Description#
Given an integer array nums
that does not contain any zeros, find the largest positive integer k
such that -k
also exists in the array.
Return the positive integer k
. If there is no such integer, return -1
.
Example 1:
Input: nums = [-1,2,-3,3]
Output: 3
Explanation: 3 is the only valid k we can find in the array.
Example 2:
Input: nums = [-1,10,6,7,-7,1]
Output: 7
Explanation: Both 1 and 7 have their corresponding negative values in the array. 7 has a larger value.
Example 3:
Input: nums = [-10,8,6,7,-2,-3]
Output: -1
Explanation: There is no a single valid k, we return -1.
Constraints:
1 <= nums.length <= 1000
-1000 <= nums[i] <= 1000
nums[i] != 0
Solutions#
Solution 1: Hash Table#
We can use a hash table $s$ to record all elements that appear in the array, and a variable $ans$ to record the maximum positive integer that satisfies the problem requirements, initially $ans = -1$.
Next, we traverse each element $x$ in the hash table $s$. If $-x$ exists in $s$, then we update $ans = \max(ans, x)$.
After the traversal ends, return $ans$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.
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| class Solution:
def findMaxK(self, nums: List[int]) -> int:
s = set(nums)
return max((x for x in s if -x in s), default=-1)
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| class Solution {
public int findMaxK(int[] nums) {
int ans = -1;
Set<Integer> s = new HashSet<>();
for (int x : nums) {
s.add(x);
}
for (int x : s) {
if (s.contains(-x)) {
ans = Math.max(ans, x);
}
}
return ans;
}
}
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| class Solution {
public:
int findMaxK(vector<int>& nums) {
unordered_set<int> s(nums.begin(), nums.end());
int ans = -1;
for (int x : s) {
if (s.count(-x)) {
ans = max(ans, x);
}
}
return ans;
}
};
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| func findMaxK(nums []int) int {
ans := -1
s := map[int]bool{}
for _, x := range nums {
s[x] = true
}
for x := range s {
if s[-x] && ans < x {
ans = x
}
}
return ans
}
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| function findMaxK(nums: number[]): number {
let ans = -1;
const s = new Set(nums);
for (const x of s) {
if (s.has(-x)) {
ans = Math.max(ans, x);
}
}
return ans;
}
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| use std::collections::HashSet;
impl Solution {
pub fn find_max_k(nums: Vec<i32>) -> i32 {
let s = nums.into_iter().collect::<HashSet<i32>>();
let mut ans = -1;
for &x in s.iter() {
if s.contains(&-x) {
ans = ans.max(x);
}
}
ans
}
}
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Solution 2#
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| use std::collections::HashSet;
impl Solution {
pub fn find_max_k(nums: Vec<i32>) -> i32 {
let mut ans = -1;
let mut h = HashSet::new();
for &n in &nums {
h.insert(n);
}
for &n in &nums {
if h.contains(&-n) && n > ans {
ans = n;
}
}
ans
}
}
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