QUESTION:
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
EXPLANATION:
easy的题目, 也就是两个for循环再加上一个判断就是可以了.
SOLUTION:
class Solution {
func countPairs(_ nums: [Int], _ k: Int) -> Int {
var result:Int = 0
for indexI in 0..<nums.count {
for indexJ in indexI+1..<nums.count {
if ((nums[indexI] == nums[indexJ])
&& (indexI*indexJ)%k == 0) {
result+=1
}
}
}
return result
}
}