[Leetcode] Max Number of K-Sum Pairs(Medium)
LeetCode 1679 - Max Number of K-Sum Pairs
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
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.
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.
How can we solve this problem?
這題就是要移除Array中2個elements加起來等於k的操作有幾次。
第一個解法，我們可以使用sorting以及two-pointer approach來解決。先將array排序，然後設置i為0,j為n-1,直接使用iteration找出nums[i]+nums[j] = k的數，然後answer+1即可。
Solution:
class Solution {
public:
int maxOperations(vector<int>& nums, int k) {
//sort
sort(nums.begin(),nums.end()); //O n log n
int ans = 0;
int i = 0,j = nums.size()-1;
while(i < j){
if(nums[i] + nums[j] == k) {
ans++;
i++;
j--;
}else if(nums[i] + nums[j] > k)j--;
else i++;
}
return ans;
}
};
第二種解法是透過map來記錄。首先，會透過x = k - nums[i]得出一個數，如果這個數不存在map裡面或者map[x] <= 0就代表沒有，就把目前的nums[i]加入到map。如果存在，就將answer+1並且從map中移除map[x]的數量(map[x]--)。
Solution:
class Solution {
public:
int maxOperations(vector<int>& nums, int k) {
unordered_map<int,int> temp;
int ans = 0;
for(int i = 0;i<nums.size();i++){
int sum = k - nums[i];
if(temp[sum] > 0){
//exist this value;
//for example:[3,1,3,4,3]
//3,1,3(we are here) 6-3=3 and 3:1 and we found a pair(3,3)=6 ,and remove the existing value
ans++;
temp[sum]--;
}else temp[nums[i]]++;
}
return ans;
}
};