    let count = 0;

    for (const index of indices) {
      if ((pair * index) % k) continue;

      count += 1;
    }

    return count;
  };

  for (let index = 0; index < n; index++) {
    const num = nums[index];

    if (numMap.has(num)) {
      const indices = numMap.get(num);

      result += getDivisibleCount(indices, index);
      indices.push(index);
      continue;
    }

    numMap.set(num, [index]);
  }

  return result;
};