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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

• Java
• C++
• Python
• Go
• RenderScript

• 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;
      }
  }
  

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

• 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
  
  

• func findNonMinOrMax(nums []int) int {
  	mi, mx := 100, 0
  	for _, x := range nums {
  		if x < mi {
  			mi = x
  		}
  		if x > mx {
  			mx = x
  		}
  	}
  	for _, x := range nums {
  		if x != mi && x != mx {
  			return x
  		}
  	}
  	return -1
  }
  

• 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
      }
  }
  

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