LeetCode in Kotlin
2176. Count Equal and Divisible Pairs in an Array
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 <= 1001 <= nums[i], k <= 100
Solution
class Solution {
fun countPairs(nums: IntArray, k: Int): Int {
var ans = 0
for (i in nums.indices) {
for (j in i + 1 until nums.size) {
if (nums[i] == nums[j] && i * j % k == 0) {
++ans
}
}
}
return ans
}
}