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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

Similar Questions:

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/count-equal-and-divisible-pairs-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    int countPairs(vector<int>& A, int k) {
        int N = A.size(), ans = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) {
                ans += A[i] == A[j] && i * j % k == 0;
            }
        }
        return ans;
    }
};