Easy
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
public class Solution {
public int countPairs(int[] nums, int k) {
int ans = 0;
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
if (nums[i] == nums[j] && (i * j) % k == 0) {
++ans;
}
}
}
return ans;
}
}