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2733. Neither Minimum nor Maximum

Description

Given an integer array nums containing distinct positive integers, find and return any number from the array that is neither the minimum nor the maximum value in the array, or -1 if there is no such number.

Return the selected integer.

 

Example 1:

Input: nums = [3,2,1,4]
Output: 2
Explanation: In this example, the minimum value is 1 and the maximum value is 4. Therefore, either 2 or 3 can be valid answers.

Example 2:

Input: nums = [1,2]
Output: -1
Explanation: Since there is no number in nums that is neither the maximum nor the minimum, we cannot select a number that satisfies the given condition. Therefore, there is no answer.

Example 3:

Input: nums = [2,1,3]
Output: 2
Explanation: Since 2 is neither the maximum nor the minimum value in nums, it is the only valid answer.

 

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i] <= 100
  • All values in nums are distinct

Solutions

Solution 1: Simulation

First, we find the minimum and maximum values in the array, denoted as $mi$ and $mx$ respectively. Then, we traverse the array and find the first number that is not equal to $mi$ and not equal to $mx$, and return it.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

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class Solution:
    def findNonMinOrMax(self, nums: List[int]) -> int:
        mi, mx = min(nums), max(nums)
        return next((x for x in nums if x != mi and x != mx), -1)

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class Solution {
    public int findNonMinOrMax(int[] nums) {
        int mi = 100, mx = 0;
        for (int x : nums) {
            mi = Math.min(mi, x);
            mx = Math.max(mx, x);
        }
        for (int x : nums) {
            if (x != mi && x != mx) {
                return x;
            }
        }
        return -1;
    }
}

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class Solution {
public:
    int findNonMinOrMax(vector<int>& nums) {
        auto [mi, mx] = minmax_element(nums.begin(), nums.end());
        for (int x : nums) {
            if (x != *mi && x != *mx) {
                return x;
            }
        }
        return -1;
    }
};

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func findNonMinOrMax(nums []int) int {
    mi, mx := slices.Min(nums), slices.Max(nums)
    for _, x := range nums {
        if x != mi && x != mx {
            return x
        }
    }
    return -1
}

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impl Solution {
    pub fn find_non_min_or_max(nums: Vec<i32>) -> i32 {
        let mut mi = 100;
        let mut mx = 0;

        for &ele in nums.iter() {
            if ele < mi {
                mi = ele;
            }
            if ele > mx {
                mx = ele;
            }
        }

        for &ele in nums.iter() {
            if ele != mi && ele != mx {
                return ele;
            }
        }

        -1
    }
}

Solution 2

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class Solution:
    def findNonMinOrMax(self, nums: List[int]) -> int:
        mi, mx = min(nums), max(nums)
        for x in nums:
            if x != mi and x != mx:
                return x
        return -1

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