{"id":7102,"date":"2020-07-16T00:58:25","date_gmt":"2020-07-16T07:58:25","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7102"},"modified":"2020-07-16T01:05:12","modified_gmt":"2020-07-16T08:05:12","slug":"leetcode-1512-number-of-good-pairs","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-1512-number-of-good-pairs\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1512. Number of Good Pairs"},"content":{"rendered":"\n

Given an array of integers nums<\/code>.<\/p>\n\n\n\n

A pair (i,j)<\/code> is called good<\/em> if nums[i]<\/code> == nums[j]<\/code> and i<\/code> < j<\/code>.<\/p>\n\n\n\n

Return the number of good<\/em> pairs.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> nums = [1,2,3,1,1,3]\nOutput:<\/strong> 4\nExplanation: <\/strong>There are 4 good pairs (0,3), (0,4), (3,4), (2,5) 0-indexed.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [1,1,1,1]\nOutput:<\/strong> 6\nExplanation: <\/strong>Each pair in the array are good<\/em>.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [1,2,3]\nOutput:<\/strong> 0\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= nums.length <= 100<\/code><\/li>
  • 1 <= nums[i] <= 100<\/code><\/li><\/ul>\n\n\n\n
    Solution 1: Brute Force<\/strong><\/h2>\n\n\n\n
    Enumerate all the pairs.<\/p>\n\n\n\n
    Time complexity: O(n^2)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int numIdenticalPairs(vector& nums) {\n    int ans = 0;\n    for (int i = 0; i < nums.size(); ++i)\n      for (int j = i + 1; j < nums.size(); ++j)\n        ans += nums[i] == nums[j];\n    return ans;\n  }\n};\n<\/pre>\n\n<\/div>
    Java<\/h2>\n
    \n\n
    \nclass Solution {\n  public int numIdenticalPairs(int[] nums) {\n    int ans = 0;\n    for (int i = 0; i < nums.length; ++i)\n      for (int j = i + 1; j < nums.length; ++j)\n        if (nums[i] == nums[j]) ++ans;\n    return ans;\n  }\n}\n<\/pre>\n\n<\/div>
    Python3<\/h2>\n
    \n\n
    \nclass Solution:\n  def numIdenticalPairs(self, nums: List[int]) -> int:\n    n = len(nums)\n    ans = 0\n    for i in range(n):\n      for j in range(i + 1, n):\n        if nums[i] == nums[j]: ans += 1\n    return ans\n<\/pre>\n<\/div><\/div>\n\n\n\n
    Solution 2: Hashtable<\/strong><\/h2>\n\n\n\n
    Store the frequency of each number so far, when we have a number x at pos j, and it appears k times before. Then we can form additional k pairs.<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(range)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int numIdenticalPairs(vector& nums) {\n    int ans = 0;\n    vector f(101);\n    for (int i = 0; i < nums.size(); ++i)\n      ans += f[nums[i]]++;\n    return ans;\n  }\n};\n<\/pre>\n\n<\/div>
    Java<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\n  public int numIdenticalPairs(int[] nums) {\n    int ans = 0;\n    int[] f = new int[101];\n    for (int i = 0; i < nums.length; ++i)      \n      ans += f[nums[i]]++;\n    return ans;\n  }\n}\n<\/pre>\n\n<\/div>
    Python3<\/h2>\n
    \n\n
    \n# Author: Huahua\nclass Solution:\n  def numIdenticalPairs(self, nums: List[int]) -> int:\n    f = defaultdict(int)\n    ans = 0\n    for x in nums:\n      ans += f[x]\n      f[x] += 1\n    return ans\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given an array of integers nums. A pair (i,j) is called good if nums[i] == nums[j] and i < j. Return the number of good pairs. Example 1: Input: nums = [1,2,3,1,1,3] Output: 4 Explanation: There are 4 good…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[70],"tags":[],"class_list":["post-7102","post","type-post","status-publish","format-standard","hentry","category-hashtable","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7102","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=7102"}],"version-history":[{"count":3,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7102\/revisions"}],"predecessor-version":[{"id":7106,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7102\/revisions\/7106"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=7102"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=7102"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=7102"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}