2176.count-equal-and-divisible-pairs-in-an-array.cpp

//  Tag: Array
//  Time: O(N^2)
//  Space: O(1)
//  Ref: -
//  Note: -
//  Video: https://youtu.be/OiRTe8BDOR0

//  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
//  
//  

class Solution {
public:
    int countPairs(vector<int>& nums, int k) {
        int n = nums.size();
        int res = 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) {
                    res += 1;
                }
            }
        }

        return res;
    }
};

class Solution {
public:
    int countPairs(vector<int>& nums, int k) {
        int n = nums.size();
        int res = 0;
        unordered_map<int, vector<int>> m;
        for (int i = 0;  i < n; ++i)
            m[nums[i]].push_back(i);

        for (auto &[x, ids] : m) {
            unordered_map<int, int> gcds;
            for (auto i : ids) {
                auto gcd_i = gcd(i, k);
                for (auto &[gcd_j, cnt] : gcds)
                    res += gcd_i * gcd_j % k ? 0 : cnt;
                gcds[gcd_i] += 1;
            }
        }
        return res;
    }
};