[ Python ] ✅✅ Simple Python Solution | 100 % Faster🥳✌👍
If You like the Solution, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n# Runtime: 54 ms, faster than 100.00% of Python3 online submissions for Find the Number of Go
# If You like the Solution, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n# Runtime: 54 ms, faster than 100.00% of Python3 online submissions for Find the Number of Good Pairs I.\n# Memory Usage: 16.6 MB, less than 100.00% of Python3 online submissions for Find the Number of Good Pairs I.\n![image](https://assets.leetcode.com/users/images/3d7b7de7-5e81-4e2f-9f10-71351793e10a_1716699466.4009428.png)\n\n\tclass Solution:\n\t\tdef numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:\n\n\t\t\tresult = 0\n\n\t\t\tfor num1 in nums1:\n\n\t\t\t\tfor num2 in nums2:\n\n\t\t\t\t\tif num1 % (num2 * k) == 0:\n\n\t\t\t\t\t\tresult = result + 1\n\n\t\t\treturn result\n\t\t\t\n\tTime Complexity : O(N ^ 2)\n\tSpace Complexity : O(1)\n# Thank You \uD83E\uDD73\u270C\uD83D\uDC4D
['Python', 'Python3']