{"id":7768,"date":"2020-12-06T11:36:21","date_gmt":"2020-12-06T19:36:21","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7768"},"modified":"2020-12-06T11:41:40","modified_gmt":"2020-12-06T19:41:40","slug":"leetcode-1679-max-number-of-k-sum-pairs","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-1679-max-number-of-k-sum-pairs\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1679. Max Number of K-Sum Pairs"},"content":{"rendered":"\n

You are given an integer array nums<\/code> and an integer k<\/code>.<\/p>\n\n\n\n

In one operation, you can pick two numbers from the array whose sum equals k<\/code> and remove them from the array.<\/p>\n\n\n\n

Return the maximum number of operations you can perform on the array<\/em>.<\/p>\n\n\n\n

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

Input:<\/strong> nums = [1,2,3,4], k = 5\nOutput:<\/strong> 2\nExplanation:<\/strong> Starting with nums = [1,2,3,4]:\n- Remove numbers 1 and 4, then nums = [2,3]\n- Remove numbers 2 and 3, then nums = []\nThere are no more pairs that sum up to 5, hence a total of 2 operations.<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [3,1,3,4,3], k = 6\nOutput:<\/strong> 1\nExplanation:<\/strong> Starting with nums = [3,1,3,4,3]:\n- Remove the first two 3's, then nums = [1,4,3]\nThere are no more pairs that sum up to 6, hence a total of 1 operation.<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= nums.length <= 105<\/sup><\/code><\/li>
  • 1 <= nums[i] <= 109<\/sup><\/code><\/li>
  • 1 <= k <= 109<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution 1: Frequency Map<\/strong><\/h2>\n\n\n\n
    For each x, check freq[x] and freq[k – x]. Note: there is a special case when x + x == k.<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int maxOperations(vector& nums, int k) {\n    unordered_map m;\n    int ans = 0;\n    for (int x : nums) ++m[x];\n    for (int x : nums) {      \n      if (m[x] < 1 || m[k - x] < 1 + (x + x == k)) continue;\n      --m[x];\n      --m[k - x];\n      ++ans;      \n    }\n    return ans;\n  }\n};\n<\/pre>\n\n<\/div>
    Python3<\/h2>\n
    \n\n
    \nclass Solution:\n  def maxOperations(self, nums: List[int], k: int) -> int:\n    m = defaultdict(int)\n    ans = 0\n    for x in nums: m[x] += 1\n    for x in nums:\n      if m[x] < 1 or m[k - x] < 1 + (x + x == k): continue\n      m[x] -= 1\n      m[k - x] -= 1\n      ans += 1\n    return ans        \n<\/pre>\n<\/div><\/div>\n\n\n\n
    Solution 2: Two Pointers<\/strong><\/h2>\n\n\n\n
    Sort the number, start from i = 0, j = n - 1, compare s = nums[i] + nums[j] with k and move i, j accordingly.<\/p>\n\n\n\n
    Time complexity: O(nlogn)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int maxOperations(vector& nums, int k) {\n    sort(begin(nums), end(nums));\n    int i = 0, j = nums.size() - 1;\n    int ans = 0;\n    while (i < j) {\n      const int s = nums[i] + nums[j];\n      if (s == k) {\n        ++ans;\n        ++i; --j;\n      } else if (s < k) {\n        ++i;\n      } else {\n        --j;\n      }\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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