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
.
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.
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.
1 <= nums.length <= 100
1 <= nums[i], k <= 100
impl Solution {
pub fn count_pairs(nums: Vec<i32>, k: i32) -> i32 {
let mut ret = 0;
for i in 0..nums.len() {
for j in i + 1..nums.len() {
if nums[i] == nums[j] && (i * j) as i32 % k == 0 {
ret += 1;
}
}
}
ret
}
}