Today I have done leetcode's April LeetCoding Challenge with cpp.

April LeetCoding Challenge 17

Description

Count Equal and Divisible Pairs in an Array

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

Solution

class Solution {
public:
  int countPairs(vector<int>& nums, int k) {
    map<int, vector<int>> positions;
    for(int i = 0; i < nums.size(); ++i) {
      positions[nums[i]].push_back(i);
    }
    int answer = 0;
    for(const auto &[_n, pos] : positions) {
      for(int i = 0; i < pos.size() - 1; ++i) {
        for(int j = i + 1; j < pos.size(); ++j) {
          answer += !((pos[i] * pos[j]) % k);
        }
      }
    }
    return answer;
  }
};

// Accepted
// 238/238 cases passed (7 ms)
// Your runtime beats 3.85 % of cpp submissions
// Your memory usage beats 6.09 % of cpp submissions (16.2 MB)