Task – Find Minimum and Maximum Element in an Array

Problem Statement

Given an array arr[], the task is to find:

The minimum element in the array
The maximum element in the array

Return both values in a list as:

[min, max]
Example 1

Input:

arr[] = [1, 4, 3, 5, 8, 6]

Output:

[1, 8]

Explanation:

Minimum element is 1
Maximum element is 8
Example 2

Input:

arr[] = [12, 3, 15, 7, 9]

Output:

[3, 15]

Explanation:

Minimum element is 3
Maximum element is 15
Constraints
1 ≤ arr.size() ≤ 10^5
1 ≤ arr[i] ≤ 10^9
Logical Approach

To solve this problem:

Assume the first element is both minimum and maximum.
Traverse the array from index 1 to end.
For each element:
If it is smaller than current minimum → update minimum
If it is greater than current maximum → update maximum
Return both values.

This approach ensures we check each element only once.

Time Complexity: O(n)
Space Complexity: O(1)
SOLUTION:

class Solution:
def getMinMax(self, arr):
minimum = arr[0]
maximum = arr[0]

    for i in range(1, len(arr)):
        if arr[i] < minimum:
            minimum = arr[i]
        if arr[i] > maximum:
            maximum = arr[i]

    return [minimum, maximum]

Explanation of the Code
minimum = arr[0]
We assume the first element is the smallest initially.
maximum = arr[0]
We assume the first element is the largest initially.
Loop starts from index 1 because index 0 is already considered.
Inside loop:
If current element is smaller than minimum, update it.
If current element is greater than maximum, update it.
Finally, return both values as a list.