Description#
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
Solutions#
Solution 1: Sorting#
We sort $nums$. Then $l$ and $r$ point to the first and last elements of $nums$ respectively, and we compare the sum $s$ of the two integers with $k$.
- If $s = k$, it means that we have found two integers whose sum is $k$. We increment the answer and then move $l$ and $r$ towards the middle;
- If $s > k$, then we move the $r$ pointer to the left;
- If $s < k$, then we move the $l$ pointer to the right;
- We continue the loop until $l \geq r$.
After the loop ends, we return the answer.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of $nums$.
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| class Solution:
def maxOperations(self, nums: List[int], k: int) -> int:
nums.sort()
l, r, ans = 0, len(nums) - 1, 0
while l < r:
s = nums[l] + nums[r]
if s == k:
ans += 1
l, r = l + 1, r - 1
elif s > k:
r -= 1
else:
l += 1
return ans
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| class Solution {
public int maxOperations(int[] nums, int k) {
Arrays.sort(nums);
int l = 0, r = nums.length - 1;
int ans = 0;
while (l < r) {
int s = nums[l] + nums[r];
if (s == k) {
++ans;
++l;
--r;
} else if (s > k) {
--r;
} else {
++l;
}
}
return ans;
}
}
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| class Solution {
public:
int maxOperations(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
int cnt = 0;
int i = 0, j = nums.size() - 1;
while (i < j) {
if (nums[i] + nums[j] == k) {
i++;
j--;
cnt++;
} else if (nums[i] + nums[j] > k) {
j--;
} else {
i++;
}
}
return cnt;
}
};
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| func maxOperations(nums []int, k int) int {
sort.Ints(nums)
l, r, ans := 0, len(nums)-1, 0
for l < r {
s := nums[l] + nums[r]
if s == k {
ans++
l++
r--
} else if s > k {
r--
} else {
l++
}
}
return ans
}
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| function maxOperations(nums: number[], k: number): number {
const cnt = new Map();
let ans = 0;
for (const x of nums) {
if (cnt.get(k - x)) {
cnt.set(k - x, cnt.get(k - x) - 1);
++ans;
} else {
cnt.set(x, (cnt.get(x) | 0) + 1);
}
}
return ans;
}
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| impl Solution {
pub fn max_operations(nums: Vec<i32>, k: i32) -> i32 {
let mut nums = nums.clone();
nums.sort();
let (mut l, mut r, mut ans) = (0, nums.len() - 1, 0);
while l < r {
match nums[l] + nums[r] {
sum if sum == k => {
ans += 1;
l += 1;
r -= 1;
}
sum if sum > k => {
r -= 1;
}
_ => {
l += 1;
}
}
}
ans
}
}
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Solution 2: Hash Table#
We use a hash table $cnt$ to record the current remaining integers and their occurrence counts.
We iterate over $nums$. For the current integer $x$, we check if $k - x$ is in $cnt$. If it exists, it means that we have found two integers whose sum is $k$. We increment the answer and then decrement the occurrence count of $k - x$; otherwise, we increment the occurrence count of $x$.
After the iteration ends, we return the answer.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of $nums$.
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| class Solution:
def maxOperations(self, nums: List[int], k: int) -> int:
cnt = Counter()
ans = 0
for x in nums:
if cnt[k - x]:
ans += 1
cnt[k - x] -= 1
else:
cnt[x] += 1
return ans
|
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| class Solution {
public int maxOperations(int[] nums, int k) {
Map<Integer, Integer> cnt = new HashMap<>();
int ans = 0;
for (int x : nums) {
if (cnt.containsKey(k - x)) {
++ans;
if (cnt.merge(k - x, -1, Integer::sum) == 0) {
cnt.remove(k - x);
}
} else {
cnt.merge(x, 1, Integer::sum);
}
}
return ans;
}
}
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| class Solution {
public:
int maxOperations(vector<int>& nums, int k) {
unordered_map<int, int> cnt;
int ans = 0;
for (int& x : nums) {
if (cnt[k - x]) {
--cnt[k - x];
++ans;
} else {
++cnt[x];
}
}
return ans;
}
};
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| func maxOperations(nums []int, k int) (ans int) {
cnt := map[int]int{}
for _, x := range nums {
if cnt[k-x] > 0 {
cnt[k-x]--
ans++
} else {
cnt[x]++
}
}
return
}
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| impl Solution {
pub fn max_operations(nums: Vec<i32>, k: i32) -> i32 {
let mut cnt = std::collections::HashMap::new();
let mut ans = 0;
for x in nums {
let m = k - x;
if let Some(v) = cnt.get_mut(&m) {
ans += 1;
*v -= 1;
if *v == 0 {
cnt.remove(&m);
}
} else if let Some(v) = cnt.get_mut(&x) {
*v += 1;
} else {
cnt.insert(x, 1);
}
}
ans
}
}
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