You are given an integer array nums and an integer k.
In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array.
Return the maximum number of operations you can perform on the array.
Example 1:
Input: nums = [1,2,3,4], k = 5 Output: 2 Explanation: Starting with nums = [1,2,3,4]: - Remove numbers 1 and 4, then nums = [2,3] - Remove numbers 2 and 3, then nums = [] There are no more pairs that sum up to 5, hence a total of 2 operations.
Example 2:
Input: nums = [3,1,3,4,3], k = 6 Output: 1 Explanation: Starting with nums = [3,1,3,4,3]: - Remove the first two 3's, then nums = [1,4,3] There are no more pairs that sum up to 6, hence a total of 1 operation.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1091 <= k <= 109
Solution 1: Frequency Map
For each x, check freq[x] and freq[k – x]. Note: there is a special case when x + x == k.
Time complexity: O(n)
Space complexity: O(n)
C++
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class Solution { public: int maxOperations(vector<int>& nums, int k) { unordered_map<int, int> m; int ans = 0; for (int x : nums) ++m[x]; for (int x : nums) { if (m[x] < 1 || m[k - x] < 1 + (x + x == k)) continue; --m[x]; --m[k - x]; ++ans; } return ans; } }; |
Python3
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class Solution: def maxOperations(self, nums: List[int], k: int) -> int: m = defaultdict(int) ans = 0 for x in nums: m[x] += 1 for x in nums: if m[x] < 1 or m[k - x] < 1 + (x + x == k): continue m[x] -= 1 m[k - x] -= 1 ans += 1 return ans |
Solution 2: Two Pointers
Sort the number, start from i = 0, j = n – 1, compare s = nums[i] + nums[j] with k and move i, j accordingly.
Time complexity: O(nlogn)
Space complexity: O(1)
C++
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class Solution { public: int maxOperations(vector<int>& nums, int k) { sort(begin(nums), end(nums)); int i = 0, j = nums.size() - 1; int ans = 0; while (i < j) { const int s = nums[i] + nums[j]; if (s == k) { ++ans; ++i; --j; } else if (s < k) { ++i; } else { --j; } } return ans; } }; |