Description
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
Solutions
Solution 1: Enumeration
We first enumerate the index j in the range [0, n), and then enumerate the index i in the range [0, j). We count the number of pairs that satisfy nums[i] = nums[j] and (i × j) \bmod k = 0.
The time complexity is O(n2), where n is the length of the array nums. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustC
class Solution: def countPairs(self, nums: List[int], k: int) -> int: ans = 0 for j, y in enumerate(nums): for i, x in enumerate(nums[:j]): ans += int(x == y and i * j % k == 0) return ans(code-box)
