Welcome to Subscribe On Youtube

2441. Largest Positive Integer That Exists With Its Negative

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.

  • 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;
    }
}

  • 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;
    }
};

  • 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)


  • 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
}

  • 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;
}


  • 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
    }
}

All Problems

All Solutions