1679. Max Number of K Sum Pairs

# 1679. Max Number of K-Sum Pairs#

## 题目 #

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 <= 105`
• `1 <= nums[i] <= 109`
• `1 <= k <= 109`

## 解题思路 #

• 读完题第一感觉这道题是 TWO SUM 题目的加强版。需要找到所有满足和是 k 的数对。先考虑能不能找到两个数都是 k/2 ，如果能找到多个这样的数，可以先移除他们。其次在利用 TWO SUM 的思路，找出和为 k 的数对。利用 TWO SUM 里面 map 的做法，时间复杂度 O(n)。

## 代码 #

``````package leetcode

// 解法一 优化版
func maxOperations(nums []int, k int) int {
counter, res := make(map[int]int), 0
for _, n := range nums {
counter[n]++
}
if (k & 1) == 0 {
res += counter[k>>1] >> 1
// 能够由 2 个相同的数构成 k 的组合已经都排除出去了，剩下的一个单独的也不能组成 k 了
// 所以这里要把它的频次置为 0 。如果这里不置为 0，下面代码判断逻辑还需要考虑重复使用数字的情况
counter[k>>1] = 0
}
for num, freq := range counter {
if num <= k/2 {
remain := k - num
if counter[remain] < freq {
res += counter[remain]
} else {
res += freq
}
}
}
return res
}

// 解法二
func maxOperations_(nums []int, k int) int {
counter, res := make(map[int]int), 0
for _, num := range nums {
counter[num]++
remain := k - num
if num == remain {
if counter[num] >= 2 {
res++
counter[num] -= 2
}
} else {
if counter[remain] > 0 {
res++
counter[remain]--
counter[num]--
}
}
}
return res
}
``````

Apr 8, 2023